Q. The concentration of the $Ag^{+}$ ions in a saturated solution of $Ag_{2}C_{2}O_{4}$ is $2.2\times 10^{- 4 \, }molL^{- 1}.$ Solubility product of $Ag_{2}C_{2}O_{4}$ is
NTA AbhyasNTA Abhyas 2022
Solution:
The dissociation of $Ag_{2}C_{2}O_{4}$ is:
$\underset{s}{Ag_{2} C_{2} O_{4}}\leftrightarrow \underset{2 s}{2 Ag^{+}}+\underset{s}{C_{2} O_{4}^{2 -}}$
The solubility product is given as:
$K_{sp}=\left[Ag^{+}\right]^{2}\left[C_{2} O_{4}^{2 -}\right]$
Given that $\left[Ag^{+}\right]=2.2x10^{- 4}molL^{- 1}$
Therefore, $\left[C_{2} O_{4}^{2 -}\right]=\frac{2 . 2 \times 10^{- 4}}{2}=1.1x10^{- 4}M$
The solubility product is given as:
$K_{sp}=\left(\right.2.2\times \left(10\right)^{- 4}\left.\right)^{2}\left(\right.1.1\times \left(10\right)^{- 4}\left.\right)=5.324\times \left(10\right)^{- 12}$
$\underset{s}{Ag_{2} C_{2} O_{4}}\leftrightarrow \underset{2 s}{2 Ag^{+}}+\underset{s}{C_{2} O_{4}^{2 -}}$
The solubility product is given as:
$K_{sp}=\left[Ag^{+}\right]^{2}\left[C_{2} O_{4}^{2 -}\right]$
Given that $\left[Ag^{+}\right]=2.2x10^{- 4}molL^{- 1}$
Therefore, $\left[C_{2} O_{4}^{2 -}\right]=\frac{2 . 2 \times 10^{- 4}}{2}=1.1x10^{- 4}M$
The solubility product is given as:
$K_{sp}=\left(\right.2.2\times \left(10\right)^{- 4}\left.\right)^{2}\left(\right.1.1\times \left(10\right)^{- 4}\left.\right)=5.324\times \left(10\right)^{- 12}$