The value of $\mathrm{K}_{\mathrm{c}}$ for the reaction $2 \mathrm{HI}(\mathrm{g}) \rightleftharpoons \mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g})$ is $1 \times 10^{-4}$. At a given time, the composition of reaction mixture is $[\mathrm{HI}]=2 \times 10^{-5} \mathrm{~mol}$, $\left[\mathrm{H}_2\right]=1 \times 10^{-5} \mathrm{~mol}$ and $\left[\mathrm{I}_2\right]=1 \times 10^{-5} \mathrm{~mol}$. In which direction will the reaction proceed?
Given that,
$$\begin{aligned} {[\mathrm{HI}] } & =2 \times 10^{-5} \mathrm{~mol} \\ {\left[\mathrm{H}_2\right] } & =1 \times 10^{-5} \mathrm{~mol} \\ {\left[\mathrm{I}_2\right] } & =1 \times 10^{-5} \mathrm{~mol} \end{aligned}$$
At a given time, the reaction quotient $Q$ for the reaction will be given by the expression
$$\begin{aligned} Q & =\frac{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]}{[\mathrm{HI}]^2} \\ & =\frac{1 \times 10^{-5} \times 1 \times 10^{-5}}{\left(2 \times 10^{-5}\right)^2}=\frac{1}{4} \\ & =0.25=2.5 \times 10^{-1} \end{aligned}$$
As the value of reaction quotient is greater than the value of $K_c$, i.e., $1 \times 10^{-4}$ the reaction will proceed in the reverse reaction.
On the basis of the equation $\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]$, the pH of $10^{-8} \mathrm{~mol} \mathrm{dm}^{-3}$ solution of HCl should be 8 . However, it is observed to be less than 7.0. Explain the reason.
Concentration $10^{-8} \mathrm{~mol} \mathrm{dm}^{-3}$ indicates that the solution is very dilute. So, we cannot neglect the contribution of $\mathrm{H}_3 \mathrm{O}^{+}$ions produced from $\mathrm{H}_2 \mathrm{O}$ in the solution. Total $\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=10^{-8}+10^{-7} \mathrm{M}$. From this we get the value of pH close to 7 but less than 7 because the solution is acidic. From calculation, it is found that pH of $10^{-8} \mathrm{~mol} \mathrm{dm}^{-3}$ solution of HCl is equal to 6.96.
pH of a solution of a strong acid is 5.0 . What will be the pH of the solution obtained after diluting the given solution a 100 times?
Given that,
$\begin{aligned} \mathrm{pH} & =5 \\ {\left[\mathrm{H}^{+}\right] } & =10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}\end{aligned}$
On diluting the solution 100 times $\left[\mathrm{H}^{+}\right]=\frac{10^{-5}}{100}=10^{-7} \mathrm{~mol} \mathrm{~L}^{-1}$
On calculating the pH using the equation $\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]$, value of pH comes out to be 7. It is not possible. This indicates that solution is very dilute.
Hence, $\quad$ Total $\mathrm{H}^{+}$ion concentration $=\mathrm{H}^{+}$ions from acid $+\mathrm{H}^{+}$ion from water
$\begin{aligned} {\left[\mathrm{H}^{+}\right] } & =10^{-7}+10^{-7}=2 \times 10^{-7} \mathrm{M} \\ \mathrm{pH} & =-\log \left[2 \times 10^{-7}\right] \\ \mathrm{pH} & =7-0.3010=6.699\end{aligned}$
A sparingly soluble salt gets precipitated only when the product of concentration of its ions in the solution ( $Q_{\mathrm{sp}}$ ) becomes greater than its solubility product. If the solubility of $\mathrm{BaSO}_4$ in water is $8 \times 10^{-4} \mathrm{~mol}$ $\mathrm{dm}^{-3}$. Calculate its solubility in $0.01 \mathrm{~mol} \mathrm{dm}^{-3}$ of $\mathrm{H}_2 \mathrm{SO}_4$.
$\begin{array}{r}\mathrm{BaSO}_4(\mathrm{~s}) \rightleftharpoons \mathrm{Ba}^{2+}(\mathrm{aq})+\mathrm{SO}_4^{2-}(\mathrm{aq}) \\ K_{\text {sp }} \text { for } \mathrm{BaSO}_4=\left[\mathrm{Ba}^{2+}\right]\left[\mathrm{SO}_4^{2-}\right]=\mathrm{s} \times \mathrm{s}=\mathrm{s}^2\end{array}$
$\begin{array}{lr}\text { But } & s=8 \times 10^{-4} \mathrm{~mol} \mathrm{dm}^{-3} \\ \therefore & K_{\mathrm{sp}}=\left(8 \times 10^{-4}\right)^2=64 \times 10^{-8}\end{array}$
In the presence of $0.01 \mathrm{MH}_2 \mathrm{SO}_4$, the expression for $K_{\text {sp }}$ will be
$$\begin{aligned} K_{\mathrm{sp}} & =\left[\mathrm{Ba}^{2+}\right]\left[\mathrm{SO}_4^{2-}\right] \\ K_{\mathrm{sp}} & =(s) \cdot(s+0.01) \quad\left(0.01 \mathrm{M} \mathrm{SO}_4^{2-} \text { ions from } 0.01 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4\right) \\ 64 \times 10^{-8} & =s \cdot(s+0.01) \\ s^2+0.01 s-64 \times 10^{-8} & =0 \end{aligned}$$
$\begin{aligned} S & =\frac{-0.01 \pm \sqrt{(0.01)^2+\left(4 \times 64 \times 10^{-8}\right)}}{2} \\ & =\frac{-0.01 \pm \sqrt{10^{-4}+\left(256 \times 10^{-8}\right)}}{2} \\ & =\frac{-0.01 \pm \sqrt{10^{-4}\left(1+256 \times 10^{-4}\right)}}{2} \\ & =\frac{-0.01 \pm 10^{-2} \sqrt{1+0.0256}}{2}=\frac{10^{-2}(-1 \pm 1.012719)}{2} \\ & =5 \times 10^{-3}(-1+1.012719)=6.4 \times 10^{-5} \mathrm{~mol} \mathrm{dm}^{-3}\end{aligned}$
pH of $0.08 \mathrm{~mol} \mathrm{~dm}^{-3} \mathrm{~HOCl}$ solution is 2.85 . Calculate its ionisation constant.
$\begin{aligned} \mathrm{pH} \text { of } \mathrm{HOCl} & =2.85 \\ \text{But,}\quad -\mathrm{pH} & =\log \left[\mathrm{H}^{+}\right] \\ \therefore\quad -2.85 & =\log \left[\mathrm{H}^{+}\right] \\ \Rightarrow \quad \overline{3} .15 & =\log \left[\mathrm{H}^{+}\right] \\ \Rightarrow \quad {\left[\mathrm{H}^{+}\right] } & =1.413 \times 10^{-3}\end{aligned}$
For weak monobasic acid $\left[\mathrm{H}^{+}\right]=\sqrt{K_a \times C}$
$\begin{aligned} \Rightarrow \quad K_a & =\frac{\left[\mathrm{H}^{+}\right]^2}{C}=\frac{\left(1.413 \times 10^{-3}\right)^2}{0.08} \\ & =24.957 \times 10^{-6}=2.4957 \times 10^{-5}\end{aligned}$