Secondary Topics in Equilibrium

Secondary Topics in Equilibrium Outline 1. Common Ions 2. Buffers 3. Titrations Review 1. Common Ions Include the common ion into the equilibrium expression Calculate the molar solubility in mol L -1 when 0.1 M AgNO 3 is introduced to a saturated solution of Ag 2 SO 4 (s). Ksp = 1.2 x 10-5 Ag 2 SO 4 (s) 2Ag + (aq) + SO 4 2- (aq) [ ] = 1.2 x 1 (Assume 0.1+ 2x 0.1) Solve for x = 0.0012 M Note: When given ph, common ions such as [H + ] and [OH - ] can be substituted for. 2. Buffers Conjugate salts with a weak acid or base. Calculating the ph of a Buffer Calculate the ph of a solution that is 0.60 M HF and 1.00 M KF. Ka for HF is 7.2 x 10-4. x = 4.32 x 10-4 M; ph = 3.36 or Preparation of a Buffer Calculate the mass of NaC 2 H 3 O 2 required to prepare a buffer of ph 4.55 when added to 0.500 L of 0.67 M acetic acid. Ka = 1.8 x 10-5 for acetic acid. From ph, [H + ] = 10^(-pH) = 2.82 x 10-5 or

3. Titrations Weak Acid with Strong Base or Weak Base with Strong Acid Example: 50mL sample of 0.1 M HC 3 H 5 O 2 is titrated with 0.1 M NaOH. Ka = 1.3 x 10-5 Titration Equation HC 3 H 5 O 2 (aq) + OH - (aq) H 2 O (l) + C 3 H 5 O - 2 (aq) Determine Equivalence Point M 1 V 1 = M 2 V 2 0.1M * 0.05 L = 0.1M * V 2 V 2 = 0.05 L = 50 ml when moles are equal. ph Calculation ½ way to Equivalence Point (25mL of NaOH for the example above) ph = pka = - log (1.3 x 10-5 ) = 4.89 (Due to, therefore log ) ph Calculation Before Equivalence Point (30mL of NaOH for the example above) HA H + + A - 0.05L*0.1M = 0.005 moles 0 moles 0.03L*0.1M = 0.003 moles OH - 0.003 moles OH - 0.002 moles HA 0.003 moles A - 0.08L 0.08L 0.025M HA 0.0375 M A - Note: Since there is a conjugate base (same moles as the strong base) and a weak acid you can use Henderson-Hasselbach. ph = pka + log = 5.06 ph Calculation At the Equivalence Point (50mL of NaOH for the example above) HA H + + A - 0.05L*0.1M = 0.005 moles 0 moles 0.05L*0.1M = 0.005 moles OH - 0.005 moles OH - 0 moles HA 0.005 moles A - 0.1L 0 M HA 0.05 M A - Note: Since there is no weak acid, it turns into a weak acid problem. Kb = = 7.69 x 10-10 = x = [OH - ] = 6.20 x 10-6 M ph = 8.79 ph Calculation After the Equivalence Point (60mL of NaOH for the example above) HA H + + A - 0.05L*0.1M = 0.005 moles 0.06L*0.1M = 0.006 moles OH - 0.001 moles OH - 0.11 L 0.0091 M OH - Note: The difference is now a strong base. poh = - log (0.0091), ph = 11.96

AgBr (s) Ag + (aq) + Br (aq) 1. A saturated solution is prepared by adding excess silver bromide to distilled water to form 1.0 L of solution at 25 C. The Ksp for silver bromide is 5.0 x 10-13. a. Write the equilibrium-constant expression for the equation above. b. Calculate the molar solubility in mol L -1 when silver bromide is placed in pure distilled water. c. Calculate the molar solubility if a 0.030 M AgNO 3 solution is added to a saturated solution of silver bromide. d. What happens to the molar solubility if a saturated solution is prepared by adding excess silver bromide to distilled water to form 2.0 L of solution at 25 C? 2. A saturated solution of lead (II) iodide, PbI 2, in solution has a Ksp of 8.7 x 10-9. a. Develop of balanced chemical equilibrium reaction. b. Calculate the molar solubility in pure water. c. Calculate the mass in grams of NaI (FW = 150) that need to be added to excess Pb 2+ to form a precipitate? d. Calculate the molar solubility in a 0.050 mol L -1 Pb(NO 3 ) 2 solution. e. Calculate the molar solubility in a 0.010 mol L -1 NaI solution. 3. Calculate the molar solubility of LaF 3 (Ksp = 2.0 x 10-19 ) in mol L -1 in a. Pure water b. What happens to the molar concentration of [La 3+ ] if 0.020 M NaF is introduced to the water? Ag 2 CrO 4 (s) 2 Ag + (aq) + CrO 4 2- (aq) 4. A saturated solution of silver chromate is prepared in distilled water at 25 C as shown in the solubility chemical equation above. The concentration of CrO 4 2- (aq) in the saturated solution is found to be 6.7 x 10-5 mol L -1. a. Write the equilibrium-constant expression. b. Calculate the molar concentration of Ag + (aq) in the solution. c. Calculate the value of the equilibrium constant, Ksp. d. In a different saturated solution of silver chromate at 25 C, 3.32 g of silver chromate (FW = 331.8) was placed in the 500 ml solution. Calculate the molar concentration in mol L -1 of Ag + (aq) in the solution. 5. A 0.085 M propionic acid, HC 3 H 5 O 2 is prepared at 25 C. Ka for propionic acid = 1.3 x 10-5. a. Write the equilibrium constant expression for propionic acid. b. Calculate the ph of this solution at equilibrium. c. Calculate the ph of a buffer solution when 0.060 M KC 3 H 5 O 2 is placed in this propionic acid. 6. A 0.10 M solution of triethylamine, (CH 3 ) 3 N, is prepared at 25 C. Kb = 6.4 x 10-5 a. Write the balanced chemical reaction for the dissolution of triethylamine. b. Calculate the ph of this solution at equilibrium. c. Calculate the ph of a solution when 0.075 M (CH 3 ) 3 NCl is added. 7. A 0.20 M solution of hydrofluoric acid at 25 C has a Ka of 6.8 x 10-4. a. Write the equilibrium constant expression for hydrofluoric acid. b. Calculate the molar concentration of F - at 25 C. c. Calculate the ph of a solution that has 0.84 grams of NaF (FW = 42) placed in a 400 ml solution of 0.20 M hydrofluoric acid.

d. A solution that is made by mixing 125 ml of 0.050 M hydrofluoric acid with 50.0 ml of 0.10 M sodium fluoride. Volumes are additive. 8. Calculate the percent ionization of the following. a. 0.0075 M butanoic acid (Ka = 1.5 x 10-5 ) b. 0.0075 M butanoic acid in a solution containing 0.085 M sodium butanoate. c. 0.0085 M lactic acid (Ka = 1.4 x 10-4 ) d. 0.0085 M lactic acid in a solution containing 0.050 M sodium lactate. 9. Calculate the mass in grams of solid salt that should be added to form the following buffers assuming no volume change when the salt is added. a. NaBrO is added to 1.00 L of 0.050 M hypobromous acid (HBrO) to form a buffer solution of ph 9.15. (Ka = 2.5 x 10-9 ) b. NaC 3 H 5 O 3 is added to 1.00 L of 0.150 M lactic acid (HC 3 H 5 O 3 ) to form a buffer solution of ph 4.00. (Ka = 1.4 x 10-4 ) 10. Calculate the mass in grams of solid salt that should be added to form the following buffers assuming no volume change when the salt is added. a. NH 4 Cl is added to 2.00 L of 0.10 M NH 3 to form a buffer solution of ph 9.00. (Kb = 1.8 x 10-5 ) b. Sodium benzoate (NaC 7 H 5 O 2 ) is added to 250.0 ml of 0.20 M benzoic acid to form a buffer solution of ph 4.50. (Ka = 6.3 x 10-5 ) 11. Calculate the ratio of HCO 3 - to H 2 CO 3 in a solution that has a ph of 6.50? (Ka 1 = 4.3 x 10-7 ) 12. Calculate the ratio of NH 4 + to NH 3 in a solution that has a ph of 10.50? (Kb = 1.8 x 10-5 ) 13. A 35.0-mL sample of 0.150 M acetic acid, HC 2 H 3 O 2 (Ka = 1.8 x 10-5 ) is titrated with 0.150 M NaOH solution. a. Determine the titration chemical equation. b. Calculate the equivalence point. c. Calculate the ph after 17.5 ml of base has been added. d. Calculate the ph after 35.0 ml of base has been added. e. Calculate the ph after 36.0 ml of base has been added. 14. A 30.0-mL sample of 0.030 M NH 3 (Kb = 1.8 x 10-5 ) is titrated with 0.025 M HCl solution. a. Determine the titration chemical equation. b. Calculate the equivalence point. c. Calculate the ph after 18.0 ml of base has been added. d. Calculate the ph after 35.0 ml of base has been added. e. Calculate the ph after 36.0 ml of base has been added. 15. A 25.0-mL sample of 0.050 M HCN (Ka = 4.9 x 10-10 ) is titrated with 0.050 M NaOH solution. a. Determine the titration chemical equation. b. Calculate the equivalence point. c. Calculate the ph after 12.5 ml of base has been added. d. Calculate the ph after 25.0 ml of base has been added. e. Calculate the ph after 26.0 ml of base has been added.

16. Calculate the ph at the equivalence point in titrating 0.200 M solutions of each of the following solutions with a 0.080 M NaOH. a. HBr b. Lactic Acid, HC 3 H 5 O 3 (Ka = 1.4 x 10-4 ) c. HCN (Ka = 4.9 x 10-10 )