Question

50 ml of 0.2 M HCl, 50 ml of 0.1 M ${\text{H}}_{\text{3}}{\text{PO}}_{\text{2}}$ and 100 ml of 0.1 M ${\text{H}}_{\text{3}}{\text{PO}}_{\text{3}}$
are mixed. The normality of the resulting solution is:

• A. 0.1 N
• B. 0.15 N
• C. 0.125 N
• D. 0.020 N
• E. 2.175 N

Answer 1

The correct option is C 0.125 N

$n_{factor}$ of $HCl=1,$ Normality $=0.2\times 1=0.2$ $N$

$n_{factor}$ of $H_{3}P{O}_2=1,$ Normality $=0.1\times 1=0.1$$N$

$n_{factor}$ of $H_{3}P{O}_3=1,$ Normality $=0.1\times 2=0.2$ $N$

Net normality $=\frac{V_1\times N_1+V_2\times N_2+......+V_n\times N_n}{V_1+V_2+V_3+.....+V_n}$

$=\frac{50\times 0.2+50\times 0.1+100\times 0.1}{200}=\frac{25}{200}=0.125$ $N$

Hence, the correct option is $C$