Equilibrium - Result Question 68
####71. Concentration of the $Ag^{+}$ions in a saturated solution of $Ag_2 C_2 O_4$ is $2.2 \times 10^{-4} mol L^{-1}$. Solubility product of $Ag_2 C_2 O_4$ is :- [2017]
(a) $2.66 \times 10^{-12}$
(b) $4.5 \times 10^{-11}$
(c) $5.3 \times 10^{-12}$
(d) $2.42 \times 10^{-8}$
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Solution:
- (c)
$ Ag_2 C_2 O_4(s) \rightarrow \underset{2 s}{2 Ag^{+}}(aq)+C_2 O_4^{2-}(aq) $
$K _{s p}=[Ag^{+}]^{2}[C_2 O_4^{2-}]$
$[Ag^{+}]=2.2 \times 10^{-4} M$
Given that:
$\therefore$ Concentration of $C_2 O_4{ }^{2-}$ ions,
$ [C_2 O_4^{2-}]=\frac{2.2 \times 10^{-4}}{2} M=1.1 \times 10^{-4} M $
$ \begin{align*} \therefore \quad K _{s p} & =(2.2 \times 10^{-4})^{2}(1.1 \times 10^{-4}) \tag{72.}\\ & =5.324 \times 10^{-12} \end{align*} $
(b) $MY \rightarrow \mid M^{+}+Y^{-}$
$K _{s p}=s^{2}=6.2 \times 10^{-13}$
$s=\sqrt{6.2^{\prime} 10^{-13}}$
$s=7.87 \times 10^{-7} mol L^{-1}$
$NY_3 \rightarrow \mid N^{3+}+3 Y^{-}$
$K _{s p}=s \times(3 s)^{3}=27 s^{4}=6.2 \times 10^{-13}$
$s=(\frac{6.2 \times 10^{-13}}{27})^{1 / 4}$
$s=3.89 \times 10^{-4} mol L^{-1}$
$\therefore \quad$ molar solubility of $NY_3$ is more than MY in water.
