Question

In what proportion $1M$ $NaOH$ and $0.5M$ $H_2{SO}_4$ are mixed so as to release maximum amount of energy and to form $100mL$ solution?

• A. $33$ and $67$
• B. $67$ and $33$
• C. $40$ and $60$
• D. $50$ and $50$

Answer 1

The correct option is D $50$ and $50$

$2NaOH+H_{2}S{O}_4⟶N{a}_{2}S{O}_4+2H_{2}O2xx$

$N_1{V}_1=N_2{V}_2⟶\left(I\right)$

Normality of $H_{2}S{O}_4=2\times Molarity=2\times 0.5=1.0N=N_1$

For $NaOH=Molarity=1N=N_2$

Thus for neutralisation,

$\frac{N_1}{N_2}=\frac{V_2}{V_1}⟹V_1=V_2$

$\because V_{total}=100ML$

$V_1=V_2=50ML$