Question

At 22 °C, an excess amount of a generic metal hydroxide, M(OH)2 , is mixed with pure water. The resulting equilibrium solution has a pH of 10.34 . What is the ?sp of the salt at 22 °C?

A solution contains 0.25 M Pb2+ and 0.44 M Al3+. Calculate the pH range that would allow Al(OH)3 to precipitate but not Pb(OH)2. The Ksp values for Al(OH)3 and Pb(OH)2 can be found in this table.

Answer 1

1) Consider the ionization of M(OH)₂ as below.

M(OH)₂ (s) <=====> M²⁺ (aq) + 2 OH^- (aq)

K_sp = [M²⁺][OH^-]²

The pH of the solution is given as pH = 10.34.

We know that pH + pOH = 14; therefore,

pOH = 14 – pH

= 14 – 10.34

= 3.66

We further know that pOH = -log [OH^-]; therefore,

log [OH^-] = antilog (-pOH)

= antilog (-3.66)

= 2.1878*10^-4 M.

As per the stoichiometric equation,

[M²⁺] = ½*[OH^-]

= ½*(2.1878*10^-4 M)

= 1.0939*10^-4 M.

Therefore,

K_sp = [M²⁺][OH^-]

= (1.0939*10^-4)*(2.1878*10^-4)² (K_sp has no unit)

= 5.2359*10^-12

≈ 5.23*10^-12 (ans, correct to two decimal places).

2) The ionization equations are

Al(OH)₃ (s) <=======> Al³⁺ (aq) + 3 OH^- (aq)

K_sp = [Al³⁺][OH^-]³ = 3.00*10^-34 ……(1)

Pb(OH)₂ (s) <======> Pb²⁺ (aq) + 2 OH^- (aq)

K_sp = [Pb²⁺][OH^-] = 1.43*10^-20 …….(2)

Put [Al³⁺] = 0.44 M in (1) and obtain [OH^-].

3.00*10^-34 = (0.44)*[OH^-]³

======> [OH^-]³ = (3.00*10^-34)/(0.44)

======> [OH^-]³ = 6.8182*10^-34

======> [OH^-] = 8.8015*10^-12

The [OH^-] necessary to just precipitate Al(OH)₃ is 8.8015*10^-12 M.

We know that

pOH = -log [OH^-]

= -log (8.8015*10^-12 M)

= 11.0554.

Again,

pH + pOH = 14; therefore,

pH = 14 – pOH

= 14 – 11.0554

= 2.9446 ≈ 2.94 (ans).

The pH necessary to just precipitate Al(OH)₃ is 2.94.

Put [Pb²⁺] = 0.25 M in (2) and obtain [OH^-].

1.43*10^-20 = (0.25)*[OH^-]²

======> [OH^-]² = (1.43*10^-20)/(0.44)

======> [OH^-]² = 5.72*10^-20

======> [OH^-] = 2.3916*10^-10

The [OH^-] necessary to just precipitate Pb(OH)₂ is 2.3916*10^-10 M.

We know that

pOH = -log [OH^-]

= -log (2.3916*10^-10 M)

= 9.6213.

Again,

pH + pOH = 14; therefore,

pH = 14 – pOH

= 14 – 9.6213

= 4.3787 ≈ 4.38 (ans).

The pH necessary to just precipitate Pb(OH)₂ is 4.38.

It is seen that at pH equal to or greater than 2.94, Al(OH)₃ precipitates while at pH equal to or greater than 4.38, Pb(OH)₂ precipitates. Since we wish to precipitate only Al(OH)₃, the pH of the solution can has to be within 2.94 and 4.38; however, the pH must be less than 4.38 to keep Pb(OH)₂ in solution. Therefore, the desired pH range is 2.94 pH < 4.38 (ans).