A solution of HCl contains 0.1920 g of an acid in 0.5 litre of a solution. The degree of dissociation is 95%. The pH of the solution is

Concentration C of HCl solution

$=\frac{0.1920}{36.5 \times 0.5} = 0.010 \mathrm{M}$

The degree of dissociation is 95%, i.e.,$\frac{95}{100}$ = 0.95

Therefore, $\left[{\mathrm{H}}^{+}\right]=0.010 \times 0.95 = 0.0095$

$$\begin{gathered} \mathrm{pH}=-\log \left[{\mathrm{H}}^{+}\right] = -\log \left(0.0095\right) \\ = -\left(-2.02\right) = 2.02\approx 2.0 \end{gathered}$$