Question

1. If 1.70 x 10-4 moles of Agt(aq) dissolves per liter of water when AglO3 is dissolved, what is the Ksp for AgIO3 ? 2. If you have a solution of Cat2 that has a concentration of 0.050 M, and you adjust the pH of the solution to 8.00, will Ca(OH)2 precipitate. (Ksp for Ca(OH)2 = 6.5 x 106) 3. Determine the concentration of fluoride ion in a saturated solution of PbF2 (in moles/liter), if the Ksp = 3.6 x 10-8 for PbF2

Answer 1

Q1. moles Ag⁺ = 1.70 x 10^-4 mol

concentration Ag⁺ = (moles Ag⁺) / (volume of solution)

concentration Ag⁺ = (1.70 x 10^-4 mol) / (1 L)

concentration Ag⁺ = 1.70 x 10^-4 M

AgIO₃ (s) $\rightarrow$ Ag⁺ (aq) + IO₃^- (aq)

[IO₃^-] = [Ag⁺] = 1.70 x 10^-4 M

K_sp AgIO₃ = [Ag⁺][IO₃^-]

K_sp AgIO₃ = (1.70 x 10^-4 M) * (1.70 x 10^-4 M)

K_sp AgIO₃ = 2.89 x 10^-8

Q2. pH = 8.00

pOH = 14 - pH

pOH = 14 - 8.00

pOH = 6.00

[OH^-] = 10^-pOH

[OH^-] = 10^-6.00

[OH^-] = 1.0 x 10^-6 M

Reaction quotient, Q = [Ca²⁺][OH^-]²

Q = (0.05 M) * (1.0 x 10^-6 M)²

Q = 5.0 x 10^-14

Since Q < K_sp, therefore, precipitate will not form.

Q3. PbF₂ dissociates as

PbF₂ (s) $\rightleftharpoons$ Pb²⁺ (aq) + 2 F^- (aq)

Let molar concentration of dissolved PbF₂ = s

[Pb²⁺] = s

[F^-] = 2s

K_sp PbF₂ = [Pb²⁺][F^-]²

3.6 x 10^-8 = (s) * (2s)²

3.6 x 10^-8 = s * 4s²

3.6 x 10^-8 = 4s³

s = (3.6 x 10^-8 / 4)^1/3

s = 2.1 x 10^-3 M

Concentration of fluoride ion = 2s

Concentration of fluoride ion = 2 * (2.1 x 10^-3 M)

Concentration of fluoride ion = 4.2 x 10^-3 M