Question

$100\, c.c.$ of $N / 10\, NaOH$ solution is mixed with $100 \,c.c$. of $N / 5 \,HCl$ solution and the whole volume is made to $1$ litre. The $pH$ of the resulting solution will be

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (B)

We know, for an aicd base mixture,
$N _{3}\left( V _{1}+ V _{2}\right)=\left| N _{1} V _{1}- N _{2} V _{2}\right|$
Where, $N_{3}\left(V_{1}+V_{2}\right)=$ Number of gram equivalents of $HCl$
$(\because$ Acid is in excess)
Now,
$g$. equivalents of $HCl =|0.1 \times 0.1-0.1 \times 0.2|=0.01$
$\therefore $ Number of moles $=\frac{\text { Number of } g \text { equivalents }}{n \text { factor of } HCl }$
$=\frac{0.01}{1}=0.01$
$\therefore \underset{0.01}{HCl} \to \underset{0.01}{H^+} + \underset{0.01}{Cl^{\ominus}}$
$\Rightarrow PH = -\log_{10} [H^+] = 2$