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CHAPTER 15 – 2nd HALF APPLIC. OF ACID & BASE EQUILIBRIA (BUFFERS & TITRATIONS) Problems to prepare students for hourly exam II. 

Buffer concepts



Buffer calculations



Titrations and Indicators

E. Tavss 4/06

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BUFFER CONCEPTS 7 CHEM 162-2006 HOURLY EXAM III + ANSWERS CHAPTER 15 – APPLICATIONS OF AQUEOUS EQUILIBRIA BUFFER CONCEPTS Consider a potential buffer solution containing 2.0 M HCN and 1.0 M NaCN (Ka for HCN = 6.2 × 10-10). Which one of the following statements is true? A. The solution is not a buffer, because [HCN] is not equal to [CN-]. B. The pH will be below 7.00, because the concentration of the acid is greater than that of the base. C. [OH-] > [H+] D. The buffer will be more resistant to pH changes from addition of strong acid than of strong base. E. The pH will equal 7.00, because the acid is very weak. A. This is a weak acid and its conjugate base in a 1:2 ratio. A buffer can be a weak acid and its conjugate base in a 1:10 ratio to a 10:1 ratio. Hence, this is a buffer. “A” is false. B. The pH depends on the Ka vs the Kcb. The Ka is 6.2 x 10-10. The Kcb is (1 x 10-14)/(6.2 x 10-10) = 1.61 x 10-5. Therefore, the acid is a weaker acid in formation of hydronium ions as compared to the strength of the base in the formation of hydroxide ions. Therefore, more hydroxide ions will form than hydronium ions, so the pH will be above 7. “B” is false. C. For the reasons stated in “B”, “C” is true. D. The buffer will be just as resistant in its reaction to strong acid or base, as long as the buffer system is not overwhelmed with strong acid or base. “D” is false. E. The pH will only equal 7 if the Ka of the acid and conjugate base were equal, i.e., 1 x 10-7. Several of the answers can also be determined quantitatively: HCN

+

H 2O

 

Initial 2.0 Change -X Equilibrium 2.0 - X ([X] x [1.0 + X])/(2.0 – X) = 6.2 x 10-10 Simplifying: ([X] x [1.0])/(2.0 ) = 6.2 x 10-10 X = 1.24 x 10-9 pH = -log(1.24 x 10-9) = 8.9

H 3O + 0 +X +X

+

CN1.0 X 1.0 + X

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15 CHEM 162-2006 HOURLY EXAM III + ANSWERS CHAPTER 15 – APPLICATIONS OF AQUEOUS EQUILIBRIA BUFFER CONCEPTS What is the net ionic equation, when a solution of potassium hydroxide is added to a solution of nitrous acid. A. B. C. D. E.

HNO2 + K+ + OH- → KNO2 + H2O HNO2 + H2O → NO2- + H3O+ HNO2 + KOH → K+ + NO2- + H2O HNO2 + OH- → NO2- + H2O H+ + OH- → H2O

 HOH + KNO2 Co mp l e t ee q u a t i o n :HNO2+KOH  HOH + K+ + NO2Ionic equation: HNO2+K++OH-  HOH + NO2Net ionic equation: HNO2+OH-

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19. CHEM162-2000 HOURLY EXAM II CHAPTER 15 - APPLIC OF ACID & BASE EQUILIB (BUFFERS & TITRATIONS) BUFFER CONCEPTS Which one of the following statements concerning buffers is false? A. A buffer can consist of a weak base and a salt containing its conjugate acid in the same solution. B. A buffer changes pH only slightly when small amounts of strong acid or strong base are added. C. The pH of a buffer solution should be within one pH unit of the pKa of the acid component of the buffer. D. The pH of a buffer will not change significantly if the buffer is diluted. E. The buffer capacity will not change significantly if the buffer is diluted.

8.

CHEM162-2001 HOURLY EXAM II + ANSWERS CHAPTER 15 - APPLIC OF ACID & BASE EQUILIB. (BUFFERS & TITRATIONS) BUFFER CONCEPTS When dissolved in 1.0 L of water, which of the following would not produce a buffer solution? A buffer solution is a weak acid and the salt of the weak acid, or a weak base and the salt of the weak base. E is a strong acid and the salt of the strong acid. Hence, it is not a buffer. A. B. C. D. E.

1.0 mol HNO2 and 1.0 mol KNO2 1.0 mol NH3 and 1.0 mol NH4Br 1.0 mol HNO2 and 2.0 mol KNO2 1.0 mol HClO and 1.0 mol KClO 1.0 mol HCl and 1.0 mol KCl

CHEM162-2002 HOURLY EXAM II + ANSWERS CHAPTER 15 - APPLIC OF ACID & BASE EQUILIB. (BUFFERS & TITRATIONS) BUFFER CONCEPTS 14. Which of the following solutions will produce a buffer? Each has a volume of 1.0 Liter. A buffer contains a weak acid and its conjugate base, or a weak base and its conjugate acid. W. A mixture of 0.20 mol CH3COOH and 0.10 mol NaOH

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A mixture of 0.20 mol CH3COOH and 0.10 mol NaOH will go almost to completion to provide 0.10 mol CH3COOH and 0.10 mol CH3COO-Na+, a buffer solution. X. A mixture of 0.10 mol CH3COOH and 0.20 mol NaOH A mixture of 0.10 mol CH3COOH and 0.20 mol NaOH will go almost to completion to provide 0.10 mol CH3COO-Na+ and 0.10 mol NaOH, not a buffer solution.

Y. A mixture of 0.20 mol NH3 and 0.10 mol HC1 A mixture of 0.20 mol NH3 and 0.10 mol HCl will go almost to completion to provide 0.10 mol NH3 and 0.10 mol NH4Cl, a buffer solution. Z. A mixture of 0.10 mol NH3 and 0.20 mol HC1 A mixture of 0.10 mol NH3 and 0.20 mol HCl will go almost to completion to provide 0.10 mol NH4Cl and 0.10 mol HCl, not a buffer solution.

A. B. C. D. E.

all will be buffers Y and Z only X and Z only W and Y only W and X only

CHEM162-2003 HOURLY EXAM II + ANSWERS CHAPTER 15 - APPLIC OF ACID & BASE EQUILIB. (BUFFERS & TITRATIONS) 9. Which of the following combinations would be a buffer? W. 1.0 mol NaCl + 1.0 mol HCl in 1.00 liter solution X. 1.0 mol KNO3 + 1.0 mol HNO3 in 1.00 liter solution Y. 1.0 mol KNO2 + 1.0 mol HNO2 in 1.00 liter solution Z. 1.0 mol NH4Cl + 1.0 mol NH3 in 1.00 liter solution A. Y only B. W and X only C. X and Y only D. all will be buffers E. Y and Z only A buffer is a weak acid and the salt of a weak acid, or a weak base and the salt of a weak base. “W” is a strong acid and the salt of a strong acid. “X” is a strong acid and the salt of a strong acid. “Y” is a weak acid and the salt of a weak acid. “Z” is a weak acid and the salt of a weak acid. CHEM162-2003 HOURLY EXAM II + ANSWERS CHAPTER 15 - APPLIC OF ACID & BASE EQUILIB. (BUFFERS & TITRATIONS) 5

BUFFER CONCEPTS 23. Given the following information: H3PO4 Ka1 = 7.5 x 10-3 Ka2= 6.2x10-8 Ka3= 4.8x 10-13

H2SO4 Ka1 is large Ka2 = 1.2 x 10-2

Which one of the following combinations would be the best choice for preparing a buffer of pH 7.0? A. H3PO4 and NaH2PO4 B. Na2HPO4 and Na3PO4 C. NaH2PO4 and Na2HPO4 D. NaHSO4 and Na2SO4 E. H2SO4 and NaHSO4  H3O+ + H2PO4Ka1: H3PO4 + H2O 

pKa1 = -log(7.5x10-3) = 2.12

 H3O+ + HPO42Ka2: H2PO4- + H2O 

pKa1 = -log(6.2x10-8) = 7.21

 H3O+ + PO43Ka3: HPO42- + H2O 

pKa1 = -log(4.8x10-13) = 12.32

 H3O+ + HSO4Ka1: H2SO4 + H2O 

pKa1 = -log(large) = ~-7

 H3O+ + SO42Ka2: HSO4- + H2O  Henderson-Hasselbalch equation: pH = pKa + log([Base]/[Acid])

pKa1 = -log(1.2x10-2) = 1.92

Use the Henderson-Hasselbalch equation for virtually all buffer problems. If we assume that the base and the acid are equal in concentration, then log([Base]/[Acid]) equals log(1). Since log(1) equals zero, then pH = pKa. In this case, we would want a buffer whose acid has a pKa of approximately 7. The acid that closest to that is H2PO4-, with pKa1 of 7.21. The conjugate base of H2PO4- is HPO42-. So the buffer system is H2PO4- is HPO42-. If we write it with the corresponding cations (spectator ions), it would be NaH2PO4 is Na2HPO4. (Correspondingly, if we want a buffer of pH 2.12, then H3PO4 and H2PO4- would be perfect because, at equal concentrations of H3PO4 and H2PO4-, log([A-]/[HA]) would be equal to log(1), which equals zero, so since pKa1 = 2.12, then pH = 2.12.)

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18.

CHEM 162-2000 FINAL EXAM CHAPTER 15 - APPLIC OF ACID & BASE EQILIBR (BUFFERS & TITR) BUFFER CONCEPTS Which of the following would not produce a buffer solution when dissolved in 500 mL of water? A. B. C. D. E.

11.

D

0.5 mol HCN and 0.2 mol KCN 0.5 mol HCN and 2.0 mol KCN 1.0 mol HCN and 0.5 mol KOH 0.5 mol HClO4 and 0.5 mol KClO4 0.5 mol CH3NH2 and 0.5 mol CH3NH3Cl

CHEM-2001 FINAL EXAM + ANSWERS CHAPTER 15 - APPLIC OF ACID & BASE EQUILIB (BUFFERS & TITRATION) BUFFER CONCEPTS Which of the following would produce a buffer solution when dissolved in 500 mL of water? A buffer solution is a weak acid and a salt of a weak acid, or a weak base and a salt of a weak base. The ratios are usually from 1:100 to 100:1. “I” is a weak acid and a salt of a weak acid, in equal amounts. “II” is a weak acid and a salt of a weak acid, in a ratio acceptable for a buffer. In “III” the strong base will completely react with the weak acid producing H2O and 0.5 mol KF. This does not satisfy the definition of a buffer. In “IV” the strong base, which is the limiting reactant, will completely react with 0.3 mol of the weak acid, producing H2O, 0.3 mol KF, and leaving 0.2 mol HF. 0.3 mol KF and 0.2 mol HF constitute a buffer. I. 0.5 mol HF and 0.5 mol KF II. 0.5 mol HF and 0.3 mol KF III. 0.5 mol HF and 0.5 mol KOH IV. 0.5 mol HF and 0.3 mol KOH A. B. C. D. E.

I only I and III only II and IV only I, II, and IV only I, II, III, and IV

Chem 162-2003 Final exam + answers

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Chapter 15A-Applications of Aqueous Equilibria (of Acids and Bases) Buffer concepts (including Henderson-Hasselbalch equation) 23. Which of the following is false for a buffer solution? A. A buffer solution can be prepared by mixing appropriate amounts of a weak base and a strong acid. B. Exclusive of water, at least one component of a buffer solution must be neutral C. A buffer solution can consist of a mixture of a weak acid and its conjugate base. D. The solution resists changes in its [H+]. E. Added H+ ions will react with the conjugate base of the weak acid in solution A. True. This is one way to make a buffer. B. False. There is no reason to believe that a buffer can’t consist of two charged species. For example, the combination of H2PO4- and HPO42- forms a buffer at pH of 7.2. C. True. That’s one of the definitions of a buffer. D. True. That’s what buffers do. E. True. That’s how buffers work against acid addition.

17.

Chem 162-2005 Exam II Applications of Aqueous Equlibria - Chapter 15A Buffers, common ions, Henderson-Hasselbalch equation Concept Which of the following would not produce a buffer solution when dissolved in 500 mL of water? A. B. C. D. E.

0.50 mol HCN and 0.20 mol KCN 0.20 mol HCN and 0.50 mol KCN 0.50 mol HClO4 and 0.50 mol KClO4 0.50 mol CH3NH2 and 0.50 mol CH3NH3Cl 1.0 mol HCN and 0.50 mol KOH

A. A combination of a weak acid and its conjugate base is a buffer. B. A combination of a weak acid and its conjugate base is a buffer. C. A combination of a strong acid and its conjugate base is not a buffer. D. A combination of a weak base and its conjugate acid is a buffer. E. Reaction between 1.0 mol of a weak acid and 0.50 mol of a strong base results in 0.5 mol of the weak acid remaining, and 0.50 mol of the conjugate base formed. This is a buffer.

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9.

Chem 162-2005 Final Exam + Answers Chapter 15 - Applications of Aqueous Equilibria (of Acids and Bases) Buffers (including Henderson-Hasselbalch equation) Concepts Consider a 100 mL buffer solution of HA and NaA of pH = 5. If the solution is diluted to 500 mL by adding water, which one of the following will change?

A. B. C. D. E.

The ratio of [HA]/[A-]in Henderson-Hasselbalch equation The pKb of the conjugate base of the acid The capacity of a given volume of the buffer The pH of the solution The pKa of the acid

 H 3O + + A HA + H2O  pH = 5 [H3O+] = 10-5 = 1 x 10-5 M HA Initial Change Equilibrium

X 0.2X

+

H2O

 

H 3O + 1 x 10-5

+

AZ 0.2Z

Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]) (A) Before dilution: [HA]/[A-] = X/Z After dilution: [HA]/[A-] = (0.2X)/(0.2Z) = X/Z, i.e., no change (B) The calculation of pKb’ is related to Kw and Ka, not to the concentration of HA or A-, so it is not related to a change in concentration of HA or A-. (C) See page 727 in Zumdahl. “The capacity of a buffered solution is determined by the magnitudes of [HA] and [A-].” As an example, if [A-] is high, then the solution will buffer the effect of a lot of strong acid, but if [A-] is low, then as soon as the [A-] is depleted, the pH will drop sharply. (D) According to the Henderson-Hasselbalch equation, if the ratio of the concentrations of A- and HA doesn’t change, and if the pKa doesn’t change, then the pH won’t change. If we don’t use the Henderson-Hasselbalch equation for the calculation, but use the conventional ICE table type of equilibrium calculations, then we’ll see that, in fact, the pH does change, However the pH change will be very small. The buffer capacity change is more meaningful than the pH change, and is therefore the better answer of the two. (E) The pKa of the acid is related to the Ka of the acid, which is unrelated to the concentration of the acid (or the concentration of the conjugate base).

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BUFFER CALCULATIONS 2 CHEM 162-2006 HOURLY EXAM III + ANSWERS CHAPTER 15 – APPLICATIONS OF AQUEOUS EQUILIBRIA BUFFER CALCULATIONS Which one of the following solutions will be the best buffer at a pH of 9.26? (Ka for HC2H3O2 is 1.8 × 10-5, Kb for NH3 is 1.8 × 10-5). A. B. C. D. E.

0.10 M HC2H3O2 and 0.10 M NaC2H3O2 5.0 M HC2H3O2 and 5.0 M NaC2H3O2 0.10 M NH3 and 0.10 M NH4Cl 5.0 M NH3 and 5.0 M NH4Cl 5.0 M HC2H3O2 and 5.0 M NH3 ANS:D

pH = pKa + log([A-]/[HA]) or pH = pKca + log([B]/[BH+]) A.p H=l o g ( 1 . 8 x 1 0 5)+l o g( 0 . 1 / 0 . 1 )=4 . 7 4 B.p H=l o g( 1 . 8 x 1 0 5 )+l o g ( 5 . 0 / 5 . 0 )=4 . 7 4 5 C.p H=l o g ( ( 1 x 1 0-14) / ( 1. 8 x 1 0 ) )+l o g ( 0 . 1 0 / 0 . 1 0)=9 . 26 5 D.p H=l o g ( ( 1 x1 0-14) / ( 1 . 8 x 1 0 ) )+l o g ( 5 . 0/ 5 . 0 )=9 . 2 6Th i swi l lb eab e t t e rb u ffe rt h a n “ C”b e c a u s ei th a sah i g he rb u ffe rc a p a c i t y . E.Ab u ffe rc a n n o tb ema d ewi t hawe a ka c i da n dawe a kb a s e , b e c a u s et he ya r et wowe a k t of or me n o u gho ft h ec o n j u g a t ea c i do rc o n j u g a t eb a s et os a t i s f yt h e1 : 10t o1 0 : 1r u l ef or ma ki n gb u ffe r s .

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6 CHEM 162-2006 HOURLY EXAM III + ANSWERS CHAPTER 15 – APPLICATIONS OF AQUEOUS EQUILIBRIA BUFFER CALCULATIONS HF is a weak acid with Ka = 7.2 × 10-4. What is the pH of a solution that contains 0.50 M HF and 0.60 M NaF? A. B. C. D. E.

1.72 3.22 3.44 5.53 8.46 HF

Initial Change Equilibrium

+

H 2O

0.50 -X 0.50-X

 

H 3O + 0 +X +X

+

F0.60 +X 0.60+X

pH = pKa + log([A-]/[HA]) Simplify, by dropping the –X and +X. pH = -log(7.2x10-4) + log(0.60/50) pH = 3.22

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21 CHEM 162-2006 HOURLY EXAM III + ANSWERS CHAPTER 15 – APPLICATIONS OF AQUEOUS EQUILIBRIA BUFFER CALCULATIONS Given the following data: H2CO3 ⇌ HCO3- + H+ Ka = 4.0 × 10-7 H2PO4- ⇌ HPO42- + H+ Ka = 6.3 × 10-8 A buffer of pH 6.4 may be prepared by using either a mixture of H2CO3 and HCO3- or a mixture of H2PO4- and HPO42-. Which one of the following statements is true concerning pH 6.4 buffers? A. B. C. D. E.

[H2CO3] > [HCO3-] and [H2PO4-] > [HPO42-] [H2CO3] = [HCO3-] and [H2PO4-] > [HPO42-] [H2CO3] = [HCO3-] and [HPO42-] > [H2PO4-] [HCO3-] > [H2CO3] and [HPO42-] > [H2PO4-] [H2CO3] > [HCO3-] and [HPO42-] > [H2PO4-]

pH = pKa + log([A-]/[HA]) For the H2CO3/HCO3- system, pKa = -log(4.0 x 10-7) = 6.40 For the H2CO3/HCO3- buffer system, according to HH: 6.4 = 6.4 + log([A-]/[HA]) Therefore, log([A-]/[HA]) = 0 Therefore, [A-]/[HA] = 1 Therefore, [A-] = [HA]. For the H2PO4-/HPO42- system, pKa = -log(6.3x10-8) = 7.20 6.4 = 7.20 + log([A-]/[HA]) For the H2PO4-/HPO42- buffer system, according to HH the base concentration must be < the acid concentration in order to lower the 7.20 pH to 6.40. [H2PO4-] > [HPO42-]. “B” meets these requirements.

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22 CHEM 162-2006 HOURLY EXAM III + ANSWERS CHAPTER 15 – APPLICATIONS OF AQUEOUS EQUILIBRIA BUFFER CALCULATIONS H2A is a diprotic acid with Ka1 = 2.5 × 10-5, Ka2 = 3.1 × 10-9. If 100.0 mL of 1.00 M NaOH is added to 200.0 mL of 1.00 M H2A, calculate the pH of the resultant solution. A. 13.4 B. 11.0 C. 8.51 D. 6.56 E. 4.60 This is a very difficult problem. I spent more time on it then on any other problem, and I’m still not certain how to do it. E.T.  H3O+ + HAKa1 = 2.5x10-5 H2A+H2O  + 2 HA + H2O  H3O + A Ka2 = 3.1x10-9 H2A Initial

+

OH-

 

H 2O +

1.00x(0.200/(0.200+0.100) 1.00x(0.100/(0.100+0.200) = 0.0667M = 0.0333M

HA0

Change Equilibrium Br i ngt oc ompl e t i on

H2O HA+ Initial 1.00x(0.200/(0.200+0.100) 1.00x(0.100/(0.100+0.200) 0 = 0.0667M = 0.0333M Change -0.0333 -0.0333 +0.0333 Equilibrium +0.0333 0 +0.0333 H 2A

+

OH-

 

pH = pKa + log([A-]/[HA]) 5 pH =l og( 2. 5 x1 0 )+l og( [ 0 . 0 3 3 3 ] /[ 0 . 0 3 3 3] )=4 . 6 0 pOH =1 4 . 0–4 . 6 0=9 . 4 0 9 . 4 0 1 0 [ OH-]=1 0 =3 . 9 8x1 0 Fr om fir s tr e a c t i on, pH=4 . 6 0 .The r e f or e , [ H O+]=2 . 5 1 x10-5 3 + HA + H 3O +  H2O  Initial Change Equilibrium

0.0333 -X 0.0333-X

2.51x10-5 +X (2.51x10-5)+X

A20 +X +X

[ H3O+]=2 . 5 1 x 10-5 p H=l o g ( 2 . 5 1 x 1 0-5)=4 . 6 0

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24 CHEM 162-2006 HOURLY EXAM III + ANSWERS CHAPTER 15 – APPLICATIONS OF AQUEOUS EQUILIBRIA BUFFER CALCULATIONS HOCl is a weak acid with Ka = 3.5 × 10-8. A buffer solution is prepared which contains 0.50 M HOCl and 0.40 M NaOCl. What is the pH after 10.0 mL of 1.0 M NaOH has been added to 100.0 mL of this buffer solution? A. B. C. D. E.

6.45 6.64 7.36 7.45 7.55 HOCl

Initial Change Equilibrium

+

OH-

 

H 2O

+

0.50

OCl0.40

Ad di n g1 0 . 0mLof1 . 0 M Na OHt o1 0 0mLoft h eb u ffe rs o l ut i o ni st h e s a mea sa d d i n g1 0 0 . 0mLo f1 . 0M Na OHt o1Lo ft h eb uffe rs o l u t i on . 1 0 0 . 0mLof1 . 0M Na OHi n1Li s0 . 1mo l ei nal i t e r , o r0 . 1 M OH. HOCl Initial Change Equilibrium

+

0.50

OH-

 

H 2O

+

0.10

OCl0.40

Si nc et hi si sar e a c t i onofas t r ongac i dwi t hawe akba s e , t her e a c t i ongoe st oc ompl e t i on.

HOCl Initial Change Equilibrium

0.50 -0.10 0.40

+

OH0.10 -0.10 0

 

H 2O

+

OCl0.40 +0.10 0.50

pH = pKa + log([A-]/[HA]) pH = -log(3.5x10-8) + log([0.50]/[0.40]) pH = 7.55

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CHEM162-2003 5TH WEEK RECITATION CHAPTER 14 - ACID AND BASE EQUILIBRIA ACID AND BASE EQUILIBRIA CALCULATIONS 27 Calculate the pH after 0.020 mol HCl is added to 1.00 L of 0.1 M sodium propanoate. (HC3H5O2, Ka = 1.3 x 10-5) (b) Approach I. This approach is not recommended, because it requires the use of a quadratic equation; i.e., X is too large to be dropped.  HP + H2O P - + H 3O +  Initial Change Equilibrium

P0.100 -X 0.100 - X

H3O+ 0.020 -X 0.020-X

HP 0 +X +X

X/((0.100 - X)(0.020 - X)) = 1/(1.3 x 10-5) = 7.69 x 104 X = 0.0199968 H3O+ = 3.2 x 10-6 pH = -log (3 x 10-6) = 5.49 Approach II. This approach is better than approach I, because it doesn’t require the use of a quadratic equation, but it is still not the best approach to solving this problem.  HP + H2O P - + H 3O +  Initial Change Equilibrium

P0.100 -0.020 0.080

H3O+ 0.020 -0.020 0

HP 0 +0.020 0.020

HP 0.020 +X 0.020+X

OH0 +X +X

 HP + OHP- + H2O  New Initial Change Equilibrium

P0.080 -X 0.080 - X

((0.020+X) x (X))...