- Home
- Standard 11
- Chemistry
6-2.Equilibrium-II (Ionic Equilibrium)
medium
The solubility of $CaC{O_3}$ in water is $3.05 \times {10^{ - 4}}\,moles/litre.$ Its solubility product will be
A
$3.05 \times {10^{ - 4}}$
B
$10$
C
$6.1 \times {10^{ - 4}}$
D
$9.3 \times {10^{ - 8}}$
Solution
$\mathop {CaC{O_3}}\limits_S \rightleftharpoons \mathop {C{a^{ + 2}}}\limits_S+ \mathop {CO_3^{ – \, – }}\limits_S$
Solubility product of $CaC{O_3}$
${K_{sp}} = {S^2}$; $S = \sqrt {{K_{sp}}} $
It is a binary electrolyte.
${S^2} = {K_{sp}}$; ${(3.05 \times {10^{ – 4}})^2} = {K_{sp}}$;
${K_{sp}} = 9.3 \times {10^{ – 8}}$
Standard 11
Chemistry
