# If the heat of neutralisation for a strong acid-base reaction is -57.1 kJ what would be the heat released when 350 $cm_3$ of 0.20 M $H_2SO_4$ is mixed with 650 $cm_3$ of 0.10 M NaOH.

(a) 37.1 kJ
(b) 3.71 kJ
(c) 3.17 kJ
(d) 0.317 kJ

Amount of $H^+$ ions = M $\times$ V = 0.20 $\times$ 350 =70 m mol
$\therefore$ Amount of $H^+$ ions = 2 $\times$ 70 = 140 m mol
Amount of NaOH = M $\times$ V = 0.10 $\times$ 650 = 65 m mol
$\therefore$ Amount of $OH^-$ ions = 65 m mol
Thus, NaOH is the limiting reactant. As such 65 m mol of $OH^-$ ions will react with 65 m mol of $H^+$ ions to give 65 m mol of water.
1 mol of $H^+$ ions react with 1 mol of $OH^-$ ions to give 1 mol of $H_2O$ and in the process 57.1 kJ of heat is produced.
Thus heat produced = $57.1 \times 65 \times 10^{-1}$ = 3.71 kJ