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6-2.Equilibrium-II (Ionic Equilibrium)
hard
$Mg ( OH )_{2}$ is precipitated, when $NaOH$ is added to a solution of $Mg ^{2+}$. If the final concentration of $Mg ^{2+}$ is $10^{-10}\, M$, the concentration of $OH ^{-}\,( M )$ in the solution is
[Solubility product for $Mg ( OH )_{2}=5.6 \times 10^{-12}$ ]
A$0.056$
B$0.12$
C$0.24$
D$0.025$
(KVPY-2017)
Solution
(c)
Given, $K_{ sp }\,\, Mg ( OH )_{2}=5.6 \times 10^{-12}$
Concentration of $Mg ^{2+}=10^{-10}\, M$
For the reaction,
$Mg ( OH )_{2} \rightleftharpoons Mg ^{2+}+2 OH ^{-}$
$K_{ sp } =\left[ Mg ^{2+}\right]\left[ OH ^{-}\right]^{2}$
$5.6 \times 10^{-12}=\left[10^{-10} M \right]\left[ OH ^{-}\right]^{2}$
$\frac{5.6 \times 10^{-12}}{10^{-10}} =\left[ OH ^{-}\right]^{2}$
$=\sqrt{5.6 \times 10^{-2}}=\left[ OH ^{-}\right]$
$\therefore \quad\left[ OH ^{-}\right] =0.24 \,M$
Note Due to common ion effect, the concentration of $OH ^{-}$gets suppressed.
Given, $K_{ sp }\,\, Mg ( OH )_{2}=5.6 \times 10^{-12}$
Concentration of $Mg ^{2+}=10^{-10}\, M$
For the reaction,
$Mg ( OH )_{2} \rightleftharpoons Mg ^{2+}+2 OH ^{-}$
$K_{ sp } =\left[ Mg ^{2+}\right]\left[ OH ^{-}\right]^{2}$
$5.6 \times 10^{-12}=\left[10^{-10} M \right]\left[ OH ^{-}\right]^{2}$
$\frac{5.6 \times 10^{-12}}{10^{-10}} =\left[ OH ^{-}\right]^{2}$
$=\sqrt{5.6 \times 10^{-2}}=\left[ OH ^{-}\right]$
$\therefore \quad\left[ OH ^{-}\right] =0.24 \,M$
Note Due to common ion effect, the concentration of $OH ^{-}$gets suppressed.
Standard 11
Chemistry
