Name _____Mr. Perfect_____________________________ Date ____F 14________ 1. A 0.50 M solution of hydrazoic acid, HN3, is 99.3 % associated in water. Calculate the acid ionization constant (Ka) for this acid. (10 pts) I C E HN3 0.50 -x 0.50-x ⇄ H+ + N30 0 +x +x x x Dissociation Plug in and solve for Ka 2. At 25 °C, the solubility of Lead(II) Chloride is 3.96 g/L. Calculate the solubility product constant (Ksp) of Lead(II) Chloride at this temperature. (5 pts) Mw = 277 g/mol I C E PbCl2(s) ⇄ Pb2+(aq) + 2Cl-(aq) --0 0 --+s +s --s 2s Solve for molar solubility Ksp = s(2s)2 = 4s3 = 4( )3 = 1.17 x 10-5 3. Calculate the solubility in g per L of a solution of Barium Sulfate. (5 pts) Ksp = 1.07 x 10-10 Mw = 233 g/mol I C E Ksp = s2 or BaSO4(s) ⇄ Ba2+(aq) + SO42-(aq) --0 0 --+s +s --s s 1.07 x 10-10 = s2 √ Solubility = 1.03x10-5 mol/L x 233 g/mol = 2.41x10-3 g/L Chemistry 102 Exam 2 Name _____Mr. Perfect_____________________________ Date ____F 14________ 4. Calculate the pH of 10.0 mL of a 0.25 M weak-base, nicotine (C10H14N2), solution upon titration with the following increments of a 0.15 M strong-acid, HCl: (20 pts) Ka = 1.0 x 10-8 Weak Base: 0.010 L x 0.25 mol/L = 0.0025 mol base a) 0.00 mL of acid (before addition = ICE Box) A- + H2O ⇄ HA + OHI 0.25 --0 0 C -x --+x +x E 0.25-x x x x = 5.0 x 10-4 = [OH-] pH = 14 – 3.30 = 10.70 pOH = -log(5.0 x 10-4) = 3.30 b) 10.00 mL of acid (Before equivalence point) 0.010 L x 0.150 mol/L = 0.0015 mol acid = [HA] 0.0025 mol base – 0.0015 mol acid = 0.0010 mol base remains = [A-] Use the Henderson-Hasselbach equation: b) 16.67 mL of acid (at equivalence point) 0.01667 L x 0.15 mol/L = 0.0025 mol acid ÷ 0.02667 L = 9.37 x 10-2 M = [HA] I C E HA ⇄ 9.37x10-2 -x 9.37x10-2-x H+ + A0 0 +x +x x x x = 3.06 x 10-7 M = [H+] pH = -log(3.06 x 10-7) = 6.51 Chemistry 102 Exam 2 Name _____Mr. Perfect_____________________________ Date ____F 14________ 5. Calculate the pH of a solution containing 0.15 M HC2H3O2, acetic acid, and 0.25 M NaC2H3O2, sodium acetate. Under these conditions, would this solution act as a buffer? Clearly state the buffer zone for this weak acid. (10 pts) Ka = 1.8 x 10-5 pKa = -log(1.8 x 10-5) = 4.74 Buffer zone = 3.74 – 5.74 6. Calculate the pH of a 0.25 M NaIO solution. (10 pts) Ka = 2.3 x 10-11 I C E -2 - x = 1.04x10 M = [OH ] IO- + H2O ⇄ HIO + OH0.25 --0 0 -x --+x +x 0.25 – x x x check (1.04x10-2/0.25) x 100 = 4.17% assumption is valid pOH = -log(1.04x10-2) = 1.98 pH = 14 – 1.98 = 12.02 7. A solution containing the following indicator, with a Ka of 0.02, turns orange at the equivalence point of a titration. Determine the range of pH values this indicator would be useful. Could this indicator be useful for a weak base-strong acid titration? Clearly state why or why not? (10 pts) HIn ⇄ H+ + In(Red) (Yellow) pKa = -log(0.02) =1.7 Appropriate Indicator range = 0.7 -2.7 = buffer zone for the indicator For a weak base: HA ⇄ H+ + A- at the equivalence point The equivalence point would be below 7, thus this is an appropriate indicator for a weak acid-strong base titration. Chemistry 102 Exam 2 Name _____Mr. Perfect_____________________________ Date ____F 14________ 8. For the following reactions, predict whether a decrease in temperature would be favorable or unfavorable. (10 pts) ∆G = ∆H - T∆S (∆G < 0 is favorable) a) 2H2O2(aq) → 2H2O(l) + O2(g) +∆S unfavorable b) PCl3(l) + Cl2(g) → PCl5(s) -∆S favorable c) CO2(s) → CO2(g) +∆S unfavorable d) H2O(g) → H2O(l) -∆S favorable 9. Complete the following table. (10 pts) [H+] [OH-] 1.91 x 10-5 5.25 x 10-10 pH 4.72 Acid or Base? Acid 6.7 x 10-3 1.49 x 10-12 2.17 acid 6.0 x 10-9 1.59 x 10-6 8.2 base 3.1 x 10-8 3.2 x 10-7 7.51 base 10. Predict if a precipitate will form from when 125 mL of 0.375 M Ca(NO3)2 is mixed with 245 mL of 0.255 M NaF. Show all work to receive full credit. The Ksp for CaF2 is 3.2 x 10-11 (10 pts) Ca(NO3)2(aq) + 2NaF(aq) → CaF2(s) + 2NaNO3(aq) CaF2(s) ⇄ Ca+2(aq) + 2F-(aq) Qsp = [Ca+2][F-]2 = (1.27 x 10-1)(1.62 x 10-1)2 = 3.33 x 10-3 Qsp (3.33 x 10-3) > Ksp (1.9 x 10-10) ; therefore a precipitation will form. Supersaturated 11. (Extra Credit) Predict the effect on the solubility (if any) for the following reaction upon addition of a base. (5 pts) Ca(OH)2(s) ⇄ Ca2+(aq) + 2OH-(aq) Addition of a base would add excess hydroxide to the products. The system will shift the equilibrium towards the reactants (shift to the left) in order to accommodate this change. “Le Châtlier’s Principle” Chemistry 102 Exam 2