A 0.1 M solution of HCl has a degree of dissociation of 0.01. What is the dissociation constant for the acid?Option: 1 $1.0 \times 10^{-4}$Option: 2 $1.0 \times 10^{-5}$Option: 3 $1.0 \times 10^{-5}$Option: 4 $1.0 \times 10^{-7}$

The degree of dissociation $\alpha$ for a weak electrolyte is given by the ratio of the concentration of the dissociated ions to the initial concentration of the electrolyte:
$\mathrm{\alpha=\frac{\left[X^{-}\right]}{[H X]}}$
Where $X^{-}$ is the anion, $\mathrm{HX}$ is the undissociated acid, and "[ ]" denotes the concentration in $\mathrm{mol} / \mathrm{L}.$
For $\mathrm{HCl}$, since it is a strong electrolyte, it undergoes complete dissociation:

$\mathrm{HX} \rightarrow \mathrm{H}^{+}+\mathrm{Cl}^{-}$
So the degree of dissociation can be calculated as:
$\alpha=\frac{[\mathrm{Cl}^{-}]}{[\mathrm{HCl}]} =0.01$
Since the initial concentration of $\mathrm{HCl} \text{ is }0.1 \mathrm{M}$, the concentration of $\mathrm{Cl}^{-}$ is also $0.1 \times 0.01=0.001 \mathrm{M}$
The dissociation constant $\left(K_a\right)$ for $\mathrm{HCl}$ is given by:
$\mathrm{K_a=\frac{[H^+][\mathrm{Cl}^{-}]} {[\mathrm{HCl}]}}$
At equilibrium, the concentration of $\mathrm{H^+}$ is equal to the degree of dissociation times the initial concentration of $\mathrm{HCl}$ :
$\mathrm{[H^+]=\alpha[H \mathrm{Cl}]=0.01 \times 0.1=0.001 \mathrm{M}}$
Substituting the values into the equation for $\left(K_a\right)$, we get:
$\mathrm{K_a=\frac{(0.001)^2}{0.099}=1.0 \times 10^{-5}}$