Question

Calculate the enthalpy change when $50mL$ of $0.01M$ $Ca{\left(OH\right)}_2$ reacts with $25mL$ of $0.01M$ $HCl$. Given that $\Delta H^o$ neutralisation of a strong acid and a strong base is $140kcal$ ${mol}^{-1}$.

• A. $14kcal$
• B. $35kcal$
• C. $10kcal$
• D. $7.5kcal$

Answer 1

Answer: (B)

$35kcal$

Number of moles of $HCl=\frac{MV}{1000}=\frac{0.01\times 25}{1000}=25\times {10}^{-5}$
$HCl⟶H^{+}+{Cl}^{-}$
$n_{H^{+}}=25\times {10}^{-5}$
Number of moles of $Ca{\left(OH\right)}_2=\frac{MV}{1000}=\frac{0.01\times 50}{1000}=50\times {10}^{-5}$
$n_{{OH}^{-}}=2\times 50\times {10}^{-5}={10}^{-3}$
In the process of neutralisation $25\times {10}^{-5}$ will be completely neutralised.
$\therefore$ $\Delta H=140\times 25\times {10}^{-5}kcal=35kcal$