Question

$500ml$ of $0.2MHCl,500ml$ of $0.2MC{H}_{3}COOH$ and $6g$ of $NaOH$ is present in a container.The $pH$ of solution at ${25}^{0}C$, if $p{K}_a$ of $C{H}_{3}COOH$ is $4.74$

• A. $4.47$
• B. $4.74$
• C. $5.20$
• D. $6.2$

Answer 1

The correct option is B $4.74$

Writing separate equations for the neutralisation of acids.

$HCl+NaOH\to NaCl+H_{2}O$

$C{H}_{3}COOH+NaOH\to C{H}_{3}COONa+H_{2}O$

Since, $HCl$ is a strong acid, therefore, it will neutralise readily.

Moles of $HCl=Molarity\times Volume=0.2\times 0.5=0.1mole$

To neutralise 0.1 mole of $HCl$, 0.1 mole of $NaOH$ is required.

Mass of $NaOH$ required $=moles\times \text{molar mass of}NaOH=0.1\times 40=4g$

Mass of $NaOH$ left after complete neutralisation of $HCl=6-4=2g$

Moles of $NaOH$ left $=\frac{2}{40}=0.05moles$

So, 0.05 moles of $NaOH$ react with 0.05 moles of $C{H}_{3}COOH$ to form 0.05 moles of salt .

Moles of $C{H}_{3}COOH$ initially $=0.2\times 0.5=0.1mole$

Therefore, moles of $C{H}_{3}COOH$ left at equilibrium $=0.1-0.05=0.05$

$pH=p{K}_a+log\frac{\left[salt\right]}{\left[acid\right]}=4.74+log\frac{\left[0.05\right]}{\left[0.05\right]}$

$pH=4.74$