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  4. 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]

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Q. $0.01$ 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 $<<1]$

JEE MainJEE Main 2021Equilibrium

Solution:

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Now, $K _{ a }=\frac{\left[ x ^{+}\right]\left[ A ^{-}\right]}{[ HA ]} \Rightarrow 2 \times 10^{-6}=\frac{0.1 \times x }{0.01}$

$\therefore x =2 \times 10^{-7}$

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