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# $K_{sp}$ of $Mg(OH)_2$ is $1 \times 10^{-12}$, 0.01 M $MG^{2+}$ will precipitate at the limiting $pH$ of

(a) 3

(b) 9

(c) 5

(d) 8

Can you answer this question?

$K_{sp}$ for $Mg(OH)_2 = [Mg^{2+}][OH^-]^2 = 10^{-12}$
$(0.01)(OH^-)^2 = 10^{-12} or [OH^-] = 10^{-5}$

$[H^+] = \frac{K_w}{[OH^-]} = \frac{10^{-14}}{10^{-5}} = 10^{-9}$
$pH = - log [H^+] = 9$

answered Nov 27, 2013