Chemical Reaction and Solution
AP Chem Exam - ‘98
1. Solve the following problem related to the solubility equilibria of some metal hydroxides in aqueous solution.
(a) The solubility of Cu(OH)2(s) is 1.72 x10–6 g/100. mL of solution at 25° C.
(i) Write the balanced chemical equation for the dissociation of Cu(OH)2(s) in aqueous solution.
Cu(OH)2 Cu 2+ + 2 OH –
(ii) Calculate the solubility (in mol/L) of Cu(OH)2 at 25 °C.
(1.72 x10–6 g/0.100 L)(1 mol/97.5 g) = 1.76 x10–7 mol/L
(iii) Calculate the value of the solubility-product constant, Ksp, for Cu(OH)2 at 25 °C.
Ksp = [Cu 2+][OH –]2 = [1.76 x10–7][3.53 x10–7]2 = 2.20 x10–20
(b) The value of the solubility-product constant, Ksp, for Zn(OH)2 is 7.7 x10–17 at 25 °C.
(i) Calculate the solubility (in moles per liter) of Zn(OH)2 at 25 °C in a solution with a pH of 9.35.
Zn(OH)2 Zn 2+ + 2 OH –
Ksp = [Zn 2+][OH –]2 pH 9.35 = pOH 4.65; [OH –] = 10–4.65 = 2.24 x10–5 M
Zn2+ = Ksp/[OH -]2 = 7.7x10-17/[2.24 x10-5]2 = 1.5 x10-7 M
(ii) At 25˚C, 50.0 mL of 0.100-M Zn(NO3)2 is mixed with 50.0 mL of 0.300-M NaOH. Calculate the molar concentration of Zn2+ (aq) in the resulting solution once equilibrium has been established. Assume that volumes are additive.
[Zn2+]init = (0.100M )(0.050 L) = 0.0500 M
[OH–]init = (0.300M)(0.050 L) = 0.150 M
X = conc. loss to get to equilibrium
Ksp = 7.7 x10–17 = (0.0500 – X)(0.150 – 2X)2
[Zn 2+] = 0.0500 – X = 3.1 x10–14M
2. An unknown compound contains only the three elements C, H, and O. A pure sample of the compound is analyzed and found to be 65.60% C and 9.44% H by mass.
(a) Determine the empirical formula of the compound.
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(C3.5H6O1)2 = C7H12O2
(b) A solution of 1.570 g of the compound in 16.08 g of camphor is observed to freeze at a temperature 15.2° C below the normal freezing point of pure camphor.
1. Solve the following problem related to the solubility equilibria of some metal hydroxides in aqueous solution.
(a) The solubility of Cu(OH)2(s) is 1.72 x10–6 g/100. mL of solution at 25° C.
(i) Write the balanced chemical equation for the dissociation of Cu(OH)2(s) in aqueous solution.
Cu(OH)2 Cu 2+ + 2 OH –
(ii) Calculate the solubility (in mol/L) of Cu(OH)2 at 25 °C.
(1.72 x10–6 g/0.100 L)(1 mol/97.5 g) = 1.76 x10–7 mol/L
(iii) Calculate the value of the solubility-product constant, Ksp, for Cu(OH)2 at 25 °C.
Ksp = [Cu 2+][OH –]2 = [1.76 x10–7][3.53 x10–7]2 = 2.20 x10–20
(b) The value of the solubility-product constant, Ksp, for Zn(OH)2 is 7.7 x10–17 at 25 °C.
(i) Calculate the solubility (in moles per liter) of Zn(OH)2 at 25 °C in a solution with a pH of 9.35.
Zn(OH)2 Zn 2+ + 2 OH –
Ksp = [Zn 2+][OH –]2 pH 9.35 = pOH 4.65; [OH –] = 10–4.65 = 2.24 x10–5 M
Zn2+ = Ksp/[OH -]2 = 7.7x10-17/[2.24 x10-5]2 = 1.5 x10-7 M
(ii) At 25˚C, 50.0 mL of 0.100-M Zn(NO3)2 is mixed with 50.0 mL of 0.300-M NaOH. Calculate the molar concentration of Zn2+ (aq) in the resulting solution once equilibrium has been established. Assume that volumes are additive.
[Zn2+]init = (0.100M )(0.050 L) = 0.0500 M
[OH–]init = (0.300M)(0.050 L) = 0.150 M
X = conc. loss to get to equilibrium
Ksp = 7.7 x10–17 = (0.0500 – X)(0.150 – 2X)2
[Zn 2+] = 0.0500 – X = 3.1 x10–14M
2. An unknown compound contains only the three elements C, H, and O. A pure sample of the compound is analyzed and found to be 65.60% C and 9.44% H by mass.
(a) Determine the empirical formula of the compound.
[pic] [pic] [pic]
[pic] [pic] [pic]
(C3.5H6O1)2 = C7H12O2
(b) A solution of 1.570 g of the compound in 16.08 g of camphor is observed to freeze at a temperature 15.2° C below the normal freezing point of pure camphor.