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6-2.Equilibrium-II (Ionic Equilibrium)
medium
The solubility of $Ba_3(AsO_4)_2$ (molar mass $= 690$) is $6.9 \times 10^{-2} \,g/100\, mL$. What is the value of its $K_{sp}$?
A
$1.08 \times 10^{-11}$
B
$1.08 \times 10^{-13}$
C
$1.0 \times 10^{-15}$
D
$6.0 \times 10^{-13}$
Solution
$s = 6.9 \times {10^{ – 2}}\,{\text{gm/100 ml }}$
$ = \frac{{6.9 \times {{10}^{ – 2}}}}{{690}} = {10^{ – 4}}\,{\text{mol/1}}00\,{\text{ml}}$
$ = {10^{ – 3}}\,{\text{mol}}/{\text{lit}}$
${{\text{K}}_{{\text{sp}}}} = 108\,{{\text{s}}^5}$
$ = 108 \times {10^{ – 15}} = 1.08 \times {10^{ – 13}}$
Standard 11
Chemistry
