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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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