# At 25°C, the solubility product of $\dpi{100} Mg \left ( OH \right )_{2}$ is 1.0 × 10-11. At which pH, will $\dpi{100} Mg^{2+}$ ions start precipitating in the form of $\dpi{100} Mg \left ( OH \right )_{2}$ from a solution of 0.001 M Mg2+ ions ? Option 1) $8$ Option 2) $9$ Option 3) $10$ Option 4) $11$

S SudhirSol

As we have learned

General expression of solubility product -

$M_{x}X_{y}\rightleftharpoons xM^{p+}(aq)+yX^{2-}(aq)$

$(x.p^{+}=y.\bar{q})$

- wherein

Its solubility product is

$K_{sp}=[M^{p+}]^{x}[X^{q-}]^{y}$

$K_{sp}$  of $Mg(OH)_2= [ Mg ^{2+}][OH^-]^2$

$[OH^-]= \sqrt{\frac{K_{sp}}{[Mg^{2+}]}}= \sqrt{\frac{1 \times 10^{-11}}{10^{-3}}} = 10^{-4 }M$

$p OH = 4$

$p H = 14-pOH= 14-4 = 10$

Option 1)

$8$

Option 2)

$9$

Option 3)

$10$

Option 4)

$11$

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