Question

50 mL of a 10 N $H_{2}S{O}_4$ solution, 25 mL of a 12 N HCl solution and 40 mL of a 5 N $HN{O}_3$ solution were mixed and the volume of this solution was made 1000 mL by adding water. The normality of $H^{+}$ ions in the resultant solution will be:

• A. 2 N
• B. 1 N
• C. 3 N
• D. 4 N

Answer 1

The correct option is B 1 N

$NV=N_1{V}_1+N_2{V}_2+N_3{V}_3$
Where,
$N_1\Rightarrow$ Normality of $H_{2}S{O}_4$ solution =10 $N;V_1\Rightarrow$ Volume of the $H_{2}S{O}_4$ solution = 50 mL
$N_2\Rightarrow$ Normality of HCl solution = $12N;V_2\Rightarrow$ Volume of the HCl solution = 25 mL
$N_3\Rightarrow$ Normality of $HN{O}_3$ solution = 5 N; $V_3\Rightarrow$ Volume of the $HN{O}_3$ solution = 40 mL
$N\Rightarrow$ Normality of $H^{+}$ in the resultant solution
V ⇒ Volume of the resultant solution
Thus,
$N=\frac{(10\times 50)+(12\times 25)+(5\times 40)}{1000}=1$ N