Question

A solution of 200 ml of 1M KOH is added to 200 ml of 1M HCl and the mixture is shaken well. The rise in temperature $T_1$ is noted. The experiment is repeated by using 100 ml each solution and increase in temperature $T_2$ is again noted. Which of the following is correct ?

• A. $T_1=T_2$
• B. $T_2=2T_1$
• C. $T_1=2T_2$
• D. $T_1=4T_2$

Answer 1

The correct option is A $T_1=T_2$

Case I : 200 ml 1M KOH + 200 ml 1M HCl
Case II: 100 ml 1M KOH + 100 ml 1 M HCl
Equivalent to $H_{2}O$ produced due to neutralisation in case II is half that of case I.
Due to this energy release, $H_2=\frac{H_1}{2}$
Case I $H_1=m_{1}S{T}_1$ . (1)
Case II $H_2=m_{2}S{T}_2\left(or\right)\frac{H_1}{2}=\frac{m_1}{2}S{T}_2$ . (2)
From (1) and (2), $T_1=T_2$