Question

Solve the following problem related to the solubility equilibria of some metal hydroxides in aqueous solution.The solubility of Cu(OH)2(s) is 1.72 x10–6 g/100. mL of solution at 25° C.(i) Write the balanced chemical equation for the dissociation of Cu(OH)2(s) in aqueous solution.Cu(OH)2 <===> Cu 2+ + 2 OH –(ii) Calculate the solubility (in mol/L) of Cu(OH)2 at 25 °C.(1.72 x10–6 g/0.100 L)(1 mol/97.5 g) = 1.76 x10–7 mol/L(iii) Calculate the value of the solubility-product constant, Ksp, for Cu(OH)2 at 25 °C.Ksp = [Cu 2+][OH –]2 = [1.76 x10–7][3.53 x10–7]2 = 2.20 x10–20

Answer 1

i) Cu(OH)2(aq) → Cu^2+(aq) + 2OH-(aq)

ii) 1.76 *10^-7 mol/L

iii) 2.2 *10^-20

Step-by-step explanation:

Step 1: Data given

The solubility of Cu(OH)2(s) is 1.72 * 10^–6 g/100 mL of solution at 25° C.

(i) Write the balanced chemical equation for the dissociation of Cu(OH)2(s) in aqueous solution.

Cu(OH)2(aq) → Cu^2+(aq) + 2OH-(aq)

(ii) Calculate the solubility (in mol/L) of Cu(OH)2 at 25 °C.

1.72 * 10^–6 g /0.100L = 1.72 *10^-5 g/L

1.72*10^-5 g/L / 97.56 g/mol = 1.76 *10^-7 mol/L

(iii) Calculate the value of the solubility-product constant, Ksp, for Cu(OH)2 at 25 °C.

For 1 mol Cu(OH)2 we have 1 mol Cu^2+ and 2 moles OH-

This means if there reacts X of Cu(OH)2 there will be produced X if Cu^2+ and 2X of OH-

Ksp = [Cu^2+]*[OH-]²

Ksp = 1.76 *10^-7 * 2*(1.76 *10^-7)²

Ksp = 2.18 *10^-20 ≈ 2.2*10^-20

Answer 2

i)
`Cu(OH)_2 \longrightarrow Cu^(+2) + 2 OH^-`

ii)
`S=1.76*10^(-7) mol/L`

iii) iii)
`Kps=2.2*10^(-20)`

Step-by-step explanation:

The dissociation equation:

i)
`Cu(OH)_2 \longrightarrow Cu^(+2) + 2 OH^-`

Where the constant of solubility is:
`Kps=[Cu^(+2)]*[OH^-]^2`

ii) The solubility of Cu(OH)2 is 1.72*10-6 g/100 mL at 25°C

The molecular weight is M=97.5 g/mol

The solubility in mol/L is:

`S=1.72*10^(-6) (g)/(100 mL)*(1000 mL)/(L)*(1 mol)/(97.5 g)`

`S=1.76*10^(-7) mol/L`

iii)
`Kps=(1.76*10^(-7))*(2*1.76*10^(-7))^2=2.2*10^(-20)`