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Given the following molar conductivities...

Given the following molar conductivities are ` 25^(@) C `, HCl , `426 Omega^(-1) cm^(2) mol^(-1)`, NaCl , `126 Omega^(-1)cm^(2) mol^(-1)`, NaC( sodium crotonate ) , `83 Omega^(-1) cm^(2) mol^(-1)` , what is the ionization constant of crotonic acid ? If the conductivity of a 0.001 M corotonic acid solution is ` 3.83 xx 10^(-5) Omega^(-1) cm^(-1) ?`

A

`10^(-5)`

B

`1.11 xx 10^(-5)`

C

`1.11 xx10^(-4)`

D

`0.01`

Text Solution

Verified by Experts

The correct Answer is:
B
`^^""^(oo) = ^^_(NaC)^(oo) + ^^_(HCl)^(oo) - ^^_(NaCl)^(oo) = 83 + 426 - 126 = 383`o
` ^^_(M) = 3.83 xx 10^(-5) xx (1000)/(0.001) = 38.3 `
` alpha = (^^)/(^^""^(oo)) =(38.3)/(383) = 0.1 = 10^(-1) , K_a = (C alpha^(2))/(1 - alpha ) = (10^(-3) xx (10^(-1))^(2) )/(1-0.1) = (10^(-3) xx (10^(-1))^(2))/(0.9) = 1.11 xx 10^(-5)`
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