The molar solubility of Cd(OH)_2 is 1.84 10^ -5 , M in water. The… - Vidyadip

MCQ

The molar solubility of $Cd(OH)_2$ is $1.84\times10^{-5}\, M$ in water. The expected solubility of $Cd(OH)_2$ in a buffer solution of $pH = 12$ is

A
$6.23\times10^{-11}\, M$
B
$1.84\times10^{-9}\, M$
C
$\frac{{2.49}}{{1.84}} \times {10^{ - 9}}\,M$
✓
$2.49\times10^{-10}\, M$

✓

Answer

Correct option: D.

$2.49\times10^{-10}\, M$

d
${K_{sp}}$ of $Cd{(OH)_2} = 4{s^3} = 4 \times {(1.84 \times {10^{ - 5}})^3}$

If $pH = 12$

$pOH = 2$

$[O{H^ - }] = {10^{ - 2}}\,M$

${K_{sp}} = [C{d^{2 + }}]{[O{H^ - }]^2}$

$4 \times {(1.84 \times {10^{ - 5}})^3} = [C{d^{2 + }}]{[O{H^ - }]^2}$

$[C{d^{2 + }}] = \frac{{4 \times {{(1.84)}^3} \times {{10}^{ - 15}}}}{{{{10}^{ - 4}}}}$

$C{d^{2 + }} = 4 \times 6.22 \times {10^{ - 11}} = 2.49 \times {10^{ - 10}}\,M$

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