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6-2.Equilibrium-II (Ionic Equilibrium)
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$Mg(OH)_2$ નો $k_{sp}\, 1 \times 10^{-12}$ તો $P^H$ ની કઇ કિંમત $0.01\,M$ માટે $Mg(OH)_2$ નું અવક્ષેપન થશે.?

A

$3$

B

$9$

C

$5$

D

$8$

Solution

$Mg(OH)_2  ⇌ Mg^{2+} + 2OH^-$

$K_{sp} = [Mg^{+2}][OH^-]^2$

$1\times 10^{-12} = [0.01][OH^-]^2$

$[OH^-]^2=1\times  10^{-10} \,\,\,  [OH^-] = 10^{-5}$

$\therefore \,\,\,\,[{H^ + }]\,\, = \,\,\frac{{{{10}^{_{ – 14}}}}}{{{{10}^{ – 5}}}} = \,\,{10^{ – 9}}$

$pH=  – log[H^+] = -log [10^{-9}] = 9$

Standard 11
Chemistry

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