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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} (Rounded off to the Nearest Integer). [Neglect volume change on adding HA. Assume degree of dissociation <<1]
Option: 1 2
Option: 2 -
Option: 3 -
Option: 4 -

Answers (1)

best_answer

0.01 moles of a weak acid HA (Ka = 2.0 x 10-6 ) is dissolved in 1.0 L of 0.1 M HCl solution.

The reaction will be -

Because 0.01 >> x , for weak acid.

Given [Neglect volume change on adding HA. Assume degree of dissociation <<1]

Now,

\mathrm{K}_{\mathrm{a}}=\frac{\left[\mathrm{X}^{+}\right]\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}

After putting the value

\Rightarrow 2 \times 10^{-6}=\frac{0.1 \times \mathrm{x}}{0.01}

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

Now,

\alpha=\frac{\text { Part dissociated }}{\text { Initial amount }}

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

Ans = 2

Posted by

Kuldeep Maurya

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