Question

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

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

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

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

d. $100 \,mL \frac{ M }{10} HCl +100 \,mL \frac{ M }{10} \, NaOH$
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 $75 mL \frac{ M }{5} HCl +25 mL \frac{ M }{5} NaOH$ will be equal to 1 .
Total volume $=75 mL +25 mL =100 mL$
Number of $mmol$ of $HCl =75 mL \times \frac{ M }{5}=15 mmol$
Number of $mmol$ of $NaOH =25 mL \times \frac{ M }{5}=5 mmol$
$5 mmol$ of $NaOH$ will neutralise $5 mmol$ of $HCl$.
$15-5=10 mmol$ of $HCl$ will remain.
$\left[ H ^{+}\right]=[ HCl ]=\frac{10 mmol }{100 mL }=0.1 M$
$pH =-\log _{10}\left[ H ^{+}\right]$
$pH =-\log _{10} 0.1 M =1$