Question

Two solutions initially at ${25}^{\circ }C$ were mixed in an insulated bottle. First solution contains 400 mL of 0.2 $N$ weak monoprotic acid solution. The other contains 100 mL of 0.8 $N$NaOH solution. After mixing the temperature rises to ${26.17}^{\circ }C$. Assuming the density of final solution $1.0gc{m}^3$ and specific heat of final solution $4.2J{g}^{-1}{K}^{-1}$ . The heat released during neutralization of 1 equivalent of weak acid with NaOH is :

• A. 30.71 kJ
• B. 40.63kJ
• C. 45.36 kJ
• D. 85.23 kJ

Answer 1

The correct option is A 30.71 kJ

$T_i={25}^{o}C$ $T_f={26.17}^{o}C$ $\Delta T={1.17}^{o}C$

$1-$ $400ml$ of $0.2N$ monoprotic acid solution.
$2-$ $100ml$ of $0.8N$ $NaOH$ solution.

For valence factor $1$, molarity$=$normality.

Density$=1.0gm/ml$

Specific heat$=C=4.2J/g-K$

Mass of solution$=$Volume$\times$density$=(400\times 100)+1=500gm$

Heat$=mC\Delta t=500\times 4.2\times 1.17=2.457kJ$

Now moles of acid $=400ml\times \frac{0.2M}{1000ml}=0.08moles$

Moles of base$=100ml\times \frac{0.8M}{1000ml}=0.08moles$.

Heat evolved in neutralisation of $1$ mole of acid $=$ (heat) $/$ moles
$=\frac{2.457}{0.08}=30.71kJ/mol$