Question

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

• A. $2\text{N}$
• B. $1\text{N}$
• C. $3\text{N}$
• D. $4\text{N}$

Answer 1

The correct option is B $1\text{N}$

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