Applications of Acid-Base Equilibria Acids and Bases Part II Applications of Acid-Base Equilibria

Common Ion Effect Calculate [H+] and the percent dissociation of HF in a solution containing 1.0 M HF (Ka = 7.2x10-4) and 1.0 M NaF. 1.0 M 1.0 M co change ceq

Imagine 1 M HF solution at equilibrium, then NaF is added. Previous Example: Calculate [H+] and the percent dissociation of HF in a solution containing 1.0 M HF (Ka = 7.2x10-4). 1 M HF 1 M HF + 1 M NaF x 4.2x10-3 7.2x10-4 pH 1.57 3.14 a% 0.42% 0.072% Le Chatelier Imagine 1 M HF solution at equilibrium, then NaF is added. What happens? Common Ion Effect

Buffered Solution is a solution that persists the change in its pH when H+ or OH- is added. Solution of weak acid and the salt of its conjugate base: HAc/NaAc Solution of weak base abd the salt of its conjugate acid: NH3/NH4+ Blood is a buffered solution.

A buffered solution contains 0. 50 M acetic acid HC2H3O2, Ka=1 A buffered solution contains 0.50 M acetic acid HC2H3O2, Ka=1.8x10-5) and 0.50 M sodium acetate (NaC2H3O2). Calculate the pH of this solution. 0.5 M 0.5 M co change ceq

Calculate the change in pH that occurs when 0 Calculate the change in pH that occurs when 0.010 mol solid NaOH is added to 1.0 L of the above buffered solution. 0.01 mol 0.01 mol n = M x V (before reaction) 0.5x1.0 =0.5 mol change -0.01 +0.01 0.49 mol 0.51 co 0.49 0.51 change -x x ceq 0.49-x 0.49+x

Compare this pH change with that which occurs when 0 Compare this pH change with that which occurs when 0.010 mol solid NaOH is added to 1.0 L water. 0.01 mol 0.01 mol

Henderson-Hasselbalch Equation

co [HA]o [A-]o change -x x ceq [HA]o-x [A-]o+x Calculate the pH of a solution containing 0.75 M lactic acid (Ka=1.4x10-4) and 0.25 M sodium lactate.

A buffered solution contains 0. 25 M NH3 (Kb=1. 8x10-5) and 0 A buffered solution contains 0.25 M NH3 (Kb=1.8x10-5) and 0.40 M NH4Cl. Calculate the pH of this solution. NH3 NH4+

Calculate the pH of the solution that results when 0 Calculate the pH of the solution that results when 0.10 mol gaseous HCl is added to 1.0 L of the buffered solution containing 0.25 M NH3 (Kb=1.8x10-5) and 0.40 M NH4Cl. 0.1 mol 0.1 mol n = M x V (before reaction) 0.25x1.0 =0.25 mol 0.40x1.0 =0.40 mol change -0.1 +0.1 0.15 mol 0.50

Acid addition pH ↓ Base addition pH ↑

Buffering Capacity It represents the amount of H+ or OH the buffer can absorb without a significant change in pH. High buffering capacity: solution absorbs large amounts of acid or base without significant change in pH. Low buffering capacity: small amounts of acid or base can produce a significant change in pH. Factors affecting the buffering capacity: 1) The concentrations of buffer components The more concentrated the components of a buffer , the greater the buffer capacity. 2) The ratio [A-]/[HA] The closer this ratio is to 1, the higher the buffering capacity.

0.01 mol solid NaOH added to 1 L buffer. The relation between buffer capacity and pH change. 0.01 mol solid NaOH added to 1 L buffer.

maximum buffer capacity Factors affecting the buffering capacity: The concentrations of buffer components 2) The ratio [A-]/[HA] The closer this ratio is to 1, the higher the buffering capacity. maximum buffer capacity

Problem A chemist needs a solution buffered at pH= 4.30 and can choose from the following acids and their sodium salts: chloroacetic acid, Ka= 1.35 x 10-3 propanoic acid, Ka= 1.3 x 10-5 3. benzoic acid, Ka= 6.4 x 10-5 4. hypochlorous acid, Ka= 3.5 x 10-8 Calculate the ratio [A-]/[HA] required for the best system to yield a pH of 4.30. pKa 2.87 4.89 4.19 7.46

Strong acid – Strong base titration HCl(aq) + NaOH(aq)  H2O + NaCl(aq) net ionic equation: H+(aq) + OH-(aq)  H2O

HNO3(aq) + NaOH(aq)  H2O + NaCl(aq) Titration of 50 mL 0.200 M HNO3 with 0.100 M NaOH A. No NaOH has been added. B. 10 mL NaOH has been added. HNO3(aq) + NaOH(aq)  H2O + NaCl(aq)

HNO3(aq) + NaOH(aq)  H2O + NaCl(aq) C. 20 mL NaOH (total) has been added. HNO3(aq) + NaOH(aq)  H2O + NaCl(aq) D. 50 mL NaOH (total) has been added.

HNO3(aq) + NaOH(aq)  H2O + NaCl(aq) So on until: E. 100 mL NaOH (total) has been added. HNO3(aq) + NaOH(aq)  H2O + NaCl(aq) Stoichiometric point or point of equivalence

HNO3(aq) + NaOH(aq)  H2O + NaCl(aq) F. 150 mL NaOH (total) has been added. HNO3(aq) + NaOH(aq)  H2O + NaCl(aq)

HNO3(aq) + NaOH(aq)  H2O + NaCl(aq) F. 200 mL NaOH (total) has been added. HNO3(aq) + NaOH(aq)  H2O + NaCl(aq)

Weak acid – Strong base titration Titration of 50 mL 0.100 M HAc with 0.100 M NaOH A. No NaOH has been added. pH of 0.10 M HAc?

HAc(aq) + OH-(aq)  H2O + Ac-(aq) B. 10 mL NaOH has been added. HAc(aq) + OH-(aq)  H2O + Ac-(aq)

HAc(aq) + OH-(aq)  H2O + Ac-(aq) C. 25 mL NaOH has been added. HAc(aq) + OH-(aq)  H2O + Ac-(aq)

HAc(aq) + OH-(aq)  H2O + Ac-(aq) D. 40 mL NaOH has been added. HAc(aq) + OH-(aq)  H2O + Ac-(aq)

HAc(aq) + OH-(aq)  H2O + Ac-(aq) E. 50 mL NaOH has been added. HAc(aq) + OH-(aq)  H2O + Ac-(aq) co change ceq

HAc(aq) + NaOH(aq)  H2O + NaAc(aq) F. 60 mL NaOH (total) has been added. HAc(aq) + NaOH(aq)  H2O + NaAc(aq) G. 75 mL NaOH (total) has been added.

HCN(aq) + OH-(aq)  H2O + CN-(aq) Hydrogen cyanide gas (HCN), a powerful respiratory inhibitor, is highly toxic. It is a very weak acid (Ka =6.2x10-10) when dissolved in water. If a 50.0-mL sample of 0.100 M HCN is titrated with 0.100 M NaOH, calculate the pH of the solution a. after 8.00 mL of 0.100 M NaOH has been added. HCN(aq) + OH-(aq)  H2O + CN-(aq)

HCN(aq) + OH-(aq)  H2O + CN-(aq) b. at the halfway point of the titration. HCN(aq) + OH-(aq)  H2O + CN-(aq) halfway point of the titration:

HCN(aq) + OH-(aq)  H2O + CN-(aq) c. at the equivalence point of the titration. HCN(aq) + OH-(aq)  H2O + CN-(aq) co change ceq pH=10.96

Titration Curves of Polyprotic Acids pH = pKa2 pOH from Kb equil’m of A2– pH = pKa1 V/2 V 3V/2 2V

A chemist dissolves 2. 00 mmol of a solid acid in 100 A chemist dissolves 2.00 mmol of a solid acid in 100.0 mL water and titrates the resulting solution with 0.0500 M NaOH. After 20.0 mL NaOH has been added, the pH is 6.00. What is the Ka value for the acid?

Weak base – Strong acid titration The pH curve for the titration of 100.0 mL of 0.050 M NH3 with 0.10 M HCl. Note the pH at the equivalence point is less than 7, since the solution contains the weak acid NH4+. .

HIn (aq)  H+ (aq) + In– (aq) Acid –Base Indicators Organic Dyes weak acid-weak base conjugate pair HIn (aq)  H+ (aq) + In– (aq) Acid form base form (one color) (another color) If you put a small amount of HIn in solution Will be one color if acidic (phenolphthalein = colorless) Another color if basic (phenolphthalein = bright magenta) Color will change when pH rises or falls

HIn (aq)  H+ (aq) + In– (aq) Color Change Bromothymol Blue HIn (aq)  H+ (aq) + In– (aq) Acid form base form (one color) (another color) If , the color of In- will be observed. If , the color of HIn will be observed.

If , the color of In- will be observed. If , the color of HIn will be observed. Color Change occurs in the range pH=pKIn1

Bromothymol Blue KIn=1.0x10-7 Color Change occurs in the range pH=pKa1 pH=71=6-8 pH<6 pH=6-8 pH>8

Colors and approximate pH range of some common acid-base indicators.

Best indicator when pH at the point of equiv. is nearest to pKIn Is this indicator suitable for your titration? titration of 50 mL of 0.1 M HC2H3O2 with 0.1 M NaOH. 100.0 mL of 0.10 M HCl with 0.10 M NaOH. To find if a given indicator is suitable, calculate pH at the point of equiv. Best indicator when pH at the point of equiv. is nearest to pKIn