Question

20 mL of a 10 N HCl solution is mixed with 10 mL of a 36 N $H_{2}S{O}_4$ solution and this solution is further diluted with water to 1 L. The normality of $H^{+}$ ions in the resultant solution will be:

• A. 0.56 N
• B. 0.50 N
• C. 0.40 N
• D. 0.35 N

Answer 1

The correct option is A 0.56 N

$N_1{V}_1+N_2{V}_2=N_3{V}_3$
Where,
$N_1\Rightarrow$ Normality of the HCl solution = 10 N
$V_1\Rightarrow$ Volume of the HCl solution = 20 mL
$N_2\Rightarrow$ Normality of the $H_{2}S{O}_4$ solution = 36 N
$V_3\Rightarrow$ Volume of the $H_{2}S{O}_4$ solution = 10 mL
$N_3\Rightarrow$ Normality of the resultant solution after mixing
$V_3\Rightarrow$ Volume of the resultant solution after mixing = 30 mL
$\Rightarrow (10\times 20)+(36\times 10)=N_3\times 30$
$\Rightarrow 200+360=30\times N_3$
$\Rightarrow N_3=\frac{560}{30}=\frac{56}{3}$ N
Let the normality of the solution after dilution be N and the final volume of the solution be V.
V = 1 L= 1000 mL
$N_3{V}_3=NV$
$\Rightarrow \frac{56}{3}\times 30=N\times 1000\Rightarrow N=0.56N$