Conjugate acid of a weak base is always stronger. What will be the decreasing order of basic strength of the following conjugate bases?
$$\mathrm{OH}^{-}, \mathrm{RO}^{-} \mathrm{CH}_3 \mathrm{COO}^{-}, \mathrm{Cl}^{-}$$
Conjugate acid of the given bases are $\mathrm{H}_2 \mathrm{O}, \mathrm{ROH}, \mathrm{CH}_3 \mathrm{COOH}$ and HCl . Order of their acidio strength is
$$\mathrm{HCl}>\mathrm{CH}_3 \mathrm{COOH}>\mathrm{H}_2 \mathrm{O}>\mathrm{ROH}$$
Hence, order of basic strength of their conjugate bases is
$$\mathrm{Cl}^{-}<\mathrm{CH}_3 \mathrm{COO}^{-}<\mathrm{OH}^{-}<\mathrm{RO}^{-}$$
Arrange the following in increasing order of pH .
$$\left.\mathrm{KNO}_3(\mathrm{aq}), \mathrm{CH}_3 \mathrm{COONa}_{(\mathrm{aq}}\right) \mathrm{NH}_4 \mathrm{Cl}(\mathrm{aq}), \mathrm{C}_6 \mathrm{H}_5 \mathrm{COONH}_4(\mathrm{aq})$$
(i) $\mathrm{KNO}_3$ is a salt of strong acid $\left(\mathrm{HNO}_3\right)$ strong base $(\mathrm{KOH})$, hence its aqueous solution is neutral; $\mathrm{pH}=7$.
(ii) $\mathrm{CH}_3 \mathrm{COONa}$ is a salt of weak acid $\left(\mathrm{CH}_3 \mathrm{COOH}\right)$ and strong base $(\mathrm{NaOH})$, hence, its aqueous solution is basic; $\mathrm{pH}>7$.
(iii) $\mathrm{NH}_4 \mathrm{Cl}$ is a salt of strong acid $(\mathrm{HCl})$ and weak base $\left(\mathrm{NH}_4 \mathrm{OH}\right)$ hence, its aqueous solution is acidic; $\mathrm{pH}<7$.
(iv) $\mathrm{C}_6 \mathrm{H}_5 \mathrm{COONH}_4$ is a salt of weak acid, $\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}$ and weak base, $\mathrm{NH}_4 \mathrm{OH}$. But $\mathrm{NH}_4 \mathrm{OH}$ is slightly stronger than $\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}$. Hence, pH is slightly $>7$.
Therefore, increasing order of pH of the given salts is,
$$\mathrm{NH}_4 \mathrm{Cl}<\mathrm{C}_6 \mathrm{H}_5 \mathrm{COONH}_4>\mathrm{KNO}_3<\mathrm{CH}_3 \mathrm{COONa}$$
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
$\begin{aligned} {\left[\mathrm{H}^{+}\right] } & =10^{-7}+10^{-7}=2 \times 10^{-7} \mathrm{M} \\ \mathrm{pH} & =-\log \left[2 \times 10^{-7}\right] \\ \mathrm{pH} & =7-0.3010=6.699\end{aligned}$