Effect of pH on solubility

Chemistry Level 2

Using K s p = [ Zn 2 + ] [ OH − ] 2 , K_{sp} = [\text{Zn}^{2+}][\text{OH}^{-}]^{2}, calculate the solubility of Zn(OH) 2 \text{Zn(OH)}_{2} in a solution buffered at pH 6.00 6.00 .

4.5 mol L − 1 4.5\text{ mol L}^{-1} 0.45 mol L − 1 0.45\text{ mol L}^{-1} 2.2 × 1 0 − 6 mol L − 1 2.2 \times 10^{-6}\text{ mol L}^{-1} 4.5 × 1 0 − 6 mol L − 1 4.5 \times 10^{-6}\text{ mol L}^{-1}

Krishna Singh by Krishna Singh , 22, India

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1 solution

K s p = [ Zn 2 + ] [ OH − ] 2 = 4.5 × 1 0 − 17 K_{sp} = [\text{Zn}^{2+}][\text{OH}^{-}]^{2} = 4.5 \times 10^{-17} ( approximately )

From buffer [ OH − ] = 1 0 − ( 14 − 6 ) = 1 0 − 8 [\text{OH}^{-}] = 10^{-(14-6)} = 10^{-8}

K s p = [ Zn 2 + ] [ OH − ] 2 K_{sp} = [\text{Zn}^{2+}][\text{OH}^{-}]^{2}

4.5 × 1 0 − 17 = [ Zn 2 + ] [ 1 0 − 8 ] 2 4.5 \times 10^{-17}= [\text{Zn}^{2+}][10^{-8}]^{2}

[ Zn 2 + ] = 0.45 mol L − 1 [\text{Zn}^{2+}]=\boxed{0.45\text{ mol L}^{-1}}

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