Ionisation constant of a weak acid (H_A) in terms of _m^ and _m i…

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MCQ

Ionisation constant of a weak acid $(H_A)$ in terms of $\Lambda _m^\infty $ and ${\Lambda _m}$ is

A
${K_a} = \frac{{C\Lambda _m^\infty }}{{\left( {{\Lambda _m} - \Lambda _m^\infty } \right)}}$
✓
${K_a} = \frac{{C\Lambda _m^2}}{{\Lambda _m^\infty \,\left( {\Lambda _m^\infty - {\Lambda _m}} \right)}}$
C
${K_a} = \frac{{C{{\left( {\Lambda _m^\infty } \right)}^2}}}{{\Lambda _m^\infty \,\left( {{\Lambda _m} - \Lambda _m^\infty } \right)}}$
D
None of these

✓

Answer

Correct option: B.

${K_a} = \frac{{C\Lambda _m^2}}{{\Lambda _m^\infty \,\left( {\Lambda _m^\infty - {\Lambda _m}} \right)}}$

b
$\quad\quad\quad\quad\quad HA ( aq ) \rightleftharpoons H ^{+}( aq )+ A ^{-}( aq )$

At eqm. $\quad c(1-\alpha) \quad\quad\quad c \alpha \quad\quad\quad c \alpha$

$K _{ a }=\frac{c \alpha^2}{1-\alpha} \text {; where } \alpha=\frac{\Lambda_{ m }}{\Lambda_{ m }^{\infty}}$

$\therefore \quad K_{ a }=\frac{c\left(\frac{\Lambda_{ m }}{\Lambda_{ m }^{\infty}}\right)^2}{\left(1-\frac{\Lambda_{ m }}{\Lambda_{ m }^{\infty}}\right)}$

$=\frac{ c \Lambda_{ m }^2}{\Lambda_{ m }^{\infty}\left(\Lambda_{ m }^{\infty}-\Lambda_{ m }\right)}$