Answer:
solubility of [tex]Fe(OH)_{3}[/tex] is [tex]2.0\times 10^{-15}(M)[/tex]
Explanation:
Solubility equilibrium of Iron(III) hydroxide:
[tex]Fe(OH)_{3}\rightleftharpoons Fe^{3+}+3OH^{-}[/tex]
If solubility of [tex]Fe(OH)_{3}[/tex] is S (M) then concentration of [tex]Fe^{3+}[/tex], [tex][Fe^{3+}][/tex] is S (M)
We know, solubility product, [tex]K_{sp}=[Fe^{3+}][OH^{-}]^{3}[/tex]
pH=6.50
or, pOH= 14-6.50
or, pOH = 7.50
or, [tex][OH^{-}]=10^{-7.50}[/tex]
So, [tex]S = [Fe^{3+}]=\frac{K_{sp}}{[OH^{-}]^{3}}[/tex] = [tex]\frac{6.3\times 10^{-38}}{(10^{-7.50})^{3}}(M)=2.0\times 10^{-15}(M)[/tex]
Hence solubility of [tex]Fe(OH)_{3}[/tex] is [tex]2.0\times 10^{-15}(M)[/tex]
