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 $\left[\mathrm{H}^{+}\right]=\sqrt{K_a \times C}$
$\begin{aligned} \Rightarrow \quad K_a & =\frac{\left[\mathrm{H}^{+}\right]^2}{C}=\frac{\left(1.413 \times 10^{-3}\right)^2}{0.08} \\ & =24.957 \times 10^{-6}=2.4957 \times 10^{-5}\end{aligned}$
Calculate the pH of a solution formed by mixing equal volumes of two solutions A and B of a strong acid having $\mathrm{pH}=6$ and $\mathrm{pH}=4$ respectively.
pH of solution $A=6$. Hence, $\left[\mathrm{H}^{+}\right]=10^{-6} \mathrm{~mol} \mathrm{~L}$
pH of solution $B=4$. Hence, $\left[\mathrm{H}^{+}\right]=10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}$
On mixing 1 L of each solution, molar concentration of total $\mathrm{H}^{+}$is halved.
$\begin{aligned} & \text{Total,}\quad {\left[\mathrm{H}^{+}\right] }=\frac{10^{-6}+10^{-4}}{2} \mathrm{~mol} \mathrm{~L}^{-1} \\ & {\left[\mathrm{H}^{+}\right] }=\frac{1.01 \times 10^{-4}}{2}=5.05 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1} \\ & {\left[\mathrm{H}^{+}\right] }=5.0 \times 10^{-5} \mathrm{~mol} \mathrm{~L} \\ & \mathrm{pH}=-\log \left[\mathrm{H}^{+}\right] \\ & \mathrm{pH}=-[\log 5+(-5 \log 10)] \Rightarrow \mathrm{pH}=-\log \left(5.0 \times 10^{-5}\right) \\ & \mathrm{pH}=5-\log 5=5-0.6990 \Rightarrow \mathrm{pH}=-\log 5+5 \\ & \mathrm{pH}=4.3010 \approx 4.3\end{aligned}$
Thus, the pH of resulting solution is 4.3.
The solubility product of $\mathrm{Al}(\mathrm{OH})_3$ is $2.7 \times 10^{-11}$. Calculate its solubility in $\mathrm{gL}^{-1}$ and also find out pH of this solution. (Atomic mass of $\mathrm{Al}=27 \mathrm{u}$ )
Let S be the solubility of Al(OH)$$_3$$.
Concentration of species at $t=0$
Concentration of various species at equilibrium
$\begin{aligned} K_{\mathrm{sp}} & =\left[\mathrm{Al}^{3+}\right]\left[\mathrm{OH}^{-}\right]^3=(\mathrm{S})(3 \mathrm{~S})^3=27 \mathrm{~S}^4 \\ S^4 & =\frac{K_{\mathrm{sp}}}{27}=\frac{2.7 \times 10^{-11}}{27}=1 \times 10^{-12} \\ S & =1 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}\end{aligned}$
(i) Solubility of $\mathrm{Al}(\mathrm{OH})_3$
Molar mass of $\mathrm{Al}(\mathrm{OH})_3$ is 78 g . Therefore,
Solubility of $\mathrm{Al}(\mathrm{OH})_3$ in $\mathrm{g}^{-1}=1 \times 10^{-3} \times 78 \mathrm{~g} \mathrm{~L}^{-1}=78 \times 10^{-3} \mathrm{~g} \mathrm{~L}^{-1}$
$=7.8 \times 10^{-2} \mathrm{~g} \mathrm{~L}^{-1}$
$\begin{aligned} \text{(ii) pH of the solution}\quad \mathrm{S} & =1 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \\ {\left[\mathrm{OH}^{-}\right] } & =3 \mathrm{~S}=3 \times 1 \times 10^{-3}=3 \times 10^{-3} \\ \mathrm{pOH} & =3-\log 3 \\ \mathrm{pH} & =14-\mathrm{pOH}=11+\log 3=11.4771\end{aligned}$