Chemical Equilibrium

Let's work through the problem step by step.The dissociation of the weak acid is given by:$\mathrm{HA} \rightleftharpoons \mathrm{H}^{+}+\mathrm{A}^{-}.$The initial concentration of HA is 0.5 M . With a degree of dissociation, $\alpha=0.01$ (or 1\%), the equilibrium concentrations are:$\; [\mathrm{HA}]=0.5(1-0.01)=0.5 \times 0.99=0.495\:\mathrm{M}$$\; [\mathrm{H}^{+}] =0.5 \times 0.01=0.005\:\mathrm{M}$$\; [\mathrm{A}^{-}] =0.005\:\mathrm{M}$The acid dissociation constant, $K_a$, is defined as:$K_a=\frac{[\mathrm{H}^{+}][\mathrm{A}^{-}]}{[\mathrm{HA}]}.$Substitute the equilibrium concentrations:$K_a=\frac{(0.005)(0.005)}{0.495}=\frac{0.000025}{0.495}.$Simplifying the expression:$K_a \approx 5 \times 10^{-5}.$Thus, the approximate value of $K_a$ (or $K_e$, as referenced) is $5 \times 10^{-5}$.The correct option is Option B.