Chemical And Ionic Equilibrium Result Question 21
21. Which of the following solutions will have $\mathrm{pH}$ close to $1.0$ ?
(1992,1 M)
(a) $100 \mathrm{~mL}$ of $(\mathrm{M} / 10) \mathrm{HCl}+100 \mathrm{~mL}$ of $(\mathrm{M} / 10) \mathrm{NaOH}$
(b) $55 \mathrm{~mL}$ of $(\mathrm{M} / 10) \mathrm{HCl}+45 \mathrm{~mL}$ of $(\mathrm{M} / 10) \mathrm{NaOH}$
(c) $10 \mathrm{~mL}$ of $(\mathrm{M} / 10) \mathrm{HCl}+90 \mathrm{~mL}$ of $(\mathrm{M} / 10) \mathrm{NaOH}$
(d) $75 \mathrm{~mL}$ of $(\mathrm{M} / 5) \mathrm{HCl}+25 \mathrm{~mL}$ of $(\mathrm{M} / 5) \mathrm{NaOH}$
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Answer:
Correct Answer: 21. ( d )
Solution:
$\begin{aligned} & 75 \mathrm{~mL} \frac{\mathrm{M}}{5} \mathrm{HCl}=15 \hspace{2mm} \mathrm{mmol} \hspace{2mm} \mathrm{HCl} \\ & 25 \mathrm{~mL} \frac{\mathrm{M}}{5} \mathrm{NaOH}=5\hspace{2mm} \mathrm{mmol} \hspace{2mm}\mathrm{NaOH}\end{aligned}$
After neutralisation, $10$ mmol HCl will be remaining in $100$ mL of solution.
Molarity of $HCl$ in the final solution $=\frac{10}{100}=0.10$
$\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]=-\log (0.10)=1$
BETA
