Question

Two solutions initially at ${25}^{\circ }C$ were mixed in an adiabatic constant pressure Calorimeter. One contains
400 ml of 0.2 M weak monoprotic acid solution. The other contain 100 ml of 0.80 M NaOH. After
mixing temperature increased to ${26.2}^{\circ }C.$ How much heat is evolved in the neutralization of 1 mole of
acid ? Assume density of solution 1.0 g / $c{m}^3$, and specific heat of solution 4.2 J / g - K. Neglect heat capacity
of the Calorimeter.

• A. $-$41.5 kJ / mole
• B. $-$31.5 kJ / mole
• C. $-$21.5 kJ / mole
• D. $-$51.5 kJ / mole

Answer 1

The correct option is B $-$31.5 kJ / mole

$T_i={25}^{O}C$
$1\to 400ml$ of $0.2M$ weak monoprotic acid solution.
$2\to 100ml$ of $0.8M$ of $NaOH$.
$T_f={26.2}^{O}C,\Delta T=T_f-T_i={1.2}^{O}C$
Mass of solution $=$ Volume of solution$\times$density$=(400+100)\times 1gm$=500gm
now heat $=mc\Delta T=500\times 4.2\times 1.2=2.52kJ$
And mole of acid used $=400ml\times \frac{0.2M}{1000ml}=0.0800mole$
Mole of base used $=100ml\times \frac{0.800M}{1000ml}=0.0800mole$
So heat evolved in neutralisation of $1mole$ of acid
$=\frac{heat}{moles}=\frac{2.520}{0.0800}=31.5kJ/mole$
As heat is evolved answer is $-31.5kJ/mole$