The pH is given here to find the OH concentation of the solution.

The pH of a solution is defined as,

`pH = -log[H^+]`

If pH = 8.5 then,

`8.5 = -log[H^+]`

`log[H^+] = -8.5`

`[H^+] = 3.162 xx 10^(-9) moldm^(-3)`

But we know for an aqueous solution,

`k_w = [H^+][OH^-]`

At room temperature `k_w = 1 xx 10^(-14) mol^2dm^(-6)`

Therefore,

`1 xx 10^(-14) mol^2dm^(-6) = 3.162 xx 10^(-9) moldm^(-3) xx [OH^-]`

Therefore [OH^-] is given by,

`[OH^-] = 3.162 xx 10^(-6) moldm^(-3)`

`Cd(OH)_2 lt----------gt Cd^(2+) + 2OH^-`

`K_(sp)` is given by,

`K_(sp) = [Cd^(2+)][OH^-]^2 = 2.5 xx 10^(-14)`

Therefore,

`[Cd^(2+)]xx(3.162xx10^(-6))^2 = 2.5 xx 10^(-14)`

`[Cd^(2+)] = 2.5 xx 10^(-3) moldm^(-3)`

**Therefore the concentration of `Cd^(2+)` is `2.5 xx 10^(-3) moldm^(-3)` .**

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