Question

The $K_{sp}$ of $M g(O H)_{2}$ is $1 \times 10^{-12} .0 .001\, M\, g(O H)_{2}$ will precipitate at the limited $pH$ :

• A. 3
• B. 9
• C. 5
• D. 8

Answer 1

Answer: (B)

First find $\left[ OH ^{-}\right]$from $K_{s p} .$
Then find $\left[ H ^{+}\right]$ by using formula
$\left[ H ^{+}\right]\left[ OH ^{-}\right]=10^{-14}$.
Finally calculate $pH$ by using following formula.
$pH =-\log \left[ H ^{+}\right]$ Given,
$K _{s p} Mg ( OH )_{2}=1 \times 10^{-12}$ Concentration
$Mg ( OH )_{2}=0.01\, M\, Mg ( OH )_{2} Mg ^{2+}+2 OH ^{-}$
$K _{s p}=\left[ Mg ^{2+}+2 OH ^{-}\right.$
or $1 \times 10^{-12}=0.01 \times\left[ OH ^{-}\right]^{2}$
or $\left[ OH ^{-}\right]^{2}=1 \times 10^{-10}$
or $\left[ OH ^{-}\right]=1 \times 10^{-5}$
$\left[ H ^{+}\right]\left[ OH ^{-}\right]=10^{-14}$
or $\left[ H ^{+}\right]\left[1 \times 10^{-5}\right]=10^{-14}$
or $\left[ H ^{+}\right]=\frac{10^{-14}}{10^{-5}}=10^{-9} pH =-\log \left[ H ^{+}\right]$
$=-\log 10^{-9}=9$
$\therefore Mg ( OH )_{2}$ will precipitate at limited pH of $9$.