Question

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 $0.20 \,M H _{2} SO _{4}$ 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

Answer 1

Answer: (B)

Amount of $H _{2} SO _{4}= M \times V$

$=0.20 \times 350=70\, M \,mol$

$\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 reagent. As such $65 \,mmol$ of $OH ^{-}$ 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 $OH ^{-}$ ions to give 1 mole of $H _{2} O$ and in the process $57.1\, kJ$ of heat is produced.

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

$=3.71 \,kJ$