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Lecture Comments (12)

1 answer

Last reply by: Živa Brun?ek
Mon Jun 27, 2016 6:20 AM

Post by Živa Brun?ek on June 27 at 06:19:45 AM

How to calculate pH when you mix two solutionswith known volumeand  with different pH, both acid?

0 answers

Post by Saadman Elman on April 26, 2015

Hi, Professor Ow, This lecture really helped me a lot. Thank you so much.

In sample no.1 you didn't give us the answer but you gave us the strategy to get to the answer. I do understand that it's very logical to assume that since we already did those steps (to get to the answer) in our previous chapter but still if i were you i would have said when all said and done the answer will be so and so WITHOUT showing those steps. Because the student really need to know if his/her answer is correct or not. So i would appreciate if you verify the answer in sample no. 1 In part a i got Ph=11.4 In part b, ph =9.80 In part c Ph=5.40 In part d i am getting Ph=1.88

1 answer

Last reply by: Professor Franklin Ow
Sun Feb 15, 2015 11:39 PM

Post by Candicee Childs on February 15, 2015

can you explain the assumption rule?

1 answer

Last reply by: Professor Franklin Ow
Thu Mar 20, 2014 3:00 AM

Post by Kristen Biermayer on March 19, 2014

How are you coming up with the neutralization equations? How do you know if you should use the acid or the base?

1 answer

Last reply by: Professor Franklin Ow
Mon Mar 17, 2014 12:10 AM

Post by Yaw Frimpong on March 16, 2014

Please correct me if I am wrong.  
At 13:41 how can pH = -log(1.8 x 10^-5) be equal to log(.50/.50) ?
log(.50/.50) = 0. Can you please clarify.
I think the "=" sign should be a "+" sign.

1 answer

Last reply by: Professor Franklin Ow
Thu Nov 7, 2013 5:08 PM

Post by Shawn Ng on October 30, 2013

Correct me if I'm wrong Professor but for the slide on 6.56, shouldn't the values in Step 2's Ice table be in Molars instead of moles?

0 answers

Post by Kimberly Micheller on August 9, 2013

In Sample Problem 1 (a.) how do you go from Kb to x to [OH-]?

Applications of Aqueous Equilibria

  • Equilibrium can be applied to various systems, including acid-base mixtures, buffers, titrations, solubility, and complex ion formation.
  • The Henderson-Hasselbalch equation can be used to approximate the pH of a buffer system.
  • Titration curves monitor pH versus volume of titrant added.
  • Solubility product constants quantify the extent to which an ionic solid is soluble in water.

Applications of Aqueous Equilibria

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  1. Intro
    • Lesson Overview
      • Calculating pH of an Acid-Base Mixture
      • Calculating pH of an Acid-Base Mixture cont'd
      • Buffers
      • Buffers cont'd
      • Buffers cont'd
      • Buffer Preparation and Capacity
      • Acid-Base Titrations
      • Acid-Base Titrations cont'd
      • Acid-Base Titrations cont'd
      • Solubility Equilibria
      • Solubility Equilibria cont'd
      • Solubility Equilibria cont'd
      • Solubility Equilibria cont'd
      • Complexation Equilibria
      • Summary
        • Sample Problem 1: Question
          • Sample Problem 1: Part a) Calculate the pH at the Beginning of the Titration
            • Sample Problem 1: Part b) Calculate the pH at the Midpoint or Half-way Point
              • Sample Problem 1: Part c) Calculate the pH at the Equivalence Point
                • Sample Problem 1: Part d) Calculate the pH After 27.50 mL of the Acid was Added
                  • Intro 0:00
                  • Lesson Overview 0:07
                  • Calculating pH of an Acid-Base Mixture 0:53
                    • Equilibria Involving Direct Reaction With Water
                    • When a Bronsted-Lowry Acid and Base React
                    • After Neutralization Occurs
                  • Calculating pH of an Acid-Base Mixture cont'd 2:51
                    • Example: Calculating pH of an Acid-Base Mixture, Step 1 - Neutralization
                    • Example: Calculating pH of an Acid-Base Mixture, Step 2 - React With H₂O
                  • Buffers 7:45
                    • Introduction to Buffers
                    • When Acid is Added to a Buffer
                    • When Base is Added to a Buffer
                  • Buffers cont'd 10:41
                    • Calculating the pH
                    • Calculating the pH When 0.010 mol NaOH is Added to 1.0 L of the Buffer
                  • Buffers cont'd 14:10
                    • Calculating the pH When 0.010 mol NaOH is Added to 1.0 L of the Buffer: Step 1 -Neutralization
                    • Calculating the pH When 0.010 mol NaOH is Added to 1.0 L of the Buffer: Step 2- ICE Table
                  • Buffer Preparation and Capacity 16:38
                    • Example: Calculating the pH of a Buffer Solution
                    • Effective Buffer
                  • Acid-Base Titrations 19:33
                    • Acid-Base Titrations: Basic Setup
                  • Acid-Base Titrations cont'd 22:12
                    • Example: Calculate the pH at the Equivalence Point When 0.250 L of 0.0350 M HClO is Titrated With 1.00 M KOH
                  • Acid-Base Titrations cont'd 25:38
                    • Titration Curve
                  • Solubility Equilibria 33:07
                    • Solubility of Salts
                    • Solubility Product Constant: Ksp
                  • Solubility Equilibria cont'd 34:58
                    • Q < Ksp
                    • Q > Ksp
                  • Solubility Equilibria cont'd 36:03
                    • Common-ion Effect
                    • Example: Calculate the Solubility of PbCl₂ in 0.55 M NaCl
                  • Solubility Equilibria cont'd 39:02
                    • When a Solid Salt Contains the Conjugate of a Weak Acid
                    • Temperature and Solubility
                  • Complexation Equilibria 41:10
                    • Complex Ion
                    • Complex Ion Formation Constant: Kf
                  • Summary 43:35
                  • Sample Problem 1: Question 44:23
                  • Sample Problem 1: Part a) Calculate the pH at the Beginning of the Titration 45:48
                  • Sample Problem 1: Part b) Calculate the pH at the Midpoint or Half-way Point 48:04
                  • Sample Problem 1: Part c) Calculate the pH at the Equivalence Point 48:32
                  • Sample Problem 1: Part d) Calculate the pH After 27.50 mL of the Acid was Added 53:00

                  Transcription: Applications of Aqueous Equilibria

                  Hi, welcome back to Educator.com.0000

                  Today's lesson in general chemistry is applications of aqueous equilibria.0003

                  We are pretty much going to build upon what we learned in the last lesson when we talked about equilibria for acid base chemistry.0008

                  We are going to see how it can easily apply to other aspects of general chemistry.0015

                  We are going to learn how to calculate the pH of an acid base mixture.0020

                  We are going to learn about a very important system in chemistry which is called buffers.0024

                  Following buffers, we are going to learn a unique experiment called acid base titrations0030

                  followed by other aspects of chemical equilibria, namely what we call solubility equilibria and compensation equilibria.0037

                  We will always wrap up the session as always with a summary followed by a sample problem.0047

                  When we did equilibria before, every equilibria we have discussed has always involved a direct reaction with water.0054

                  For example, it has always been the acid HA plus H2O.0064

                  Or it has always been a base plus H2O.0069

                  But what happens when we react directly a Bronsted-Lowry acid with a Bronsted-Lowry base?0073

                  When this happens, you don't get equilibrium immediately.0079

                  Instead the very first step that occurs is going to be a neutralization reaction.0082

                  Neutralization involves complete consumption of one or both of the reactants.0088

                  We always get the generic reaction where we have an acid HA reacting0093

                  with a base aqueous going on to form HB1+ aqueous and A- aqueous.0099

                  Once again when we do these neutralization reactions, we use a single arrow only0108

                  because one or both of the reactants is going to be consumed completely.0116

                  After neutralization occurs, then we can apply whatever we learned last time.0126

                  Basically whatever remains is going to be reacting with water in a typical aqueous equilibrium.0134

                  In that case, it is either going to be the acid HA aqueous plus H2O liquid.0143

                  Or it is going to be the base reacting with H2O liquid.0153

                  Those are going to be our main reactions after neutralization has occurred.0163

                  Let's go ahead and take a look at a representative problem and jump right into a calculation.0167

                  What is the pH of the solution that results from adding 30.0 milliliters0173

                  of 0.1 molar sodium hydroxide to 45.0 milliliters of 0.100 molar acetic acid?0178

                  Now we have a strong base reacting with a weak acid here.0187

                  Step one, step one is to always write out the neutralization reaction.0194

                  Let's go ahead and write that out.0204

                  It is going to be hydroxide aqueous reacting with the acetic acid, CH3CO2H aqueous.0206

                  That is going to go on to form water liquid and acetate CH3CO21- aqueous.0216

                  We can go ahead and proceed to fill in the ICE table as usual.0226

                  Here hydroxide is what we want, moles.0230

                  That is 0.0300 liters times 0.100 molar.0233

                  That is going to give us 0.00300 moles of hydroxide.0240

                  The acetic acid we have 0.0450 liters times 0.100 molar.0246

                  That is going to give us 0.0045 moles of acetic acid.0257

                  Acetate is going to be zero.0263

                  Remember we said that one of these reactants or both is going to be completely consumed.0266

                  We can see that this is a 1:1 ratio here.0271

                  If we have amounts of both reactants, that is like a limiting reactant problem which we discussed a long time ago.0278

                  Here because it is 1:1 ratio, the reactant that is present in smaller quantity is going to be consumed entirely.0284

                  Here if this is I-C-E, this is then going to be -0.0300 moles.0291

                  This is -0.00300 moles; this is going to be +0.00300 moles.0300

                  The hydroxide gets completely consumed.0310

                  Most of the acid is going to be eaten up here and consumed or left with only 0.0015 moles.0314

                  This is going to be 0.00300 moles.0320

                  Now step two, step two, we are then going go ahead and proceed on to do...0326

                  Whatever is left standing is going to react with water; step two, react with water.0335

                  Now this is going to be like our previous lecture.0341

                  We are going to go ahead and select the acid, CH3CO2H aqueous.0346

                  That is going to react with water going on to form CH3CO21- aqueous and H3O1+ aqueous.0352

                  We carry everything down.0367

                  The final values of table one become the initial values of table two.0368

                  This is now 0.0015 moles; this is 0.00300 moles; this is 0.0372

                  This is going to be ?x, +x, and +x.0380

                  At equilibrium, 0.0015 moles minus x, 0.00300 moles plus x, and x.0386

                  The k for acetic acid is something that should be given to you or you look it up.0395

                  That is 1.8 times 10-5 which is relatively small.0400

                  All of this reduces to 0.00300 times x divided by 0.0015.0404

                  Because the volume is the same for each of these moles, the liters are really going to cancel out.0416

                  I can directly just use the moles.0424

                  We get an x value of 9.0 times 10-6 molar which is the hydronium ion concentration at equilibrium.0427

                  The pH is going to be 5.05.0438

                  Again always make sure your answer makes sense.0443

                  Here we have an acidic pH which we should have because all we have leftover is acid and no more base.0446

                  Again this is how you calculate the pH of an acid base mixture.0455

                  Let's now go ahead and examine how to control a change in pH.0460

                  When a solution contains both a weak acid/base and its conjugate in appropriate amounts,0467

                  there is going to be a resistance to small changes in pH.0474

                  These are solutions that we call buffers.0477

                  A buffer is going to contain again a weak acid HA plus its conjugate base A-.0482

                  Or it is going to contain a weak base B and its conjugate acid BH+.0493

                  These are typically going to be in the form of salts; usually in the form of a salt.0501

                  Anytime you see these two mentioned in the same sentence, they simultaneously occur in the same mixture.0515

                  We have what is called a buffer; once again buffers resist small changes in pH.0523

                  The reason why they work is because of the following.0531

                  If you disrupt the buffer by say adding acid, the solution is going to resist a drop in pH.0534

                  The reason is because if we add acid to the buffer, this acid is going to react with any base present.0540

                  Because we have both acid and base present as a buffer, this is good to go.0549

                  We are going to form HA aqueous and H2O liquid.0555

                  Or it can react with H3O1+ aqueous and base aqueous.0561

                  We are going to form BH plus aqueous and H2O liquid.0568

                  Right away you see how the buffer can resist change in pH because water is formed0574

                  as a product which helps to counteract5 the effect of the added acid; dilutes H3O1+ added.0581

                  The other way to disrupt the buffer of course is by base, adding base to the buffer.0594

                  The solution is going to resist a rise in pH.0598

                  The reason is because if we add base, say hydroxide, that will then react with possibly HA if it is an acid in solution.0601

                  Or the base can then go ahead and react with the conjugate acid if it is a basic buffer.0611

                  In either case, we see that we still form water.0619

                  Again it is the formation of water that helps to counteract the effect of added base; dilutes OH- added.0626

                  Let's go ahead and jump right into a calculation.0643

                  A buffer is prepared by mixing 0.50 molar of acetic acid with 0.50 molar0648

                  of sodium acetate where the k is given to you for acetic acid.0653

                  Let's go ahead and calculate the pH first.0657

                  Let's go ahead and write out the equilibrium.0663

                  CH3CO2H aqueous plus H2O liquid goes on to form CH3CO21- aqueous and hydronium aqueous.0664

                  This is 0.50 molar; this is 0.50 molar; this is 0.0682

                  This is going to be ?x, +x, +x.0688

                  This is going to be 0.50 minus x, 0.50 plus x, and x.0692

                  Ka is given to us to be 1.8 times 10-5.0698

                  This is then approximately 0.50 times x divided by 0.50.0702

                  You see that we can get an x of 1.8 times 10-5 molar.0708

                  The pH is going to be 4.74.0717

                  There is a shortcut it turns out.0724

                  Anytime we have a buffer, it turns out we don't need to set up an ICE table problem.0726

                  It turns out that we do have a nice shortcut.0732

                  This is what we call the Henderson Hasselbalch equation.0735

                  The Henderson Hasselbalch equation is the following.0753

                  It is pH is equal to the pKa plus the log base 10 of0756

                  the concentration of the base initial divided by the concentration of the acid initial.0761

                  This is going to be for an acidic buffer.0767

                  If you have a basic buffer, it is still pH is equal to pKa.0772

                  But this time it is going to be the log of the initial concentration0776

                  of base divided by the initial concentration of the conjugate acid.0778

                  The nice thing about this equation is that we don't need equilibrium values for the conjugate base and the acid.0785

                  We only care about initial values.0791

                  The Henderson Hasselbalch equation, you must know it is only an approximation.0793

                  But it does work out many times.0799

                  We can go ahead and plug in our values here for the previous problem we just did.0802

                  We are going to get pH is equal to ?log of 1.8 times 10-5.0807

                  That is equal to the log of 0.50 divided by 0.50.0814

                  When all is said and done, we get 4.74.0819

                  As you can see, this matches up very nicely with the previous problem where we did the ICE table.0822

                  Once again when you see a buffer problem, think Henderson Hasselbalch equation.0831

                  Next let's go ahead and see what happens to the pH when we disrupt this buffer by adding strong base.0837

                  Now we are going to add 0.010 moles of sodium hydroxide to exactly 1 liter of the buffer.0843

                  Once again because we are adding a strong acid or a base, step one is to do the neutralization step.0853

                  Because we are adding a strong base hydroxide, this is going to go0861

                  ahead and react with the weak acid that is already in solution.0866

                  That is going to go on to form acetate and water.0871

                  This is going to be 0.010 molar; this is 0.50 molar.0881

                  This is going to be 0.50 molar; we don't care about water.0886

                  Because again this is a 1:1 ratio, remember what we always said.0892

                  Anytime there is neutralization, either one or both of the acid and base are going to be consumed.0895

                  My 1:1 ratio tells me I am going to choose a smaller amount that is going to get consumed completely.0901

                  This is -0.010; this is -0.010 molar; this is +0.010 molar.0907

                  This goes to 0; this goes to 0.49 molar; this goes to 0.51 molar.0916

                  We can go ahead then and step two...0923

                  Remember we would set up ICE table like we taught you at the beginning of this session.0926

                  But because this is a buffer, we can use Henderson Hasselbalch directly which is always nice.0931

                  It saves us a lot of time.0937

                  pH is equal to the pKa plus the log of acetate initial divided by the initial concentration of acetic acid.0942

                  Remember how we said last time, the final values of table one become initial values of table two.0960

                  That is still going to apply.0964

                  This 0.51 goes right there; this 0.49 goes right there.0967

                  When all is said and done, we expect a pH that is going to be0972

                  greater than 4.74 because we added strong base but not too much greater.0977

                  When we do this, we get pH of 4.76.0982

                  As you can see, the buffer was successful in resisting this change in pH.0986

                  We only went up 0.02 units.0993

                  The next thing we are going to discuss is how to make a buffer and what we call buffer capacity.0998

                  Exactly 1.25 liters of a buffer solution is prepared from mixing 23.5 grams of sodium dihydrogen phosphate1003

                  and 15.0 milliliters of 14.7 molar phosphoric acid where the Ka is given to you.1014

                  Calculate the pH.1019

                  Because this is a buffer, again we can use Henderson Hasselbalch equation directly.1020

                  The pH is equal to the pKa plus the log of the conjugate base initial divided the acid initial.1025

                  The hardest part of this is just to figure out which one is the conjugate base and which one is the acid.1036

                  This is going to be equal to pKa plus the log of...1041

                  the conjugate is going to be the H2PO41- concentration.1045

                  Of course the acid is going to be the phosphoric acid initial concentration.1052

                  We can go ahead and again use this to calculate the pH of this buffer by inserting everything directly in.1062

                  This 23.5 grams of course we need to get to moles first.1073

                  This 14.7 molar is going to go right there.1080

                  Remember the final volume is exactly the same.1084

                  The 1.25 liters is same for both A- and HA which means the volume term is going to cancel.1089

                  All you need is just really the moles.1106

                  Here is 15 milliliters of 14.7 molar phosphoric acid.1110

                  Then you can convert the 23.5 grams of the sodium dihydrogen phosphate directly to moles.1114

                  A buffer can only do so much.1122

                  A butter can accommodate so much extra base or extra acid.1124

                  It is what we call buffer capacity.1129

                  A buffer is considered to be effective when the pH is within 1 unit of the pKa.1132

                  In other words, pH is equal to +/-1 of the pKa.1138

                  What that means by going from the Henderson Hasselbalch equation is that the ratio of A- to HA is less than 10.1145

                  Anything outside this range, the buffer will not be effective.1153

                  Why?--because you have too little of either the acid or base to counteract any effect from disturbing the buffer.1157

                  That is our brief discussion on buffer preparation and buffer capacity.1167

                  Following buffers is going to be now a specific experiment.1176

                  This is what we call an acid base titration.1180

                  In an acid base titration, an acid or base of known concentration is used to determine the concentration of an unknown acid or base.1183

                  This is known as standardization of the unknown.1192

                  The solution that is known is what we call the analyte.1195

                  Excuse me... the solution of known concentration is what we call the titrant.1201

                  The solution that is unknown, what we are trying to standardize, what we are1209

                  trying to find the concentration of, this is what we call the analyte.1213

                  The basic setup is going to be the following.1219

                  We are going to have a pretty long thin cylindrical tube.1224

                  This is usually 50.00 milliliter; this is what we call a buret.1229

                  The burets are very very precise.1236

                  You notice I use two sig figs after the decimal.1238

                  This is always two within 0.01 milliliter.1241

                  There is going to be an Erlenmeyer flask on the bottom.1250

                  Usually the buret is going to have the titrant.1253

                  Usually the flask is going to have the analyte.1262

                  With this flask with the analyte inside the flask, there is also going to be called what is called indicator.1266

                  Indicator solution is going to change color.1274

                  When it changes color, this is what we call the equivalence point, also known as the stoichiometric point.1281

                  At this point, we have complete neutralization.1299

                  When we have complete neutralization, the moles of HA acid is equal to the moles of the base.1308

                  You see that we can use mole to mole ratio here between acid and base to backtrack and get the concentration of the unknown.1319

                  But let's go ahead and go through a typical problem first.1329

                  Calculate the pH at the equivalence point when so much HClO is titrated with 1 molar KOH.1333

                  Anytime you see volume and molarity, you have heard me say this before.1342

                  That is always a good 1-2 combo; you are going to get moles.1347

                  Let's go ahead and do that.1349

                  0.250 liters times 0.0350 molar, that is going to give us 0.00875 moles of HClO.1350

                  Because we are at the equivalence point, this is also the same number1365

                  of moles with the KOH by definition; by definition of equivalence point.1371

                  Because we now have both of our amounts, let's go ahead and set up the neutralization reaction.1389

                  HClO plus OH1-, that goes on to form H2O liquid and ClO1- aqueous.1394

                  This is 0.00875 moles; this is 0.00875 moles; this is going to be 0.1406

                  Remember last time how when we have an acid and base neutralization reaction,1415

                  we said that either one or both of the reactants is going to get consumed entirely.1419

                  This is a good example where both of them gets consumed entirely.1424

                  Why?--because we have a 1:1 ratio.1427

                  Because this is the equivalence point, we have equal amounts.1430

                  This is going to go -0.00875 moles; this is going to be -0.00875 moles.1433

                  This is going to be +0.00875 moles.1444

                  This goes to 0; this goes to 0; this goes to 0.00875 moles.1448

                  Final values of table one become initial values of table two.1457

                  But this is not a buffer anymore; why?--because we have zero amount of HClO.1460

                  Let's go ahead and do our second setup; it is the last man standing.1465

                  The only thing remaining is ClO1- aqueous plus H2O liquid going on to form HClO aqueous plus OH1- aqueous.1471

                  When all is said and done, we are going to be using Kb.1486

                  We are going to use Kb to get the concentration of hydroxide.1491

                  From there, you can go ahead and get the pH.1496

                  But don't forget, the important thing is to take this moles amount and get to molarity first.1500

                  When you get to molarity, you need to divide by the total volume; very very important.1509

                  This is a typical type of question to calculate the pH at an equivalence point of a titration.1523

                  During a titration, something that is done that is very typical is to monitor the pH yielding what is known as a titration curve.1532

                  When we look at a typical titration curve, it is going to be the following: pH versus volume of titrant added.1541

                  We can have different types of titration curves.1556

                  Each one is going to be a different experiment.1558

                  We can have a curve that looks like this.1561

                  We can have a curve that looks like this.1569

                  In each of these cases here, the pH is going to be greater than 7 here.1575

                  Here the pH is going to be less than 7 here.1583

                  Of course the final situation is pH equal to 7.1587

                  Of course you can have the mirror image right here, pH equal to 7.1599

                  Let's go ahead and discuss what each of these means right now.1607

                  The first situation, here the pH is acidic.1610

                  The only thing you have initially is your acid.1615

                  This is going to be a weak acid titrated with strong base.1619

                  Here you see that the initial pH when no titrant has been added is basic.1629

                  This is going to be a weak base titrated with strong acid.1635

                  Let's go back one.1645

                  When the weak acid is titrated with a strong base, the pH is always going to be basic at the equivalence point.1647

                  When the weak base is titrated with strong acid, the pH is going to be acidic at the equivalence point.1655

                  Here pH is neutral.1662

                  When the pH is neutral at the equivalence point, you have one of two possibilities.1665

                  Either a strong acid titrated by strong base or you have a strong base titrated by strong acid.1670

                  A titration curve typically can be broken up into four regions.1693

                  I am going to choose the weak acid titrated with a strong base as our typical example.1699

                  Here is region one, region two; region one, region two, region three, and region four.1706

                  At the end of this lesson, you are going to be able to calculate the pH along any point of the titration curve.1719

                  You are basically in one of four regions.1725

                  Each region is going to have little own strategy.1727

                  Let's go ahead and go over them.1730

                  In region one, which is basically the y-intercept; in other words, zero base added.1732

                  The equilibria that you would use is an equilibrium from last lecture which is the1747

                  weak acid HA reacting with water going on to form H3O1+ aqueous and A- aqueous.1755

                  It is a typical ICE table setup problem; you use Ka to solve for pH.1768

                  Region number two, you notice that in region number two, the pH doesn't change that much.1777

                  It is really a plateau.1784

                  The pH doesn't change that much because we have a functioning buffer.1787

                  Region two is what we call the buffer region.1790

                  Again you don't need an ICE table setup to calculate the pH in the buffer region.1797

                  You simply use the Henderson Hasselbalch equation.1802

                  pH is equal to pKa plus the log of the concentration of A- initial divided by the concentration of HA initial.1806

                  There is special point of interest in the buffer region.1818

                  It is right here; that is what we call the midpoint.1821

                  That is what we call the midpoint of the titration.1828

                  At the midpoint of the titration, we have this very special relationship where pH is equal to the pKa.1831

                  In other words, the concentration of A- initial equals the concentration of HA initial.1836

                  It is called the midpoint because we are exactly halfway done with the titration in order to get to the equivalence point.1847

                  In other words, the moles of hydroxide added is going to be equal to half the moles of HA present.1853

                  Again the moles of hydroxide added is half the moles of HA present.1866

                  We are halfway to the stoichiometric point; halfway to equivalence point.1870

                  Region number three, in region number three, we get a sudden spike in pH.1885

                  That is because the buffer no longer is functioning.1894

                  We are beyond the capacity of the buffer.1897

                  The point of interest is of course the equivalence point.1901

                  We went over already how to calculate the pH at the equivalence point.1908

                  Remember it is going to be here.1912

                  Remember the acid is completely gone; the base is completely gone.1916

                  All you have left is the conjugate base A- reacting with water going on to form HA aqueous and hydroxide aqueous.1919

                  For this one, you are going to use Kb.1933

                  As you can see, that is why pH is going to be greater than 7.1938

                  Finally region number four, region number four, you have nothing but excess hydroxide added, nothing more than that.1941

                  That is going to be a typical pH is equal to ?log of H3O1+ equation.1957

                  We are going to go through a problem later on at the end of this lecture discussing that.1967

                  That is our basic introduction to titration curves.1973

                  The four regions, each require a different strategy to compute the pH.1979

                  Let's now take a look at another application of aqueous equilibria.1984

                  It is what we call solubility equilibria.1988

                  Remember a long time ago at the beginning of the first couple of presentations, we went over our table of solubility rules.1991

                  We said that some salts were relatively soluble; we said that some salts are insoluble.2000

                  It turns out that a really all solid salts are going to be soluble in water to a measurable degree.2006

                  Some of course more so than others.2014

                  Salts that are very soluble will have the following equilibrium lie far to the right.2016

                  We can go ahead and imagine an ionic solid of the formula MX solid2021

                  dissolving into the respective cation and anion, M+ aqueous and X- aqueous.2029

                  The equilibrium expression for this type of reaction would be the concentration of M+ times the concentration of X-.2041

                  We give this equilibrium constant a very specific name.2051

                  This is what we call the Ksp or the solubility product constant.2055

                  Basically as you can see, the larger this constant, the more ions you get.2063

                  In other words, the more soluble the solid is; solubility increases with Ksp.2069

                  Salts that are relatively insoluble such as carbonate will have the equilibrium lie to the left instead.2082

                  These are going to have smaller Ksp values.2090

                  We can use Ksp to predict if a solution will form a precipitate2099

                  of the ionic solid given initial amount of the cation and anion.2103

                  This is done by simply assessing the reaction quotient Q, something we discussed in a previous lecture.2108

                  We compare it to Ksp.2114

                  Just like Le Chatelier principle, we are going to use the same principles here.2116

                  If Q is less than Ksp, we have the reaction shifting to the right because we don't have enough product.2121

                  Because it shifts to the right, that is away from the solid.2129

                  Precipitation will not occur.2132

                  However if we find and calculate that Q is greater than Ksp,2135

                  the reaction will shift to the left because we have too much product.2140

                  Because we are going toward the solid, this is the same as saying that precipitation will occur.2144

                  Now that we have gone through solubility equilibria and the solubility product constant Ksp,2154

                  let's go ahead and examine some factors that affect solubility.2159

                  When a solid salt is dissolved in a solution that already contains either the cation or2164

                  anion of the salt, its solubility will be found to be lower than in pure water.2169

                  This is what we call the common ion effect.2176

                  Let's go ahead and look at an example.2178

                  The solubility of lead(II) chloride in water is 3.9 grams per liter.2181

                  Calculate its solubility in 0.55 molar sodium chloride solution.2186

                  We have solubility in water to be 3.9 grams per liter.2192

                  We are asked to calculate the solubility in salt water.2197

                  Let's go ahead and set it up.2203

                  Remember we must always write out the proper equilibrium.2205

                  In this case, because it deals with solubility, it is going to show the ionic solid dissolving water into cation and anion.2209

                  PbCl2 is going to go ahead and form Pb2+ aqueous plus two Cl1- aqueous.2218

                  You see that we do have an initial amount of sodium chloride.2229

                  It is 0.55 molar and zero Pb2+; this is going to be +.2233

                  We don't use x anymore; instead we use s as a convenience to indicate solubility.2241

                  This is going to be +2s; this is going to be s.2247

                  This is going to be 0.55 plus 2s.2250

                  The Ksp for the lead(II) chloride is going to be given to you.2253

                  Or you look it up in your textbook; that is 1.7 times 10-5.2257

                  That is going to be equal to the concentration of Pb2+ times the concentration of Cl1- squared.2262

                  That is going to be equal to s times 0.55 plus 2s squared.2269

                  Because Ksp is relatively small, the approximation rule still applies.2275

                  That is s times 0.55; we can go ahead and solve for s.2281

                  That is going to be equal to 2.81 times 10-5 moles per liter.2287

                  That is in salt water; in pure water, the solubility was 3.9 grams per liter.2298

                  You can easily convert grams per liter to moles per liter.2305

                  We get 0.015 moles per liter here.2308

                  As you can see, the solubility in salt water is much less than the solubility in pure water.2313

                  In other words, the common ion effect reduces the solubility of an ionic salt.2321

                  Common ion effect reduces solubility.2329

                  The common ion effect is one factor that affects solubility.2344

                  Another factor is when the solid salt contains the conjugate of a weak acid.2348

                  It is found that these salts are going to be more soluble when in acidic solution.2353

                  The typical example is going to be for example a metal carbonate, MgCO3 for example.2359

                  MgCO3 solid goes on to form Mg2+ aqueous plus CO32- aqueous.2366

                  Something that can drive the reaction to the right is addition of acid.2379

                  Why?--because H3O1+ aqueous will want to go ahead.2384

                  It is going to react with the conjugate base of carbonic acid.2390

                  CO32- aqueous plus H3O1+ aqueous is going to go ahead and reform part of the carbonic acid.2398

                  We are going to first bicarbonate aqueous and H2O liquid.2410

                  As you can see, because hydronium is going to react with carbonate, it reduces the amount of carbonate.2417

                  That is going to drive the equilibrium to the right.2423

                  CO32- will be consumed by acid driving the dissolution of magnesium carbonate.2427

                  Finally the last factor that influences solubility is of course temperature.2443

                  This is going to be true for all ionic solids.2449

                  Solubility is always enhanced with increasing temperature.2451

                  Basically the reason is because the thermal energy is going to become sufficient enough that2456

                  you can overcome the electrostatic attraction between the cation and anion in the ionic solid.2462

                  The last type of equilibria is what we call complexation equilibria.2473

                  Aqueous solutions that involve transition metal cations can also form from their aqueous cation and anion counterparts.2477

                  Can also form... excuse me... from their aqueous cation and anion counterparts.2488

                  Let's go ahead and show a representative equilibrium showing the formation of what we call a complex ion.2494

                  For example, Fe3+ aqueous can react with six equivalents of carbon monoxide gas.2499

                  That can go ahead and form Fe(CO)63+ aqueous.2512

                  We are going to get into these types of compounds in a later presentation.2523

                  But this is what we call a complex ion.2529

                  Basically it involves a transitional metal and what we call a ligand.2535

                  In this case, the ligand is going to be carbon monoxide.2543

                  The extent to which the complex ion forms can be described by its equilibrium constant.2547

                  In this case, K is equal to the concentration of Fe(CO)63+ divided by the2552

                  concentration of Fe3+ times the concentration of carbon monoxide raised to the sixth power.2562

                  Because this is involving complex ion formation, we don't just call this Keq anymore.2571

                  But instead this is what we call Kf or the complex ion formation constant.2576

                  The main difference between Ksp and Kf is that they are opposite.2584

                  Ksp represents a breakup of a compound.2588

                  Kf represents the formation of a compound from its constituent ions.2594

                  While Ksp values tend to be very small, Kf values actually tend to be2599

                  just the opposite--quite large, several orders of magnitude, much much much larger.2605

                  Again that is what we call complexation equilibria.2611

                  Let's go ahead and summarize this very important part from general chemistry.2616

                  Equilibrium can be applied to various systems including acid base mixtures, buffers, titrations, solubility, and complex ion formation.2620

                  We found a very nice shortcut that we call the Henderson Hasselbalch equation.2631

                  This can be used to approximate the pH of a buffer system.2636

                  We found that during a titration, we can monitor the pH to yield or generate what is called a titration curve.2641

                  Finally we also saw that solubility product constants can quantify the extent to which an ionic solid is going to be soluble in water.2650

                  Let's now go ahead and jump into a representative sample problem.2660

                  This sample problem is quite long.2665

                  I have chosen an acid base titration on purpose because this is where students tend to trip up.2667

                  I want to go through how to calculate the pH along a titration curve in each of the four regions that we pointed out.2675

                  Exactly 1.36 grams of trimethyl amine is going to be dissolved in 250 milliliters of water.2683

                  Then 23.6 milliliters of hydrochloric acid was required to reach the equivalence point.2691

                  Basically we are asked to calculate the following.2698

                  The pH at the beginning of the titration which is region one.2700

                  The pH at the midpoint or halfway point... sorry about that... this should read or.2707

                  Because we are at halfway point, this is going to be region two.2718

                  The pH at the equivalence point which is region three.2724

                  Finally the pH after 27.50 milliliters of acid was added.2728

                  You notice that this 27.50, that is greater than the 23.60 which means we are after the equivalence point.2734

                  We are out of region three; we are into region number four.2742

                  Let's go ahead and do this.2748

                  When we go ahead and calculate the pH at the beginning of the titration, remember no titrant has been added yet.2751

                  Initially we only have the trimethyl amine; this is a weak base.2757

                  Let's get the 1.36 grams into molarity right away, 1.36 grams of the trimethyl amine.2765

                  We are going to go ahead and divide this by its molar mass which is approximately 59 grams.2777

                  That is going to give us 0.023 moles of the trimethyl amine.2783

                  We are going to go ahead; we are going to divide this, 0.023 moles.2792

                  We are going to divide it by the total volume at that point which is 0.250 liters.2800

                  We get the initial molarity of the base to be 0.092 molar.2804

                  Now it is just a typical equilibrium problem with direct reaction with water.2812

                  Let's go ahead and write out the equilibrium then.2819

                  Trimethyl amine aqueous plus H2O liquid goes on to form HN1+(CH3)3 and hydroxide aqueous.2821

                  Let's go ahead and plug in all our values.2841

                  That is 0.092 molar and 0 and 0; -x, +x, and +x.2843

                  This is going to be 0.02 minus x, x, and x.2851

                  Of course we are going to use Kb to go ahead and get x.2855

                  X is then going to give us the concentration of OH-.2862

                  From there, we can go ahead and get the pH.2867

                  Of course we expect a basic pH greater than 7 because we only have base present in solution.2870

                  No titrant has been added in the form of strong acid yet.2878

                  That is part A; part B is very simple, the pH at the midpoint.2883

                  Because we can use the Henderson Hasselbalch equation, the pH at the midpoint is simply equal to pKa.2891

                  That is a very quick calculation there; not bad.2900

                  Part A is done; part B is done; region one and region two.2903

                  Let's go ahead and do part C now, the pH at the equivalence point, region three.2907

                  At region three, at the equivalence point, the moles of the base is going to be equal to the moles of the added acid.2915

                  Let's go ahead and get the moles of the acid then.2933

                  0.02360 liters times 0.974 molar, that is going to be equal to 0.0230 moles of HCl.2936

                  The acid is monoprotic; we don't have to worry about diprotic or triprotic.2951

                  Then it is a 1:1 ratio with this amine.2957

                  This is going to be equal therefore to 0.0230 moles of the trimethyl amine that we had initially.2960

                  Let's go ahead and write out the neutralization reactions.2970

                  That is going to be HCl aqueous plus the trimethyl amine aqueous.2973

                  That is going to give us HN1+(CH3)3 and then Cl1- aqueous.2983

                  Let's go ahead and plug in all our values--0.0230 moles, 0.0230 moles, and then 0 and 0.2992

                  Again this neutralization reaction, 1:1 ratio, we have the identical amounts of each reactant.3004

                  All the reactants are going to be consumed entirely; this is -0.0230 moles, -0.0230 moles.3010

                  This is going to be +0.0230 moles and +0.0230 moles.3020

                  This goes as 0; this goes as 0.3029

                  This is 0.0230 moles; this is 0.0230 moles; the neutralization step is done.3033

                  Now we can go ahead and proceed on to write the equilibrium with direct reaction with water.3043

                  You have to choose: is it going to be the conjugate acid or is it going to be Cl-?3049

                  We already know from a previous lecture that Cl- is the conjugate of HCl.3056

                  It will not react with water to reform HCl.3063

                  Therefore you are going to select the conjugate acid of the amine.3066

                  HN(CH3)31+ aqueous direct reaction with water goes on to form H3O1+ aqueous and the trimethyl amine weak base.3071

                  This is going to be 0.0230 moles, 0, and 0.3092

                  Remember final values of table one become initial values of table two.3100

                  This is ?x, +x, and +x; this is 0.0230 minus x, x, and x.3103

                  You are going to use Ka for this expression.3113

                  Ka is going to be equal to the following expression: x squared over 0.0230 moles minus x.3116

                  Again we need to get moles into molarity.3128

                  This is simply going to be 0.0230 moles divided by the total volume at the equivalence point which is going to be 0.2736 liters.3132

                  When all is said and done, you can go ahead and solve for x.3143

                  X will give you the hydronium ion concentration which will then get you to pH.3148

                  As you can see, this titration was a weak base with a strong acid.3154

                  We had better expect the pH to be acidic or less than 73159

                  as we saw in the previous slide where we introduced the graphs.3163

                  That is going to be the case because we only have hydronium being formed with no hydroxide.3168

                  That is region three.3175

                  Finally let's go ahead and solve how to do region four.3177

                  In region number four, we have nothing but excess titrant added; region four, extra HCl added in this case.3181

                  What you need to calculate is the amount of HCl remaining after neutralization.3194

                  Need to determine amount of HCl leftover after neutralization has occurred; after neutralization.3201

                  We know that 23.6 milliliters was required for neutralization.3222

                  Here we are given a total of 27.5 milliliters; we are just going to subtract.3227

                  That tells us that 3.9 milliliters of HCl still remains after neutralization.3234

                  Because this is a strong acid, we can use the pH equation directly.3243

                  Right now we need to get this into molarity.3248

                  We are going to say 0.0039 liters of the HCl.3255

                  We are going to multiply that by the molarity which is 0.974.3259

                  That is going to give us moles.3265

                  To get the molarity, we are going to divide by the total volume at that point.3267

                  The total volume at this point is actually 0.2775 liters.3271

                  That is going to give us our concentration of HCl which because HCl is monoprotic, gives us the concentration of H3O1+.3279

                  From there, we can go ahead and calculate the pH.3289

                  Again because this is region four of this titration curve, the pH had better be3293

                  much less than 7 because this is the titration of a weak base by a strong acid.3299

                  At this point, we have nothing but strong acid remaining.3304

                  Again this as you can see, each of the four different regions of3311

                  a titration curve require its own different strategy for solving the pH.3315

                  I want to thank you for your time; I will see next time on Educator.com.3321