Given the following molar conductivities are25^(@) 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 ? Ifthe conductivity of a 0.001 M corotonic acid solution is3.83 xx 10^(-5) Omega^(-1) cm^(-1) ?

Given the following molar conductivities are25^(@) C , HCl, 426 Omega^

1.	Given the following molar conductivities are25^(@) 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 ? Ifthe conductivity of a 0.001 M corotonic acid solution is3.83 xx 10^(-5) Omega^(-1) cm^(-1) ?
Answer» `10^(-5)` `1.11 xx 10^(-5)` `1.11 xx10^(-4)` `0.01` Solution :`^^""^(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)`