 # Acid-Base Equilibria and Solubility Equilibria

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Acid-Base Equilibria and Solubility Equilibria
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A buffer solution is a solution of: A weak acid or a weak base and
The salt of the weak acid or weak base Both must be present! A buffer solution has the ability to resist changes in pH upon the addition of small amounts of either acid or base. Consider an equal molar mixture of CH3COOH and CH3COONa Add strong acid H+ (aq) + CH3COO- (aq) CH3COOH (aq) Add strong base OH- (aq) + CH3COOH (aq) CH3COO- (aq) + H2O (l) 16.3

A Buffered Solution . . . resists change in its pH when either H+ or OH are added. 1.0 L of 0.50 M H3CCOOH M H3CCOONa pH = 4.74 Adding mol solid NaOH raises the pH of the solution to 4.76, a very minor change. Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Which of the following are buffer systems? (a) KF/HF
(b) KBr/HBr, (c) Na2CO3/NaHCO3 (a) HF is a weak acid and F- is its conjugate base buffer solution (b) HBr is a strong acid not a buffer solution (c) CO32- is a weak base and HCO3- is its conjugate acid buffer solution 16.3

Henderson-Hasselbalch Equation

Buffered Solution Characteristics
Buffers contain relatively large amounts of weak acid and corresponding base. Added H+ reacts to completion with the weak base. Added OH reacts to completion with the weak acid. The pH is determined by the ratio of the concentrations of the weak acid and weak base. Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system. What is the pH after the addition of 20.0 mL of M NaOH to 80.0 mL of the buffer solution? Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Buffering Capacity . . . represents the amount of H+ or OH the buffer can absorb without a significant change in pH. Buffer solution M HNO2/0.1M KNO2 Buffer solution M HNO2/1.0M KNO2 Both solutions buffer to the same pH Buffer solution 2 has a larger capacity Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Titrations In a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete. Equivalence point – the point at which the reaction is complete (stoichiometry) Indicator – substance that changes color at (or near) the equivalence point Endpoint – point where indicator changes color Slowly add base to unknown acid UNTIL The indicator changes color (pink) 4.7

Strong Acid-Strong Base Titrations
NaOH (aq) + HCl (aq) H2O (l) + NaCl (aq) OH- (aq) + H+ (aq) H2O (l) 16.4

Weak Acid-Strong Base Titrations
CH3COOH (aq) + NaOH (aq) CH3COONa (aq) + H2O (l) CH3COOH (aq) + OH- (aq) CH3COO- (aq) + H2O (l) At equivalence point (pH > 7): CH3COO- (aq) + H2O (l) OH- (aq) + CH3COOH (aq) 16.4

pH = pKa at halfway point

Strong Acid-Weak Base Titrations
HCl (aq) + NH3 (aq) NH4Cl (aq) H+ (aq) + NH3 (aq) NH4Cl (aq) At equivalence point (pH < 7): NH4+ (aq) + H2O (l) NH3 (aq) + H+ (aq) 16.4

pH = pKa at halfway point

Strong Acid – Strong Base Titration
Step 1 - A stoichiometry problem - reaction is assumed to run to completion - then determine remaining species. Step 2 - Determine pH based on concentration of excess reactant (strong acid or base leftover) Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Consider the titration of 80.0mL of 0.10 M Sr(OH)2 by 0.4M HCl. Calculate the pH after the following volumes of HCl have been added. 0.0 mL 20.0 mL 30.0 mL 40.0 mL 80.0 mL Copyright©2000 by Houghton Mifflin Company. All rights reserved.

0 mol (equivalence point)
Sr(OH)2 + 2HCl  SrCl2 + 2H2O Moles OH- Moles H+ Moles of Excess Total Volume (mL) pH Calculations 0.016 0.016 mol OH- 80 0.008 0.008 mol OH- 100 0.012 0.004 mol OH- 110 0 mol (equivalence point) 120 0.032 0.016 mol H+ 160

Weak Acid - Strong Base Titration Weak Base – Strong Acid Titration
Step 1 - A stoichiometry problem - reaction is assumed to run to completion - then determine remaining species. Step 2 - An equilibrium problem - determine position of weak acid or base equilibrium and calculate pH. Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Consider the titration of 100.0mL of 0.10 M HC2H3O2 by 0.2M KOH. Calculate the pH after the following volumes of KOH have been added. 0.0mL 20.0 mL 25.0 mL 40.0 mL 50.0 mL 100.0 mL Copyright©2000 by Houghton Mifflin Company. All rights reserved.

KOH + HC2H3O2  KC2H3O2 + H2O KOH HC2H3O2  KC2H3O H2O Before 0.004 0.01 NA Change -0.004 +0.004 After 0.006 Copyright©2000 by Houghton Mifflin Company. All rights reserved.

KOH + HC2H3O2  KC2H3O2 + H2O KOH HC2H3O2  KC2H3O H2O Before 0.005 0.01 NA Change -0.005 +0.005 After Copyright©2000 by Houghton Mifflin Company. All rights reserved.

KOH + HC2H3O2  KC2H3O2 + H2O KOH HC2H3O2  KC2H3O H2O Before 0.008 0.01 NA Change -0.008 +0.008 After 0.002 Copyright©2000 by Houghton Mifflin Company. All rights reserved.

KOH + HC2H3O2  KC2H3O2 + H2O KOH HC2H3O2  KC2H3O H2O Before 0.02 0.01 NA Change -0.01 +0.01 After Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Acid-Base Indicator . . . marks the end point of a titration by changing color. commonly weak acids (HIn) that change color when they dissociate (In-) Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Acid-Base Indicators HIn (aq) H+ (aq) + In- (aq) [HIn]  10
Color of acid (HIn) predominates  10 [HIn] [In-] Color of conjugate base (In-) predominates 16.5

The titration curve of a strong acid with a strong base.
16.5

Which indicator(s) would you use for a titration of HNO2 with KOH ?
Weak acid titrated with strong base. At equivalence point, will have conjugate base of weak acid. At equivalence point, pH > 7 Use cresol red or phenolphthalein 16.5

Solubility Product For solids dissolving to form aqueous solutions.

Solubility Equilibria
AgCl (s) Ag+ (aq) + Cl- (aq) Ksp = [Ag+][Cl-] Ksp is the solubility product constant MgF2 (s) Mg2+ (aq) + 2F- (aq) Ksp = [Mg2+][F-]2 Ag2CO3 (s) Ag+ (aq) + CO32- (aq) Ksp = [Ag+]2[CO32-] Ca3(PO4)2 (s) Ca2+ (aq) + 2PO43- (aq) Ksp = [Ca2+]3[PO43-]2 Dissolution of an ionic solid in aqueous solution: Q < Ksp Unsaturated solution No precipitate Q = Ksp Saturated solution Q > Ksp Supersaturated solution Precipitate will form 16.6

Solubility Product Bi2S3(s)  2Bi3+(aq) + 3S2(aq) “Solubility” = s = concentration of Bi2S3 that dissolves, which equals 1/2[Bi3+] and 1/3[S2]. Note:Ksp is constant (at a given temperature) s is variable (especially with a common ion present) Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Molar solubility (mol/L) is the number of moles of solute dissolved in 1 L of a saturated solution.
Solubility (g/L) is the number of grams of solute dissolved in 1 L of a saturated solution. 16.6

What concentration of Ag is required to precipitate ONLY AgBr in a solution that contains both Br- and Cl- at a concentration of 0.02 M? (AgBr Ksp= 7.7 x 10-13) Copyright©2000 by Houghton Mifflin Company. All rights reserved.

The Common Ion Effect and Solubility
The presence of a common ion decreases the solubility of the salt. What is the molar solubility of AgBr in (a) pure water and (b) M NaBr? 16.8

pH and Solubility Mg(OH)2 (s) Mg2+ (aq) + 2OH- (aq)
The presence of a common ion decreases the solubility. Insoluble bases dissolve in acidic solutions Insoluble acids dissolve in basic solutions Mg(OH)2 (s) Mg2+ (aq) + 2OH- (aq) Ksp = [Mg2+][OH-]2 = 1.2 x 10-11 Ksp = (s)(2s)2 = 4s3 (Lower [OH-]) At pH less than 10.45 4s3 = 1.2 x 10-11 Increase solubility of Mg(OH)2 s = 1.4 x 10-4 M [OH-] = 2s = 2.8 x 10-4 M At pH greater than 10.45 (Raise [OH-]) pOH = pH = 10.45 Decrease solubility of Mg(OH)2 16.9