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Q.

At 250 C, the ionic mobility of CH3COO–, H+ are respectively 4.1 × 10-4, 3.63 × 10-3 cm /sec. The conductivity of 0.001M CH3COOH is 5 × 10-5 S.cm-1. Dissociation constant of CH3COOH is

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a

1.8 \times {10^{ - 5}}

 

 

b

3\, \times \,{10^{ - 5}}

 

 

c

3\, \times \,{10^{ - 6}}

 

 

d

 

3\, \times \,{10^{ - 4}}

 

answer is A.

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Detailed Solution

\lambda _{eq}^o = \lambda _ + ^o + \lambda _ - ^o = 96500\left( {{u_ + } + {u_ - }} \right)

 

 

u+ = ionic mobility of cation

u- = ionic mobility of anion.

\lambda _{eq\left( {AcOH} \right)}^o = 96500 \times {10^{ - 4}}\left( {4.1 + 36.3} \right) = 390mho\,c{m^2}mo{l^{ - 1}}

 

{\lambda _m} = \kappa \times \frac{{1000}}{M}

 

{\lambda _m} = 5 \times {10^{ - 5}} \times \frac{{1000}}{{{{10}^{ - 3}}}} = 50mho\,c{m^2}\,mo{l^{ - 1}}

 

\alpha = \frac{{{\lambda _c}}}{{{\lambda ^o}}} = \frac{{50}}{{390}}

 

{K_a} = \frac{{C{\alpha ^2}}}{{1 - \alpha }}

 

=\frac{10^{-3}[\frac{50}{390}]^{2}}{1-[\frac{50}{390}]}\;=1.8\times 10^{-5}

 

 

 

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