Question

Following four solutions are prepared by mixing different volumes of NaOH and HCl of different concentrations, pH of which one of them will be equal to $1$?

• A. $55$mL $\frac{M}{10}$HCl $+45$mL $\frac{M}{10}$NaOH
• B. $75$mL $\frac{M}{5}$HCl$+25$mL $\frac{M}{5}$NaOH
• C. $100$mL $\frac{M}{10}$HCl$+100$mL $\frac{M}{10}$NaOH
• D. $60$mL $\frac{M}{10}$HCl$+40$mL $\frac{M}{10}$NaOH

Answer 1

Answer: (B)

$75$mL $\frac{M}{5}$HCl$+25$mL $\frac{M}{5}$NaOH

$75$mL $\frac{M}{5}$HCl$+25$mL

$\frac{M}{5}$NaOH

25mL $\frac{M}{5}$NaOH will neutralise 25 mL

$\frac{M}{5}$HCl

$75-25=50$ mL $\frac{M}{5}$HCl will remain.

Total volume will be $75+25=100$ mL.

50 mL $\frac{M}{5}$HCl is diluted to 100 mL.

$\left[H^{+}\right]=\left[HCl\right]=\frac{M}{5}\times \frac{50}{100}=\frac{M}{10}$

$pH=-lo{g}_{10}\left[H^{+}\right]=-lo{g}_{10}\frac{M}{10}=1$