What is the maximum solubility of Fe(OH)2 (Ksp = 1.6 x 10^-14) in a solution with a pH of 8.3?

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The solubility product for `Fe(OH)_2` is given by,

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

`K_(sp) = [Fe^(2+)][OH^-]^2`

We can find the `OH^-` concentration by using the given pH value.

`pH = -log[H^+]`

`8.3 = -log[H^+]`

`log[H^+] = -8.3`

`[H^+] = 5.011 xx 10^(-9) moldm^(-3)`

But we know for water, the disassociation constant at room temperature, `k_w` is `1 xx 10^(-14) mol^2dm^(-6).`

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

`1 xx 10^(-14) = 5.011 xx 10^(-9) xx [OH^-]`

`[OH^-] = 1.996 xx 10^(-6) moldm^(-3)`

But we know,

`K_(sp) = [Fe^(2+)][OH^-]^2`

`1.6 xx 10^(-14) = [Fe^(2+)] xx (1.996 xx 10^(-6))^2`

`[Fe^(2+)] = 4.016 xx 10^(-3) moldm^(-3)`

Therefore the concentration of `[Fe^(2+)]` in a saturated solution of `pH = 8.3 is 4.016 xx 10^(-3) moldm^(-3)` .

Therefore total number of moles `Fe^(2+)` dissolved in 1 liter is `4.016 xx 10^(-3) mol` . Therefore total number of `Fe(OH)_2` moles dissolved in 1 liter is `4.016 xx 10^(-3)` mol.

The molecular weight of `Fe(OH)_2 = 55.8 + 2 xx(16+1)` g per mol

`= 89.8 g` per mol

Therefore total mass dissolved in 1 liter

`= 4.016 xx 10^(-3) mol xx 89.8 g per mol`

`= 0.36` g.

**Therefore solubility of `Fe(OH)_2` in a solution with a pH of 8.3 is 0.36 g per liter.**

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