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$0.01$ moles of a weak acid $HA \left( K _{ a }=2.0 \times 10^{-6}\right)$ is dissolved in $1.0\, L$ of $0.1\, M\, HCl$ solution. The degree of dissociation of $HA$ is ............. $\times 10^{-5}$
(Round off to the Nearest Integer).
[Neglect volume change on adding $HA$. Assume degree of dissociation $<< 1]$
$6$
$3$
$2$
$7$
Solution
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\quad\quad\quad HA \rightleftharpoons \quad H ^{+}+\quad A ^{-}$
Initial conc. $\quad0.01 M\quad \, 0.1 M \quad\quad 0$
Equ. conc. $\,(0.01- x )\quad(0.1+ x )\quad xM$
Now, $K_{a}=\frac{\left[x^{+}\right]\left[A^{-}\right]}{[H A]} \Rightarrow 2 \times 10^{-6}=\frac{0.1 \times x}{0.01}$
$\therefore \quad x=2 \times 10^{-7}$
Now, $\alpha=\frac{x}{0.01}=\frac{2 \times 10^{-7}}{0.01}=2 \times 10^{-5}$
