First slide

Applications of Kohlraush law

Question

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

Moderate

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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