Question

The pH of solution containing 1 mol Cu(OH)2 in 1L water is 7.66.

Has all the Cu(OH)2 dissolve in water? Ksp=4.8 × 10^-20.

Answer 1

pH of the solution containing Cu(OH)₂ = 7.66.

We know that pH + pOH = 14; therefore,

pOH = 14 – pH = 14 – 7.66 = 6.34

Therefore, [OH^-] = antilog (6.34) = 4.57*10^-7 mol/L

Write down the dissociation of Cu(OH)₂ in water:

Cu(OH)₂ (aq) <=====> Cu²⁺ (aq) + 2 OH^- (aq)

The ion product is given by

Q = [Cu²⁺][OH^-]²

We know we have 1 mole Cu(OH)₂ in 1.0 L water; therefore, molar concentration of Cu(OH)₂ = 1 mol/1 L = 1 M.

[Cu²⁺] = molar solubility of Cu(OH)₂ = 1 M

Therefore,

Q = (1).(4.57*10^-7)² = (1).(2.08849*10^-13) = 2.08849*10^-13 ≈ 2.1*10^-13

We can see that the ion product Q > K_sp. Therefore, the solution is super-saturated with Cu(OH)₂ and the excess Cu(OH)₂ will precipitate out of the solution.

Ans: All the Cu(OH)₂ has dissolved in the solution; infact the solution contains more Cu(OH)₂ than it can hold at the said temperature. The solution is therefore, super-saturated and the excess Cu(OH)₂ will precipitate out.