JEE Advance - Chemistry (2022 - Paper 2 Online - No. 1)
Concentration of $\mathrm{H}_{2} \mathrm{SO}_{4}$ and $\mathrm{Na}_{2} \mathrm{SO}_{4}$ in a solution is $1 \mathrm{M}$ and $1.8 \times 10^{-2} \mathrm{M}$, respectively. Molar solubility of $\mathrm{PbSO}_{4}$ in the same solution is $\mathrm{X} \times 10^{-\mathrm{Y}} \mathrm{M}$ (expressed in scientific notation). The value of $Y$ is ________.
[Given: Solubility product of $\mathrm{PbSO}_{4}\left(K_{s p}\right)=1.6 \times 10^{-8}$. For $\mathrm{H}_{2} \mathrm{SO}_{4}, K_{a l}$ is very large and $\left.K_{a 2}=1.2 \times 10^{-2}\right]$
Answer
6
Explanation
Given,
Conentration of $\mathrm{H}_2 \mathrm{SO}_4=1 \mathrm{M}$
Conentration of $\mathrm{Na}_{2} \mathrm{SO}_{4}$ = $1.8 \times 10^{-2} \mathrm{M}$
From the above two equations, we get
$$ \left[\mathrm{H}^{+}\right]=1 \mathrm{M} \text { and }\left[\mathrm{SO}_4^{2-}\right]=1.8 \times 10^{-2} $$
$$ K_c=\frac{1.8 \times 10^{-2} \times 1}{1}=1.8 \times 10^{-2} $$
and it is given that $\mathrm{K}_{a_2}\left(\mathrm{Q}_c\right)=1.2 \times 10^{-2} \mathrm{M}$
Since, $\mathrm{K}_{a_2}$ (i.e., $\mathrm{Q}_c$ ) $<\mathrm{K}_c$
$\therefore$ Rather than dissociation of $\mathrm{HSO}_4^{-}$into $\mathrm{H}^{+}$and $\mathrm{SO}_4^{2-}$ ions, association between already present $\mathrm{H}^{+}$and $\mathrm{SO}_4^{2-}$ will take place.
Assuming ' $x$ ' mol/L of $\mathrm{SO}_4^{2-}$ and $\mathrm{H}^{+}$combines to form $\mathrm{HSO}_4^{-}$
$\begin{aligned} & \mathrm{K}_{a 2}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{SO}_4^{2-}\right]}{\left[\mathrm{HSO}_4^{-}\right]} \\\\ & 1.2 \times 10^{-2}=\frac{(1-x)\left(1.8 \times 10^{-2}-x\right)}{(1+x)} \\\\ & \because x<<1, \text { so }(1+x) \approx 1 \text { and }(1-x) \approx 1 \\\\ & 1.2 \times 10^{-2}=1.8 \times 10^{-2}-x \\\\ & x=\left(1.8 \times 10^{-2}\right)-\left(1.2 \times 10^{-2}\right) \\\\ & x=0.6 \times 10^{-2} \mathrm{M} \\\\ & \text { So, }\left[\mathrm{SO}_4^{2-}\right]=1.8 \times 10^{-2}-x \\\\ & {\left[\mathrm{SO}_4^{2-}\right] }=\left(1.8 \times 10^{-2}\right)-\left(0.6 \times 10^{-2}\right) \\\\ & {\left[\mathrm{SO}_4^{2-}\right] }=1.2 \times 10^{-2} \mathrm{M}\end{aligned}$
Given, $\quad \mathrm{K}_{s p}=1.6 \times 10^{-8}$
$$ \therefore $$ $y\left(1.2 \times 10^{-2}+y\right)=1.6 \times 10^{-8}$
Since, $y<<1$, So $1.2 \times 10^{-2}+y \approx 1.2 \times 10^{-2}$
So, $y \times 1.2 \times 10^{-2}=1.6 \times 10^{-8}$
$$ \Rightarrow $$ $y =\frac{1.6 \times 10^{-8}}{1.2 \times 10^{-2}}$
$$ \Rightarrow $$ $ y =1.33 \times 10^{-6} $
$ X \times 10^{-Y} \mathrm{M} =1.33 \times 10^{-6} \mathrm{M} $
So, $\mathrm{Y}=6$
Hence, the value of $Y$ is 6 .
Conentration of $\mathrm{H}_2 \mathrm{SO}_4=1 \mathrm{M}$
Conentration of $\mathrm{Na}_{2} \mathrm{SO}_{4}$ = $1.8 \times 10^{-2} \mathrm{M}$
From the above two equations, we get
$$ \left[\mathrm{H}^{+}\right]=1 \mathrm{M} \text { and }\left[\mathrm{SO}_4^{2-}\right]=1.8 \times 10^{-2} $$
$$ K_c=\frac{1.8 \times 10^{-2} \times 1}{1}=1.8 \times 10^{-2} $$
and it is given that $\mathrm{K}_{a_2}\left(\mathrm{Q}_c\right)=1.2 \times 10^{-2} \mathrm{M}$
Since, $\mathrm{K}_{a_2}$ (i.e., $\mathrm{Q}_c$ ) $<\mathrm{K}_c$
$\therefore$ Rather than dissociation of $\mathrm{HSO}_4^{-}$into $\mathrm{H}^{+}$and $\mathrm{SO}_4^{2-}$ ions, association between already present $\mathrm{H}^{+}$and $\mathrm{SO}_4^{2-}$ will take place.
Assuming ' $x$ ' mol/L of $\mathrm{SO}_4^{2-}$ and $\mathrm{H}^{+}$combines to form $\mathrm{HSO}_4^{-}$
$\begin{aligned} & \mathrm{K}_{a 2}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{SO}_4^{2-}\right]}{\left[\mathrm{HSO}_4^{-}\right]} \\\\ & 1.2 \times 10^{-2}=\frac{(1-x)\left(1.8 \times 10^{-2}-x\right)}{(1+x)} \\\\ & \because x<<1, \text { so }(1+x) \approx 1 \text { and }(1-x) \approx 1 \\\\ & 1.2 \times 10^{-2}=1.8 \times 10^{-2}-x \\\\ & x=\left(1.8 \times 10^{-2}\right)-\left(1.2 \times 10^{-2}\right) \\\\ & x=0.6 \times 10^{-2} \mathrm{M} \\\\ & \text { So, }\left[\mathrm{SO}_4^{2-}\right]=1.8 \times 10^{-2}-x \\\\ & {\left[\mathrm{SO}_4^{2-}\right] }=\left(1.8 \times 10^{-2}\right)-\left(0.6 \times 10^{-2}\right) \\\\ & {\left[\mathrm{SO}_4^{2-}\right] }=1.2 \times 10^{-2} \mathrm{M}\end{aligned}$
Given, $\quad \mathrm{K}_{s p}=1.6 \times 10^{-8}$
$$ \therefore $$ $y\left(1.2 \times 10^{-2}+y\right)=1.6 \times 10^{-8}$
Since, $y<<1$, So $1.2 \times 10^{-2}+y \approx 1.2 \times 10^{-2}$
So, $y \times 1.2 \times 10^{-2}=1.6 \times 10^{-8}$
$$ \Rightarrow $$ $y =\frac{1.6 \times 10^{-8}}{1.2 \times 10^{-2}}$
$$ \Rightarrow $$ $ y =1.33 \times 10^{-6} $
$ X \times 10^{-Y} \mathrm{M} =1.33 \times 10^{-6} \mathrm{M} $
So, $\mathrm{Y}=6$
Hence, the value of $Y$ is 6 .
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