Find out ionisation constant of a weak acid (HA) in terms of _m^o and _m^c ? (Given '' '' can not be ignored w.r.t. 1 )

B. K_a = , C , ( _m^c) ^2 _m^o ( _m^o , - , _m^c )

Find out ionisation constant of a weak acid (HA) in terms of _m^o…

MCQ

Find out ionisation constant of a weak acid $(HA)$ in terms of $\Lambda _m^o$ and $\Lambda _m^c$ ? (Given $''\alpha ''$ can not be ignored w.r.t. $1$ )

A
${K_a} = \,\frac{{C\,\Lambda _m^0}}{{\left( {\Lambda _m^c\, - \,\Lambda _m^o} \right)}}$
✓
${K_a} = \,\frac{{C\,{{(\Lambda _m^c)}^2}}}{{\Lambda _m^o\left( {\Lambda _m^o\, - \,\Lambda _m^c} \right)}}$
C
${K_a} = \,\frac{{C\,{{(\Lambda _m^o)}^2}}}{{\Lambda _m^o\left( {\Lambda _m^o\, - \,\Lambda _m^c} \right)}}$
D
None of these

✓

Answer

Correct option: B.

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

b
$\therefore {{\text{K}}_{\text{a}}}=\frac{\text{C}{{\alpha }^{2}}}{(1-\alpha )}$ ; $\alpha =\frac{\wedge _{\text{m}}^{\text{c}}}{\wedge _{\text{m}}^{o}}$

${{\text{K}}_{\text{a}}}=\frac{\text{C}\times {{\left( \frac{\wedge _{\text{m}}^{\text{c}}}{\wedge _{\text{m}}^{{}^\circ }} \right)}^{2}}}{\left[ 1-\frac{\wedge _{\text{m}}^{\text{c}}}{\wedge _{\text{m}}^{\text{e}}} \right]}$ $=\frac{\text{C}\times {{\left( \wedge _{\text{m}}^{\text{c}} \right)}^{2}}}{\wedge _{\text{m}}^{o}\left( \wedge _{\text{m}}^{o}-\Lambda _{\text{m}}^{\text{c}} \right)}$