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A weak acid HA has \mathrm{Ka = 1.4\times10^{-5}}. Calculate its per cent dissociation in a solution which contains 0.1 mole of HA per 2 litres of solution

Option: 1

2.8%


Option: 2

1.67%


Option: 3

28%


Option: 4

Data insufficient


Answers (1)

best_answer

Given,

\mathrm{[HA] = \frac{0.1}{2}= 0.05\ M}

For weak acid HA

        \mathrm{HA\ \ \rightleftharpoons \ \ H ^{+}+\ \ A ^{-}}

            \mathrm{c}                 \mathrm{0}              \mathrm{0}

   \mathrm{c(1-\alpha)}        \mathrm{c \alpha}            \mathrm{c \alpha}

\mathrm{K_{a}=\frac{c \alpha \cdot c \alpha}{c(1-\alpha)}=\frac{c \alpha^{2}}{(1-\alpha)}}

As dissociation constant is very less than 1, so 1-\alpha \approx 1

\mathrm{\therefore K_{a} = c\alpha ^{2}}

\Rightarrow 1.4\times 10^{-5}=0.05\times \alpha ^{2}

\Rightarrow \alpha =1.67\times 10^{-2}

The degree of dissociation in % will be 1.67.

Hence, the correct answer is Option (2)

Posted by

Ritika Kankaria

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