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Period ___ Date ___/___/___ Acids, Bases, and Buffers AP PROBLEMS Do your work on another sheet of paper. 1983 C (a) Specify the properties of a buffer solution. Describe the components and the composition of effective buffer solutions. (b) An employer is interviewing four applicants for a job as a laboratory technician and asks each how to prepare a buffer solution with a pH close to 9. Archie A. says he would mix acetic acid and sodium acetate solutions. Beula B. says she would mix NH4Cl and HCl solutions. Carla C. says she would mix NH4Cl and NH3 solutions. Dexter D. says he would mix NH3 and NaOH solutions. Which of these applicants has given an appropriate procedure? Explain your answer, referring to your discussion in part (a). Explain what is wrong with the erroneous procedures. (No calculations are necessary, but the following acidity constants may be helpful: acetic acid, Ka= 1.8x10-5; NH4+, Ka = 5.6x10-10) 1982 A A buffer solution contains 0.40 mole of formic acid, HCOOH, and 0.60 mole of sodium formate, HCOONa, in 1.00 liter of solution. The ionization constant, Ka, of formic acid is 1.8x10-4. (a) Calculate the pH of this solution. (b) If 100. milliliters of this buffer solution is diluted to a volume of 1.00 litre with pure water, the pH does not change. Discuss why the pH remains constant on dilution. (c) A 5.00 milliliter sample of 1.00 molar HCl is added to 100. milliliters of the original buffer solution. Calculate the [H3O+] of the resulting solution. (d) A 800.-milliliter sample of 2.00-molar formic acid is mixed with 200. milliliters of 4.80-molar NaOH. Calculate the [H3O+] of the resulting solution. 1977 The value of the ionization constant, Ka, for hypochlorous acid, HOCl, is 3.110-8. (a) Calculate the hydronium ion concentration of a 0.050 molar solution of HOCl. (b) Calculate the concentration of hydronium ion in a solution prepared by mixing equal volumes of 0.050 molar HOCl and 0.020 molar sodium hypochlorite, NaOCl. (c) A solution is prepared by the disproportionation reaction below. Cl2 + H2O  HCl + HOCl Calculate the pH of the solution if enough chlorine is added to water to make the concentration of HOCl equal to 0.0040 molar.

1975 A (a) A 4.00 gram sample of NaOH(s) is dissolved in enough water to make 0.50 liter of solution. Calculate the pH of the solution. (b) Suppose that 4.00 grams of NaOH(s) is dissolved in 1.00 liter of a solution that is 0.50 molar in NH3 and 0.50 molar in NH4+. Assuming that there is no change in volume and no loss of NH3 to the atmosphere, calculate the concentration of hydroxide ion, after a chemical reaction has occurred. [Ionization constant at 25ºC for the reaction NH3 + H2O  NH4+ + OH-; K = 1.8x10-5] 1973 A sample of 40.0 milliliters of a 0.100 molar HC2H3O2 solution is titrated with a 0.150 molar NaOH solution. Ka for acetic acid = 1.8x10-5 (a) What volume of NaOH is used in the titration in order to reach the equivalence point? (b) What is the molar concentration of C2H3O2- at the equivalence point? (c) What is the pH of the solution at the equivalence point? Answer 1983: (a) A buffer solution resists changes in pH upon the addition of an acid or base. Preparation of a buffer: (1) mix a weak acid + a salt of a weak acid; or (2) mix a weak base + salt of a weak base; or (3) mix a weak acid with about half as many moles of strong base; or (4) mix a weak base with about half as many moles of strong acid; or (5) mix a weak acid and a weak base. (b) Carla has the correct procedure, she has mixed a weak base, NH3, with the salt of a weak base, NH4Cl. Archie has a buffer solution but a pH of around 5. Beula doesn’t have a buffer solution, her solution consists of a strong acid and a salt of a weak base. Dexter does not have a buffer solution, since his solution consists of a weak base plus a strong base. Answer 1982: (a) using the Henderson-Hasselbalch equation = 3.92 {other approaches possible} (b) The pH remains unchanged because the ratio of the formate and formic acid concentration stays the same. (c) initial concentrations

concentrations after H+ reacts with HCOO- 0.38M + 0.05M = 0.43M HCOOH 0.57M - 0.05M = 0.52M HCOO- (d) 0.800L  2.00M HCOOH = 1.60 mol 0.200L  4.80M NaOH = 0.96 mol OH- at equil., (1.60 - 0.96) = 0.64 mol HCOOH and 0.96 mol HCOO- Answer 1977: (a) HOCl + H2O <=> H3O+ + OCl- X = [H3O+] = 4.0x10-5M (b) HOCl + H2O <=> H3O+ + OCl- ; X << 0.010 X = [H3O+] = 8.0x10-8M (c) Cl2 + H2O --> HCl + HOCl [HOCl] = [HCl] = 0.0040M HCl as principal source of H3O+ pH = -log[H3O+] = 2.40 Answer 1975: (a) (b) [NH4+] = 0.50M - X; [NH3] = 0.50M + X using the Henderson-Hasselbalch equation [OH-] = 3.7 x 10-10 M Answer 1973: (a) MaVa=MbVb (0.100M)(40.0 mL) = (0.150M)(Vb) Vb = 26.7 mL (b) acetate ion is a weak base with Kb=Kw/Ka = 1.010-14/1.810-5 = 5.610-10 [CH3COO-]eq = 0.600M -X [OH-] = [CH3COOH] = X correction: x=5.77 x 10-6 0.0600M-9.6610-5M = 0.0599M [CH3COO-]eq (c) pOH = -log[OH-] = -log(5.77x10-6) = 5.24 pH = 14 – pOH = 8.76

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