Question

Which of the following solutions will have $pH$ close to $1.0?$

• A. $100mL$ of $M/10HCl+100mL$ of $M/10NaOH$
• B. $55mL$ of $M/10HCl+45mL$ of $M/10NaOH$
• C. $10mL$ of $M/10HCl+90mL$ of $M/10NaOH$
• D. $75mL$ of $M/10HCl+25mL$ of $M/10NaOH$

Answer 1

Answer: (D)

(a) is exact neutralisation. Hence $pH=7$

(b) After neutralisation, $M/10HCl$ left $=10mL$

Total volume $=100mL$

Dilution $=10$ times.

$\therefore \left[H^{+}\right]=10^{-2}$ or $pH=2$

(c) After neutralisation, $M/10NaOH$ left $=80mL$

Total volume $=100mL,pH>7$

(d) After neutralisation, $M/10HCl$ left $=50mL$

Total volume $=100mL$

Dilution $=2$ times

$\therefore \left[H^{+}\right]=\frac{1}{2\times 10}=\frac{10^{-1}}{2}M$ or $pH=1.3$