Question

50 mL of a 0.1 M magnesium hydroxide solution on titrating with a solution of a mixture of 0.1 M solution of $HCl$ and unknown $H_{2}S{O}_4$ gave a titre value of 50 mL. The molarity of the $H_{2}S{O}_4$ solution is:

• A. 0.05 M
• B. 0.1 M
• C. 0.25 M
• D. 0.35 M

Answer 1

The correct option is A 0.05 M

Let the molarity of $H_{2}S{O}_4$ solution be $x$ M
So, $n_1\times M_1\times V_1=n_2\times M_2\times V_2+n_3\times M_3\times V_3$
Where, $n_1=\text{n-factor of}Mg(OH{)}_2=2$
$M_1=\text{molarity of}Mg(OH{)}_2=0.1$ M
$V_1=\text{volume of}Mg(OH{)}_2=50$ mL
$n_2=\text{n-factor of}HCl=1$
$M_2=\text{molarity of}HCl=0.1M$
$V_2=\text{volume of}HCl=150$ mL
$n_3=\text{n-factor of}{H}_{2}S{O}_4=2$
$V_3=\text{volume of}{H}_{2}S{O}_4=50$ mL
$M_3=\text{molarity of}{H}_{2}S{O}_4=xM$
$\therefore 2\times 0.1\times 50=1\times 0.1\times 50+2\times x\times 50$
$\Rightarrow x=0.05$
$\therefore$The molarity of $H_{2}S{O}_4$ solution = 0.05 M