ANSWERS AT THE END
Advanced Placement Chemistry: 1991 Free Response Questions
1) The acid ionization constant, Ka, for propanoic acid, C2H5COOH, is 1.3 x 10¯5.
(a) Calculate the hydrogen ion concentration, [H+], in a 0.20-molar solution of propanoic acid.
(b) Calculate the percentage of propanoic acid molecules that are ionized in the solution in (a).
(c) What is the ratio of the concentration of propanoate ion, C2H5COO¯, to that of propanoic acid in a
buffer solution with a pH of 5.20 ?
(d) In a 100-milliliter sample of a different buffer solution, the propanoic acid concentration is
0.50-molar and the sodium propanoate concentration is 0.50-molar. To this buffer solution,
0.0040 mole of solid NaOH is added. Calculate the pH of the resulting solution.
Advanced Placement Chemistry: 1993 Free Response Questions
1) CH3NH2 + H2O <===> CH3NH3+ + OH¯
Methylamine, CH3NH2, is a weak base that reacts according to the equation above. The value of the
ionization constant, Kb, is 5.25 x 10¯4. Methylamine forms salts such as methylammonium nitrate,
(CH3NH3+) (NO3¯).
(a) Calculate the hydroxide ion concentration, [OH¯], of a 0.225-molar solution of methylamine.
(b) Calculate the pH of a solution made by adding 0.0100 mole of a solid methylammonium nitrate to
120.0 milliliters of a 0.225-molar solution of methylamine. Assume that no volume change occurs.
(c) How many moles of either NaOH or HCl (state clearly which you choose) should be added to the
solution in (b) to produce a solution that has a pH of 11.00? Assume that no volume change occurs.
(d) A volume of 100. milliters of distilled water is added to the solution in (c). How is the pH of the
solution affected? Explain.
Advanced Placement Chemistry: 1994 Free Response Questions
1) MgF2(s) <===> Mg2+(aq) + 2 F¯(aq)
In a saturated solution of MgF2 at 18° C, the concentration of Mg2+ is 1.21 x 10¯3 molar. The
equilibrium is represented by the equation above.
(a) Write the expression for the solubility-product constant, Ksp, and calculate its value at 18° C.
(b) Calculate the equilibrium concentration of Mg2+ in 1.000 liter of saturated MgF2 solution at 18°C to
which 0.100 mole of solid KF has been added. The KF dissolves completely. Assume the volume
change is negligible.
(c) Predict whether a precipitate of MgF2 will form when 100.0 milliliters of a 3.00 x 10¯3 molar
Mg(NO3)2 solution is mixed with 200.0 milliliters of a 2.00 x 10¯3 molar NaF solution at 18°C.
Calculations to support your prediction must be shown.
(d) At 27°C the concentration of Mg2+ in a saturated solution of MgF2 is 1.17 x 10¯3 molar. Is the
dissolving of MgF2 in water an endothermic or an exothermic process? Give an explanation to support
your conclusion.
Advanced Placement Chemistry: 1996 Free Response Questions
2)
HOCl <===> OCl¯ + H+
Hypochlorous acid, HOCl, is a weak acid commonly used as a bleaching agent. The acid-dissociation
constant, Ka, for the reaction represented above is 3.2 x 10¯8.
(a) Calculate the [H+] of a 0.14-molar solution of HOCl.
(b) Write the correctly balanced net ionic equation for the reaction that occurs NaOCl is dissolved in
water and calculate the numerical value of the equilibrium constant for the reaction.
(c) Calculate the pH of a solution made by combining 40.0 milliliters of 0.14-molar HOCl and 10.0
milliliters of 0.56-molar NaOH.
(d) How many millimoles of solid NaOH must be added to 50.0 milliliters of 0.20-molar HOCl to
obtain a buffer solution that has a pH of 7.49? Assume that the addition of the solid NaOH results in a
negligible change in volume.
(e) Household bleach is made by dissolving chlorine gas in water, as represented below.
Cl2(g) + H2O --> H+ + Cl¯ + HOCl(aq)
Calculate the pH of such a solution if the concentration of HOCl in the solution is 0.065 molar.
1991 Answers
One point penalty applied for sig figure error, applied once per problem
No credit earned for answer with no numerical justification
1.a three points
Ka= [H+] [ C2H5COO¯]/ [C2H5COOH]
1.3x10-5 = x2/0.20
X= [H+] = 1.6 x 10 -3
b) one point: % dissoc =[H+] / [C2H5COOH] * 100
1.6 x 10-3 /0.20 == 0.80 %
c) two points: pH = 5.20 = -log(H+)= 6.3 x10-6
Ka = [H+] [ C2H5COO¯]/ [C2H5COOH] =
d)three pts 0.1 L* 0.35 mol/l = .035 mol H prop
1 pt
0.10 L * 0.050 mol/L = 0.05 mol Prop
1 pt
.035 mol acid -.004 mol NaOH = 0.031 mol acid left unreacted
.05 mol prop- + .004 mol prop- = 0.054 mol prop- in soln
1.3 E -5 =[ H+ ]* [0.054 mol/.1 L] //[ .031 mol/.1 L] 1 pt
[ H+ ]= 7.5 E -6 pH = 5.13
Or
pH = pKa + log [base]/[acid]
5.20= 4.89 + log [prop-]/ [Hprop]= 4.89 + log [0.054/0.031]=
5.13
1993 Answers
1a) three points
CH3NH2 + H2O <===> CH3NH3+ + OH¯
Kb = ([CH3NH3+] [OH¯]) ÷ [CH3NH2] = 5.25 x 10¯4
CH3NH2
CH3NH3+
I
0.225
0
C
-x
+x
E
0.225 - x
x
4
5.25 x 10¯ = [(x) (x)] / (0.225 - x)
neglect the minus x to get 5.25 x 10¯4 = x2 / 0.225
x = [square root]((5.25 x 10¯4) (0.225))
[OH¯] = 1.09 x 10¯2
(Note: quadratic gives 1.06 x 10¯2)
b) three points
[CH3NH3+] = 0.0100 mol / 0.120 L = 0.0833 M
5.25 x 10¯4 = [(0.0833 + x) (x)] / (0.225 - x) = 0.0833x / 0.225
x = [OH¯] = 1.42 x 10¯3 mol/L
pOH = 2.85
pH = 11.15
OH¯
0
+x
x
alternate solution using the Henderson-Hasselbalch Equation
pH = pKa = log ([base] / [acid])
pKw = pKa + pKb
pKa = 10.7
pH = 10.7 + log (0.225 / 0.0833)
pH = 11.15
The solution using the pOH form is left to you, gentle reader.
c) two points
HCl must be added.
5.25 x 10¯4 = [(0.0833 + x) (0.0010)] / (0.225 - x)
1.18 x 10¯4 - 5.25 x 10¯4x = 8.33 x 10¯5 + 1.0 x 10¯3x
x = 0.0228 mol/L
0.0228 mol/L x 0.120 L = 2.74 x 10¯3 mol HCl
alternate solution based on Henderson-Hasselbalch Equation
11.00 = 10.72 + log ([base] / [acid])
log ([base] / [acid]) = 0.28
[base] / [acid] = 1.906 = (0.225 - x) / (0.0833 + x)
x = 0.0227 mol / L
0.0227 mol / L x 0.120 L = 2.73 x 10¯3 mol HCl
d) one point
The [CH3NH3+] / [CH3NH2] ratio does not change in the buffer solution with dilution. Therefore, no
effect on pH.
1994 Answers:
Answer:
(a) Ksp = [Mg2+][F-]2 = (1.2110-3)(2.4210-3)2
= 7.0910-9
(b) X = concentration loss by Mg2+ ion
2X = concentration loss by F- ion
[Mg2+] = (1.2110-3 - X) M
[F-] = (0.100 + 2.4210-3 - 2X) M
since X is a small number then (0.100 + 2.4210-3 - 2X)  0.100
Ksp = 7.0910-9 = (1.2110-3 - X)(0.100)2
X = 1.209291410-3
[Mg2+]= 1.2110-3-1.2092910-3 = 7.0910-7M
(c) [Mg2+] = 3.0010-3M  100.0 mL/300.0 mL = 1.0010-3M
[F-] = 2.0010-3 M  200.0 mL/300.0 mL = 1.3310-3 M
trial Ksp = (1.0010-3)(1.3310-3)2 = 1.7810-9
trial Ksp < = 7.0910-9,  no ppt.
(d) @ 18ºC, 1.2110-3 M MgF2 dissolves
@ 27ºC, 1.1710-3 M MgF2 dissolves
MgF2  Mg2+ + 2 F- + heat
IN THIS EXTREMELY RARE CASE, dissolving is exothermic. (See that it dissolves less at
higher temps!) If heat is increased it forces the equilibrium to shift left (according to LeChatelier’s
Principle) and less MgF2 will dissolve.
1996 Answers:
2)
a) two points total ; one point for correct substitutions; one point for computation
[H+] = [OCl¯] = square root (0.14 x 3.2 x 10¯8) = 6.7 x 10¯5 M
since Ka
= ( [H+][OCl¯] ) / [HOCl]
= [H+]2 / cHOCl
(b) two points total: one point each
OCl¯ + H2O <===> HOCl + OH¯ (or NaOCl + H2O ---> Na+ + HOCl + OH¯)
Kb = Kw / Ka = 1 x 10¯14 ÷ 3.2 x 10¯8 = 3.1 x 10¯7
(c) two points total; one for concentrations and one for pH calc.
Concentrations before reaction:
[HOCl] = [(0.0400) (0.14)] / 0.050 = 0.11 M
[OH¯] = [(0.0100) (0.56)] / 0.050 = 0.11 M
Thus reaction is essentially complete and exactly equals a solution of NaOCl and [OCl¯] = 0.11 M (or
reaction is at equivalence point).
Then
[OH¯] = [HOCl]
Kb = [OH¯]2 / 0.11 = 3.1 x 10¯7
[OH¯] = square root [(0.11) (3.1 x 10¯7)] = 1.8 x 10¯4
pOH = 3.73
pH = 14 - 3.73 = 10.27
(d) two points; one for half-neutralized; one for mmol calcs.
pH = 7.49 therefore [H+] = 3.2 x 10¯8
pH = pKa , or [H+] = Ka.
So [OCl¯] / [HOCl] = 1 , or solution must be half neutralized.
initial mmol HOCl = 50.0 x 0.20 = 10.0 mmol
mmol NaOH required = 10.0 ÷ 2 = 5.0 mmol
(e) one point
From equation, 1 mol H+ produced for each 1 mole of HOCl produced, thus [H+] = [HOCl] = 0.065
therefore pH = 1.19