Question

Concentrated strong acid is added to a solid mixture of 0.015 mole samples of $Fe(OH{)}_2$ and $Ca(OH{)}_2$ placed in one litre of water. At what value of pH will the dissolution of each hydroxide be complete? (Assume negligible volume change)
$K_{sp}\left[Fe\right(OH{)}_2]=7.9\times {10}^{-15}$ and $K_{sp}\left[Cu\right(OH{)}_2]=1.6\times {10}^{-19}$

• A. 6.67
• B. 7.86
• C. 8.86
• D. 7.26

Answer 1

Answer: (B)

7.86

Let, $\left[F{e}^{2+}\right]=xM$ and $\left[O{H}^{-}\right]=yM$.

Therefore, $\left[C{a}^{2+}\right]=0.015-xM$.

$K_{sp,Fe(OH{)}_2}=\left[F{e}^{2+}\right]\left[O{H}^{-}\right]$

$7.9\times {10}^{-15}=x{y}^2$ .....(1)

$K_{sp,Ca(OH{)}_2}=\left[C{a}^{2+}\right]\left[O{H}^{-}\right]$

$1.6\times {10}^{-19}=(0.015-x)y^2$ .....(2)

Dividing equation (2) by eqation (1) ,

$\frac{1.6\times {10}^{-19}}{7.9\times {10}^{-15}}=\frac{(0.015-x)y^2}{x{y}^2}$

$2.025\times {10}^{-5}=\frac{0.015-x}{x}$

$2.025\times {10}^{-5}x=0.015-x$

$x=0.014999$

Substitute value of x in equation (1),

$7.9\times {10}^{-15}=x{y}^2$

$7.9\times {10}^{-15}=0.014999\times y^2$

$y=7.26\times {10}^{-7}M$

$pOH=-log\left[O{H}^{-}\right]=-log(7.26\times {10}^{-7})=6.14$

$pH=14-pOH=14-6.14=7.86$