Calculate `K_(a)` of acetic acid it its `0.05N` solution has equivalent conductances of `7.36 mho cm^(2)` at `25^(@)C (lambda_(CH_(3)COOH)^(oo) = 390.70)`

Calculate `K_(a)` of acetic acid it its `0.05N` solution has equivalen

1.	Calculate `K_(a)` of acetic acid it its `0.05N` solution has equivalent conductances of `7.36 mho cm^(2)` at `25^(@)C (lambda_(CH_(3)COOH)^(oo) = 390.70)`
Answer» Correct Answer - `1.76 xx 10^(-5)` mole/litre Degree of dissociation `(x) =(Lamda_(c))/(Lamda_(0))=(7.36)/(390.7) = 0.0188` For equilibrium `{(0.05,0,0,"Initial conc (moles/litre)"):}` `CH_(3)COOH=CH_(3)COO^(-)+H^(+)` `0.05(1-x) 0.05 x 0.05 x` Equilibrium concentration (for `CH_(3)COOH,0.05 N = 0.05 M`) `K_(a)=(0.05 x xx 0.05 x)/(0.05(1-x))` since `x` is very small `K_(a)=0.05 x^(2)= 0.05 xx(0.0188)^(2)=1.76 xx10^(-5) "mole/L"`.

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Calculate `K_(a)` of acetic acid it its `0.05N` solution has equivalent conductances of `7.36 mho cm^(2)` at `25^(@)C (lambda_(CH_(3)COOH)^(oo) = 390.70)`

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