JEE MAIN - Chemistry (2010 - No. 6)
At 25°C, the solubility product of Mg(OH)2 is 1.0 $$\times$$ 10–11. At which pH, will Mg2+ ions start precipitating in the form of Mg(OH)2 from a solution of 0.001 M Mg2+ ions?
9
10
11
8
Explanation
$$Mg{\left( {OH} \right)_2}\,\rightleftharpoons\,M{g^{ + + }} + 2O{H^ - }$$
$${K_{sp}} = \left[ {M{g^{ + + }}} \right]{\left[ {O{H^ - }} \right]^2}$$
$$1.0 \times {10^{ - 11}} = {10^{ - 3}} \times {\left[ {O{H^ - }} \right]^2}$$
$$\left[ {O{H^ - }} \right] = \sqrt {{{{{10}^{ - 11}}} \over {{{10}^{ - 3}}}}} = {10^{ - 4}}$$
$$\therefore$$ $$\,\,\,pOH = 4$$
$$\therefore$$ $$\,\,\,pH + pOH = 14$$
$$\therefore$$ $$\,\,\,pH = 10$$
$${K_{sp}} = \left[ {M{g^{ + + }}} \right]{\left[ {O{H^ - }} \right]^2}$$
$$1.0 \times {10^{ - 11}} = {10^{ - 3}} \times {\left[ {O{H^ - }} \right]^2}$$
$$\left[ {O{H^ - }} \right] = \sqrt {{{{{10}^{ - 11}}} \over {{{10}^{ - 3}}}}} = {10^{ - 4}}$$
$$\therefore$$ $$\,\,\,pOH = 4$$
$$\therefore$$ $$\,\,\,pH + pOH = 14$$
$$\therefore$$ $$\,\,\,pH = 10$$
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