Question

calculate the ksp value for Fe(OH)3 given that that solubility is 2.7x10^-10mol/L

Answer 1

So,

We could use the following reaction to find the Ksp value:

$Fe(OH)_{3(s)}\leftrightarrow1Fe^{3+}_{(aq)}+3(OH)^-_{(aq)}$

We know that the solubility of Fe(OH)3 is:

$\frac{2.7\cdot10^{-10}molFe(OH)_3}{L}$

As you can notice, we could find the concentrations of each ion formed in the products using the stoichiometry of the reaction:

$\begin{gathered} \lbrack Fe^{3+}\rbrack=\frac{2.7\cdot10^{-10}molFe(OH)_3}{L}\cdot\frac{1molFe^{3+}}{1^{}molFe(OH)_3}^{} \\ \lbrack Fe^{3+}\rbrack=\frac{2.7\cdot10^{-10}molFe^{3+}_{}}{L} \end{gathered}$

And now, we could do the same thing to find the concentration of the [OH]- ions:

$\begin{gathered} \lbrack OH\rbrack^-=\frac{2.7\cdot10^{-10}molFe(OH)_3}{L}\cdot\frac{3mol\lbrack OH^-\rbrack^{}}{1^{}molFe(OH)_3} \\ \\ \lbrack OH^-\rbrack=\frac{8.1\cdot10^{-10}mol\lbrack OH^-\rbrack}{L} \end{gathered}$

Now that we obtained each ion concentration, we could use the formula for Ksp:

$\begin{gathered} K_{sp}=\lbrack Fe^{3+}\rbrack^1\cdot\lbrack OH^-\rbrack^3 \\ K_{sp}=(2.7\cdot10^{-10}_{})^1\cdot(8.1\cdot10^{-10})^3 \\ K_{sp}=1.43489\cdot10^{-37}\approx0 \end{gathered}$

Thus, the ksp value for Fe(OH)3 given that that solubility is 2.7x10^-10mol/L is:

$K_{sp}=1.43489\cdot10^{-37}\approx0$