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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}

Answers (1)


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}}

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