Chemical and Ionic Equilibrium 2 Question 21
21. Which of the following solutions will have $\mathrm{pH}$ close to 1.0 ?
(1992, 1M)
(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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Solution:
- $75 \mathrm{~mL} \frac{\mathrm{M}}{5} \mathrm{HCl}=15 \mathrm{mmol} \mathrm{HCl}$
$25 \mathrm{~mL} \frac{\mathrm{M}}{5} \mathrm{NaOH}=5 \mathrm{mmol} \mathrm{NaOH}$ After neutralisation, $10 \mathrm{mmol} \mathrm{HCl}$ will be remaining in $100 \mathrm{~mL}$ of solution.
Molarity of $\mathrm{HCl}$ in the final solution $=\frac{10}{100}=0.10$
$$ \mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]=-\log (0.10)=1 $$
