The ionization constant of propanoic acid is `1.32 xx 10^(-5)`. Calculate the degree of ionization of the acid in its 0.05M solution and also its pH. What will be its degree of ionization if the solution is 0.01M in HCl also?

Explanation

` {:(, CH_3CH_2COOH ,hArr, CH_3CH_2COO^(-) , +, H^(+) ),(" Initial conc." , 0.05 ,, - ,,- ),("Conc. at eq", 0.5[1-alpha],, 0.05 alpha , ,0.5alpha):}`
` therefore " " K_a= ([CH_3CH_2COO^(-) ][H^(+)])/([CH_3CH_2COOH]) `
` or " " , 32 xx 10 ^(-5) = (0.05 alpha xx 0.05alpha)/(0.05 (1-alpha) ) ~~ ((0.05alpha)^(2))/(0.05) =( 0.05alpha^(2))/(1)`
` or " "alpha = ((1.32xx10^(-5))/(0.05) )^(1//2) = 1.63 xx 10 ^(_2)`
` therefore " "pH = - log_(10) [H^(+)] =- log_(10) (0.05 xx alpha ) `
` " " = - log_(10) [0.05 xx 1.63 xx 10^(-2) ]=3.09`
In the presence of 0.01 M HCl , suppose the degree of ionisation of the acid is ` alpha ` Therefore
` [CH_3CH_2COOH]=0.05 (1-alpha .)~~ 0.05`
` [CH_3CH_2COO^(-)]= 0.05 xx alpha . and [H^(+)] = 0.05 alpha. +0.01`
` " " K_ a= ([CH_3CH_2 COO^(-) ][H^(+) ])/( [CH_3CH_2COOH]) `
` or " " 1.32 xx 10 ^(-5) = (0.05alpha xx(0.05alpha .+0.01))/(0.05)`
` or " " 0.05alpha ^(2) + 0.01 alpha.- 1.32 xx 10^(-5) =0 `
` alpha . = (- 0.01 +- sqrt( (0.01)^(2) + 4 xx 0.05 xx 1.32 xx 10^(-5)))/(2xx 0.05)`
` = 1.31 xx 10^(-3)`