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 :

Explanation:

Chemical reaction equation for [tex]Cr(OH)_{3}[/tex] is as follows.

        [tex]Cr(OH)_{3} \rightarrow Cr^{3+} + 3OH^{-}[/tex]

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

             pH = [tex]-log [H^{+}][/tex]

             antilog (-6) = [tex][H^{+}][/tex]

               [tex][H^{+}] = 10^{-6}[/tex]

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

                   = [tex]\frac{10^{-14}}{10^{-6}}[/tex]

                   = [tex]10^{-8}[/tex]

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

            [tex]Cr(OH)_{3} \rightarrow Cr^{3+} + 3OH^{-}[/tex]

                                     s     [tex]10^{-8}[/tex]

Therefore, calculate the value of [tex]K_{sp}[/tex] for this reaction as follows.

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

                [tex]1.6 \times 10^{-30} = s \times (10^{-8})^{-30}[/tex]

                        s = [tex]1.6 \times 10^{-6} mol/L[/tex]        

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

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