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

Answer: (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 $NaOH=25mL\times \frac{M}{5}=5mmol$
$5mmol$ of $NaOH$ will neutralise $5mmol$ of $HCl$.
$15-5=10mmol$ of $HCl$ will remain.
$\left[H^{+}\right]=[HCl]=\frac{10mmol}{100mL}=0.1M$
$pH=-{\log }_{10}\left[H^{+}\right]$
$pH=-{\log }_{10}0.1M=1$