Question

$150mL$ of $0.5N$ $HCl$ solution at ${25}^{o}C$ was mixed with $150mL$ of $0.5N$ $NaOH$ solution at same temperature. Calculate heat of neutralization of $HCl$ with $NaOH$, if temperature was recorded to be ${29}^{o}C$. (${\rho }_{H_{2}O}=1g/mL$)

Answer 1

Total mass of solution = $150+150=300$ g

Q = Total heat produced

= $300\times (25-25)$cal

= $300\times 0$ = 300cal

Heat of neutralization = $Q/150\times 1000\times 1/0.5=-13.68$ Kcal

Since heat is liberated, heat of neutralization should be negative. So heat of neutralization =$-13.68$ Kcal.

Hence, $-13.68$kcal is the answer.