Question

Chromium(III) hydroxide has Ksp 1.6 x 10-30. What is the molar solubility of Cr(OH)3 in a solution whose pH is maintained at 6.00?

Answer 1

Explanation:

Chemical reaction equation for $Cr(OH)_{3}$ is as follows.

        $Cr(OH)_{3} \rightarrow Cr^{3+} + 3OH^{-}$

As it is given that pH is 6. So, the concentration of hydrogen ions will be as follows.

             pH = $-log [H^{+}]$

             antilog (-6) = $[H^{+}]$

               $[H^{+}] = 10^{-6}$

 and $[OH^{-}] = \frac{10^{-14}}{[H^{+}]}$

                   = $\frac{10^{-14}}{10^{-6}}$

                   = $10^{-8}$

Now, let us assume that the solubility is "s". Therefore,

            $Cr(OH)_{3} \rightarrow Cr^{3+} + 3OH^{-}$

                                     s     $10^{-8}$

Therefore, calculate the value of $K_{sp}$ for this reaction as follows.

                $K_{sp} = [Cr^{3+}][10^{-8}]^{3}$

                $1.6 \times 10^{-30} = s \times (10^{-8})^{-30}$

                        s = $1.6 \times 10^{-6} mol/L$        

Thus, we can conclude that the molar solubility of $Cr(OH)_{3}$ is $1.6 \times 10^{-6} mol/L$.