Question

If the heat of neutralisation for a strong acid-base reaction is -57.1 kJ. What would be the heat released when 350 $c{m}^3$of 0.20 M $H_{2}S{O}_4$ is mixed with 650 cm${}^3$ of 0.10 M $NaOH$.

• A. $8.5kJ$
• B. $17.1kJ$
• C. $5.7kJ$
• D. $3.71kJ$
• E. $2.85kJ$

Answer 1

Answer: (D)

$3.71kJ$

Moles of $H_{2}S{O}_4=M\times V=0.20\times 350=70mmol$

$\therefore$ Moles of $H^{+}$ ions $=2\times 70=140mmol$

Moles of $NaOH=M\times V=0.10\times 650=65mmol$

$\therefore$ Moles of $O{H}^{-}$ ions $=65mmol$

Thus, $NaOH$ is the limiting reactant. As such 65 mmol of $O{H}^{-}$ ions will react with 65 mmol of $H^{+}$ ions to give 65 mmol of water.

1 mol of $H^{+}$ ions react with 1 mol of $O{H}^{-}$ ions to give 1 mol of $H_{2}O$ and in the process 57.1 kJ of heat is produced.

Thus heat produced $=\frac{-65\times 57.1}{1000}=3.71kJ$