Following four solutions are prepared by mixing different volumes of $\mathrm{NaOH}$ and $\mathrm{HCl}$ of…

Following four solutions are prepared by mixing different volumes of $\mathrm{NaOH}$ and $\mathrm{HCl}$ of different concentrations, $\mathrm{pH}$ of which one of them will be equal to 1 ?
  1. $55 \mathrm{~mL} \frac{\mathrm{M}}{10} \mathrm{HCl}+45 \mathrm{~mL} \frac{\mathrm{M}}{10} \mathrm{NaOH}$
  2. $75 \mathrm{~mL} \frac{\mathrm{M}}{5} \mathrm{HCl}+25 \mathrm{~mL} \frac{\mathrm{M}}{5} \mathrm{NaOH}$
  3. $100 \mathrm{~mL} \frac{\mathrm{M}}{10} \mathrm{HCl}+100 \mathrm{~mL} \frac{\mathrm{M}}{10} \mathrm{NaOH}$
  4. $60 \mathrm{~mL} \frac{\mathrm{M}}{10} \mathrm{HCl}+40 \mathrm{~mL} \frac{\mathrm{M}}{10} \mathrm{NaOH}$

Solution

$75 \mathrm{~mL} \frac{\mathrm{M}}{5} \mathrm{HCl}+25 \mathrm{~mL} \frac{\mathrm{M}}{5} \mathrm{NaOH}$ $25 \mathrm{~mL} \frac{\mathrm{M}}{5} \mathrm{NaOH}$ will neutralise $25 \mathrm{~mL} \frac{\mathrm{M}}{5} \mathrm{HCl}$ $75-25=50 \mathrm{~mL} \frac{\mathrm{M}}{5} \mathrm{HCl}$ will remain. Total volume will be $75+25=100 \mathrm{~mL}$ $50 \mathrm{~mL} \frac{\mathrm{M}}{5} \mathrm{HCl}$ is diluted to $100 \mathrm{~mL}$ $\left[\mathrm{H}^{+}\right]=[\mathrm{HCl}]=\frac{\mathrm{M}}{5} \times \frac{50}{100}=\frac{\mathrm{M}}{10}$ $ \mathrm{pH}=-\log _{10}\left[\mathrm{H}^{+}\right]=-\log _{10} \frac{\mathrm{M}}{10}=1 $

Asked in: JEE Main 2018 (15 Apr Shift 2 Online)

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