Question

Solubility of calcium carbonate is $0.0305$gm/litre. ${{\text{K}}_{{\text{sp}}}}$for ${\text{CaC}}{{\text{O}}_3}$ is.
A. $93.05 \times {10^{ - 4}}$
B. $9.3 \times {10^{ - 8}}$
C. $0.000305$
D. $93 \times {10^{ - 8}}$

Answer 1

Hint: ${{\text{K}}_{{\text{sp}}}}$ represents the solubility product constant at equilibrium. When a solid substance dissolves into water it dissociates into ion. The product of solubilities of the ions with the number of each ion in power is known as solubility product.

Formula used: ${{\text{K}}_{{\text{sp}}}}\, = \,{\left[ {xS} \right]^x}\,\,{\left[ {yS} \right]^y}$

Complete step by step answer:The solubility product constant represents the product of concentrations of constituting ions of an ionic compound at equilibrium each raised number of ions as power.
An ionic compound dissociates into the water as follows:
${{\text{A}}_{\text{x}}}{{\text{B}}_{\text{y}}}\, \rightleftarrows \,\,{\text{x}}{{\text{A}}^{{\text{ + Y}}}}\,{\text{ + }}\,{\text{y}}{{\text{B}}^{ - x}}$
Where,
${\text{AB}}$ is an ionic compound.
The general representation for the solubility product of an ionic compound is as follows:
${{\text{K}}_{{\text{sp}}}}\, = \,{\left( {x{A^{ + y}}\,} \right)^x}\, \times \,\,{\left( {y{A^{ - x}}\,} \right)^y}$
${{\text{K}}_{{\text{sp}}}}\, = \,{\left[ {xS} \right]^x}\,\,{\left[ {yS} \right]^y}$
Where,
${{\text{K}}_{{\text{sp}}}}$is the solubility product constant.
${\text{S}}$is the solubility of each ion.
$x$ and $y$ represents the number of ions.
The compound calcium carbonate dissociates into the water as follows:
${\text{CaC}}{{\text{O}}_3}\, \rightleftarrows \,{\text{C}}{{\text{a}}^{{\text{2 + }}}}{\text{ + }}\,{\text{CO}}_3^{2 - }$
Calcium carbonate in water produces one calcium ion and one carbonate ion.
The solubility product for calcium carbonate is represented as follows:
${{\text{K}}_{{\text{sp}}}}\, = \,\left[ {{\text{C}}{{\text{a}}^{{\text{2 + }}}}} \right]\,\,\left[ {{\text{CO}}_3^{2 - }} \right]$
${{\text{K}}_{{\text{sp}}}}\, = \,\left( {{\text{S}}\,} \right)\, \times \,\,\left( {{\text{S}}} \right)$
${{\text{K}}_{{\text{sp}}}}\, = \,\,{{\text{S}}^2}$
calcium carbonate is $0.0305$gm/litre. ${{\text{K}}_{{\text{sp}}}}$for ${\text{CaC}}{{\text{O}}_3}$ is.
On substituting $0.0305$ gm/litre for S.
${{\text{K}}_{{\text{sp}}}}\, = \,\,{\left( {0.0305} \right)^2}$
${{\text{K}}_{{\text{sp}}}}\, = \,\,9.3 \times {10^{ - 8}}$
So, the ${{\text{K}}_{{\text{sp}}}}$for ${\text{CaC}}{{\text{O}}_3}$ is $9.3 \times {10^{ - 8}}$.

Therefore, option (B) $9.3 \times {10^{ - 8}}$, is correct.

Note: The formula of solubility product constant depends upon the number of ions produced by the ionic compounds in the water. Molar solubility represents the number of ions dissolved per liter of solution. Here, solubility represents the number of ions dissolved in a given amount of solvent. The compound having ions in $1:1$ ratio, the solubility product is determined by ${{\text{S}}^{\text{2}}}$. The compound having ions in $1:2$ ratio, the solubility product is determined by ${\text{4}}{{\text{S}}^3}$.