Question

A solution of 500 mL of 0.2 M KOH and 500 mL of 0.2 M HCl is mixed and stirred ; the rise in temperature is $T_1$. The experiment is repeated using 250 mL each of solution, the temperature rise is $T_2$. Which of the following is true ?

• A. $T_1=T_2$
• B. $T_1=2T_2$
• C. $T_1=4T_2$
• D. $T_2=9T_1$

Answer 1

The correct option is A $T_1=T_2$

The chemical equation is

$KOH+HCl\to KCl+H_{2}O$

For 1st solution

Molarity of KOH = 0.2 M

i.e. in 1 L (1000 mL) no. of moles of KOH = 0.2

Therefore, in 500 mL moles of KOH = 0.1

Molarity of HCl = 0.2 M

Moles of HCl = 0.1

Heat of neutralisation $Q=mC\Delta t$

In these reactions the solutions are treated as that of pure water therefore volume of the solution in mL is equal to the weight in grams (density of water = 1 gm/cc)

Total volume of the solution = 1000 ml

Therefore weight of the solution = 1000 gm

Thus,

$Q=1000C{T}_1$

or $T_1=\frac{Q}{1000C}$ .............. (1)

Similarly, For 2nd solution

No. of moles of KOH = 0.05

No. of moles of HCl = 0.05

SInce the moles of KOH and HCl in this solution are half to that of the 1st solution

Therefore, the heat of neutralisation will be half

Heat of neutralisation = Q/2

The mass now is 250 + 250 = 500 g

Thus

$Q/2=500C{T}_2$

or $T_2=\frac{Q}{1000C}$ ...................(2)

From (1) and (2)

$T_1=T_2$