Question

Calculate the molar solubility of _____ salt in pure water. The Kip value is _____. The instructors will choose a salt such as CaCO_3, Pbl_2, Hg_2Cl_2, Ag_3PO_4, or Ca_3(PO_4)_2. The instructors will give a numerical value for K_sp appropriate to the salt. Your full-credit answer must show (1) the balanced equilibrium reaction equation, (2) the K_ip equilibrium constant expression written in terms of "x", (3) essential mathematical work, and (4) the molar solubility as the final answer. You can use an ICE table or the "shortcut" method to derive the Kip expression. For practice, use the five listed salt formulas and assume a K_sp value of 1.6 times 10^-8.

Answer 1

Given Ksp for sample = 1.6 * 10^-8

(1)

CaCO₃ (s) $\rightleftharpoons$ Ca²⁺ (aq.) + CO₃^2- (aq.)

If solubility of CaCO₃ is S mol/L, then [Ca²⁺] = [CO₃^2-] = S mol/L

Ksp = [Ca²⁺][CO₃^2-]

1.6 * 10^-8 = (S)(S)

S = solubility of CaCO₃ = 1.26 * 10^-4 M

(2)

PbI₂ (s) $\rightleftharpoons$ Pb²⁺ (aq.) + 2 I^- (aq.)

If Solublity of PbI₂ = S mol/L, then, [Pb²⁺] = S mol/L, [I^-] = 2S mol/L

Ksp = [Pb²⁺][I^-]²

1.6 * 10^-8 = (S)(2S)²

S = 1.59 * 10^-3 M

(3)

Hg₂Cl2 (s) $\rightleftharpoons$ Hg₂²⁺ (aq.) + 2 Cl^- (aq.)

If Solublity of Hg₂Cl₂ = S mol/L, then, [Hg₂²⁺] = S mol/L, [Cl^-] = 2S mol/L

Ksp = [Hg₂²⁺][Cl^-]²

1.6 * 10^-8 = (S)(2S)²

S = 1.59 * 10^-3 M

(4)

Ag3PO4 (s) $\rightleftharpoons$ 3 Ag+ (aq.) + PO43- (aq.)

If solubility of Ag₃PO₄ is S mol/L, then, [Ag⁺] = 3S mol/L and [PO₄^3-] = S mol/L

Ksp = [Ag⁺]³[PO₄^3-]

1.6 * 10^-8 = (3S)³(S)

S = 4.93 * 10^-3 M

(5)

Ca₃(PO₄)₂ (s) $\rightleftharpoons$ 3 Ca²⁺ (aq.) + 2 PO₄^3- (aq.)

If the solubility = S mol/L, then, [Ca²⁺] = 3S mol/L and [PO₄^3-] = 2S mol/L

Ksp = [Ca²⁺]³[PO₄^3-]²

1.6 * 10^-8 = (3S)³(2S)²

S = 1.71 * 10^-2 M