Question

Calculate the enthalpy change when $50mL$ of $0.01M$ $Ca(OH{)}_2$ reacts with $25mL$ of $0.01M$ $HCl$. Given that $\Delta H^o$ of neutralization of a strong acid and a strong base is $14kcalmo{l}^{-1}$

• A. $6.4cal$
• B. $9.4cal$
• C. $3.5cal$
• D. $1.4cal$

Answer 1

The correct option is C $3.5cal$

First we will calculate the no. of moles:
Number of moles of $HCl=\frac{Molarity\times Volume}{1000}=\frac{0.01\times 25}{1000}=25\times {10}^{-5}$

Now HCl dissociates as
$HCl\to H^{+}+C{l}^{-}$

So, $n_{H^{+}}=n_{HCl}=25\times {10}^{-5}$

Number of moles of $Ca(OH{)}_2=\frac{Molarity\times Volume}{1000}=\frac{0.01\times 50}{1000}=50\times {10}^{-5}$
$n_{O{H}^{-}}=2\times n_{Ca(OH{)}_2}=2\times 50\times {10}^{-5}=100\times {10}^{-5}$

In the process of neutralisation $25\times {10}^{-5}$ mole $H^{+}$ will be completely neutralised since it is the limiting reagent
$\therefore \Delta H$= Molar enthalpy of neutralisation $\times$ Number of moles
$=14\times 25\times {10}^{-5}kcal$=$0.0035kcal=3.5cal$

Theory:

Standard Enthalpy of Neutralization ${\Delta }_{neut}{H}^o$

Heat released when one mole of $H^{+}$ in dilute solution combines with one mole of $O{H}^{-}$ to give rise to undissociated water at $1bar,298K$.

The amount of heat released in formation of one mole of water when an acid is neutralised by a base.