Question

By adding $20 \,mL \,0.1 \,N \,HCl$ to $20 \,mL \,0.001 \,N \,KOH$, the pH of the obtained solution will be

• A. 2
• B. 1.3
• C. 0
• D. 7

Answer 1

Answer: (B)

$20 \,mL$ of $0.1\, N \,HCl =\frac{0.1}{1000} \times 20 \,g \,eq$
$=2 \times 10^{-3} g \,eq$
$20 \,mL $ of $ 0.001\, KOH =\frac{0.001}{1000} \times 20\, g\, eq $
$=2 \times 10^{-5} g \,eq $
$ \therefore HCl $ left unneutralised
$=2\left(10^{-3}-10^{-5}\right)$
$=2 \times 10^{-3}(1-0.01) $
$=2 \times 0.99 \times 10^{-3}$
$=1.98 \times 10^{-3} g \,eq $.
Volume of solution $=40\, mL$
$ \therefore [ HCl ] =\frac{1.98 \times 10^{-3}}{40} \times 1000\, M $
$=4.95 \times 10^{-2} $
$ \therefore pH =2-\log 4.95 $
$=2-0.7 =1.3 $