Question

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

• A. 37.1 KJ
• B. 3.71 KJ
• C. 3.17 KJ
• D. 0.317 KJ

Answer 1

Answer: (B)

Amount of $H_{2}S{O}_4=M\times V$

$=0.20\times 350=70Mmol$

$\therefore$ Amount of $H^{+}$ ions $=2\times 70=140Mmol$

Amount of $NaOH=M\times V$

$=0.10\times 650=65Mmol$

$\therefore$ Amount of $O{H}^{-}$ ions $=65Mmol$

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

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

Thus; heat produced $=57.1\times 65\times 10^{-3}$

$=3.71kJ$