The ratio of the solubility of $Mg (OH) X_{2}$ in a buffer of pH $9.0$ to its solubility in a buffer of pH $11.0$ (buffers do not otherwise react with Mg(OH)2) is

Answer: A

s = K_{s p} / [OH X^{-}]^{2}

. At pH 9,

[OH X^{-}] = 1 0^{- 5}

; at pH 11,

[OH X^{-}] = 1 0^{- 3}

. Ratio

= (1 0^{- 3} /1 0^{- 5})^{2} = 1 0^{4}

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