Answer:
i) [tex]Cu(OH)_2 \longrightarrow Cu^{+2} + 2 OH^-[/tex]
ii) [tex]S=1.76*10^{-7} mol/L[/tex]
iii) iii) [tex]Kps=2.2*10^{-20}[/tex]
Explanation:
The dissociation equation:
i) [tex]Cu(OH)_2 \longrightarrow Cu^{+2} + 2 OH^-[/tex]
Where the constant of solubility is: [tex]Kps=[Cu^{+2}]*[OH^-]^2[/tex]
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:
[tex]S=1.72*10^{-6} \frac{g}{100 mL}*\frac{1000 mL}{L}*\frac{1 mol}{97.5 g}[/tex]
[tex]S=1.76*10^{-7} mol/L[/tex]
iii) [tex]Kps=(1.76*10^{-7})*(2*1.76*10^{-7})^2=2.2*10^{-20}[/tex]
Answer:
i) Cu(OH)2(aq) → Cu^2+(aq) + 2OH-(aq)
ii) 1.76 *10^-7 mol/L
iii) 2.2 *10^-20
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
