The ionization constant of acetic acid is `1.74 xx 10^(-5)`. Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of acetate ion in the solution and its pH.

Explanation

` {:( ,CH_3COOH ,hArr, CH_3COO^(-) , +, H^(+) ) , (" Initial conc." , 0.05 M ,, -,,- ),(" Conc. at eq. ", 0.05-x ,, x,,x):}`
According to the law of equilibrium
` " " K= ([CH_3COO^(-) ][H^(+)])/([CH_3COOH]) `
or ` " " 1.74 xx 10 ^(-5) = (x.x)/(0.05-x)=(x^(2))/(0.05)`
` " " ( because ` x is very small)
` or " " x= (1.74 xx 10 ^(-5) xx 0.05 )^(1//2) = 9.33 xx 10^(-4)`
` therefore ` Degree of dissociation = ` (x)/(" Total no.of moles ") `
` " " [CH_3COO^(-) ]=x = 9.33 xx 10 ^(-4) " mol " L^(-1) `
` " " [H^(+)] =x =9.33 xx 10 ^(-4) " mol " L^(-1)`
` therefore " " pH =- log _(10 ) [H^(+) ] = - log_( 10 ) (9.33 xx 10 ^(-4)) =3.03 `