Chemical Reaction and Solution

2470 Words Feb 25th, 2013 10 Pages
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
…show more content…
∆Hcomb = ∑∆H˚ƒ(products) – ∑∆H˚ƒ(reactants)
- 3058 kJ/mol = [6 (- 393.5) + 3 (- 285.85)] – [X + 0] = - 161 kJ

(c) Calculate the value of the standard free-energy change, ∆G˚, for the combustion of phenol at 25° C.

∆S˚ = ∑S˚(products) – ∑S˚(reactants)
[6 (213.6) + 3 (69.91)] – [144.0 + 7 (205.0)] = – 87.67 J

∆G˚ = ∆H˚ – T∆S˚ = – 3058 – 298(- 0.08767) kJ = – 3032 kJ

(d) If the volume of the combustion container is 10.0 L, calculate the final pressure in the container when the temperature is changed to 110.° C. (Assume no oxygen remains unreacted and that all products are gaseous.)

(2.000 g phenol)(mol phenol/94.113 g)(7 mol O2/mol phenol) = 0.1488 mol O2 mol of gaseous product = 6/7 (0.1489) + 3/7 (0.1489) = 0.1913 mol
P = [pic] = 0.601 atm

4. (a) Solutions of tin(II) chloride and iron(III) chloride are mixed.

Sn 2+ + Fe 3+ Sn 4+ + Fe 2+

(b) Solutions of cobalt(II) nitrate and sodium hydroxide are mixed.

Co 2+ + OH – Co(OH)2

(c) Ethene gas is burned in air.

C2H4 + O2 CO2 + H2O

(d) Equal volumes of equimolar solutions of phosphoric acid and potassium hydroxide are mixed.

H3PO4 + OH – H2PO4 – + H2O

(e) Solid calcium sulfite is heated in a vacuum.

CaSO3 CaO + SO2

(f) Excess hydrochloric acid is added to a solution of diamminesilver(I) nitrate.

H + + Cl – + Ag(NH3)2 + AgCl + NH4 +
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