Question

Following solutions were prepared by mixing different volumes of $NaOH$ and $HCl$ of different concentrations:

a. $60mL\frac{M}{10}HCl+40mL\frac{M}{10}NaOH$

b. $55mL\frac{M}{10}HCl+45mL\frac{M}{10}NaOH$

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

d. $100mL\frac{M}{10}HCl+100mL\frac{M}{10}NaOH$

$pH$ of which one of them will be equal to $1$?

• A. $d$
• B. $b$
• C. $c$
• D. $a$

Answer 1

The correct option is C $c$

$pH$ of $75mL\frac{M}{5}HCl+25mL\frac{M}{5}NaOH$ will be equal to $1$.

Total volume $=75mL+25mL=100mL$

Number of mmol of $HCl$ $=75mL\times \frac{M}{5}=15mmol$

Number of mmol of $NaOH$ $=25mL\times \frac{M}{5}=5mmol$

5 mmol of $NaOH$ will neutralise 5 mmol of $HCl$.

$15-5=10$ mmol of $HCl$ will remain.

$\left[H^{+}\right]=\left[HCl\right]=\frac{10mmol}{100mL}=0.1M$

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

$pH=-lo{g}_{10}0.1M=1$