Find the pHpH of a solution prepared from 1.0 LL of a 0.15 MM solution of Ba (OH) 2Ba (OH) 2 and excess Zn (OH) 2 (s) Zn (OH) 2 (s). The KspKsp of Zn (OH) 2Zn (OH) 2 is 3*10-153*10-15 and the KfKf of Zn (OH) 2-4Zn (OH) 42 - is 2*10152*1015.

pH = 13.09

Explanation:

Zn (OH) 2 - -> Zn+2 + 2OH - Ksp = 3X10^-15

Zn+2 + 4OH - - -> Zn (OH) 4-2 Kf = 2X10^15

K = Ksp X Kf

= 3*2*10^-15 * 10^15

= 6

Concentration of OH⁻ = 2[Ba (OH) ₂] = 2 * 0.15 = 3 M

Zn (OH) ₂ + 2OH⁻ (aq) - -> Zn (OH) ₄²⁻ (aq)

Initial: 0 0.3 0

Change: - 2x + x

Equilibrium: 0.3 - 2x x

K = Zn (OH) ₄²⁻/[OH⁻]²

6 = x / (0.3 - 2x) ²

6 = x / (0.3 - 2x) (0.3 - 2x)

6 (0.09 - 1.2x + 4x²) = x

0.54 - 7.2x + 24x² = x

24x² - 8.2x + 0.54 = 0

Upon solving as quadratic equation, we obtain;

x = 0.089

Therefore,

Concentration of (OH⁻) = 0.3 - 2x

= 0.3 - (2*0.089)

= 0.122

pOH = - log[OH⁻]

= - log 0.122

= 0.91

pH = 14-0.91

= 13.09