Question

Solubility of $Mg ( OH )_{2}$ in $0.1 \,M \,CH _{3} COONa$ is $\left[K_{s p}\right.$ of $Mg ( OH )_{2}=1.8 \times 10^{-11}, K_{a}\left( CH _{3} COOH \right)=1.8 x$ $\left.10^{-5}, P k_{a}=4.74\right]$

• A. 0.33 M
• B. 0.66 M
• C. $1.0 \times 10^{-2} M$
• D. $1.8 \times 10^{-3} M$

Answer 1

Answer: (A)

Aqueous solution of $CH _{3} COONa$ is basic due to hydrolysis of $CH _{3} COO ^{-} $
$CH _{3} COO ^{-}( aq )+ H _{2} O \rightleftharpoons CH _{3} COOH ( aq )+ OH ^{-}( aq )$
$pH =7+\frac{p K_{a}}{2}+\frac{\log C}{2}$
$=7+\frac{4.74}{2}+\frac{\log 0.1}{2}=8.87$
$\therefore pOH =14-8.87=5.13$
$[ OH ]=10^{- pOH }=10^{-5.13}=10^{-6} 10^{0.87}$
$=7.4 \times 10^{-6} M$
Let, solubility of $Mg ( OH )_{2}= x$ in $0.1\, M \,CH _{3} COO Na $
$Mg ( OH )_{2}( s ) \rightleftharpoons Mg ^{2+}( aq )+2 OH ^{-}( aq )$
$K_{s p}=\left[ Mg ^{2+}\right][ OH ]^{2}$
$1.8 \times 10^{-11}=( x )\left(7.4 \times 10^{-6}\right)^{2}$
$x=0.33\, M$