0.01 moles of a weak acid HA ( K _ a =2.0 10^ -6 ) is dissolved in 1.0 , L of 0.1 , M , HCl solution. The degree of dissociation of HA is ............. 10^ -5 (Round off to the Nearest Integer). [Neglect volume change on adding HA. Assume degree of dissociation << 1]

0.01 moles of a weak acid HA ( K _ a =2.0 10^ -6 ) is dissolved i…

MCQ

$0.01$ 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 $<< 1]$

A
$6$
B
$3$
✓
$2$
D
$7$

✓

Answer

Correct option: C.

$2$

c
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\quad\quad\quad HA \rightleftharpoons \quad H ^{+}+\quad A ^{-}$

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, $\alpha=\frac{x}{0.01}=\frac{2 \times 10^{-7}}{0.01}=2 \times 10^{-5}$

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