Question

Ksp of $\mathrm{AgCl}$ is $1\times {10}^{-10}$. Its solubility in $0.1{\mathrm{KNO}}_3$ will be

• A. ${10}^{-5}\mathrm{mol}{\mathrm{lit}}^{-1}$
• B. $>{10}^{-5}\mathrm{mol}{\mathrm{lit}}^{-1}$
• C. $<{10}^{-5}\mathrm{mol}{\mathrm{lit}}^{-1}$
• D. None of the above

Answer 1

The correct option is A

${10}^{-5}\mathrm{mol}{\mathrm{lit}}^{-1}$

The explanation for the correct option:

Option (A) ${10}^{-5}\mathrm{mol}{\mathrm{lit}}^{-1}$

Step 1: Given data

Ksp of $\mathrm{AgCl}=1\times {10}^{-10}$

Step 2:Calculating Solubility and Solubility product

Dissociation of $\mathrm{AgCl}$ is given as $\underset{\mathrm{Silver}\mathrm{chloride}}{\mathrm{AgCl}\left(\mathrm{s}\right)}\to \underset{\mathrm{Silver}\mathrm{ion}}{{\mathrm{Ag}}^{+}}+\underset{\mathrm{Chloride}\mathrm{ion}}{{\mathrm{Cl}}^{-}}$

Solubility product$=$ Solubility of Silver ion $\times$solubility of chloride ion

Let the Solubility of chloride and silver ions be “s”

$\mathrm{Ksp}=\mathrm{s}\times \mathrm{s}\Rightarrow {\mathrm{s}}^2$

$1\times {10}^{-10}={\mathrm{s}}^2$

$\therefore \mathrm{s}={10}^{-5}$

Dissociation of $\underset{\mathrm{Potassium}\mathrm{nitrate}}{{\mathrm{KNO}}_3\left(\mathrm{s}\right)}\to \underset{\mathrm{Potassium}\mathrm{ion}}{{\mathrm{K}}^{+}}+\underset{\mathrm{Nitrate}\mathrm{ion}}{{{\mathrm{NO}}_3}^{-}}$

Since there are no common ions present at the addition of ${\mathrm{KNO}}_3$, the solubility remains unaffected.

Therefore, Solubility of $\mathrm{AgCl}$ in ${\mathrm{KNO}}_3={10}^{-5}\mathrm{mol}{\mathrm{lit}}^{-1}$

Hence, option (A) is correct.