The pH of a 0.1 M `CH_(3)COOH` which is `2%` ionised in aqueous solution is

Options

A. `2.0`
B. `1.0`
C. `log((0.1)/(0.001))`
D. 2.7

Answer

D

Explanation

`{:(,CH_(3)COOH,hArr,CH_(3)COO^(-),+,H^(+)),("Eqm. conc.",0.1(1-0.02),,0.1 xx 0.02 ,,0.1 xx 0.02):}`
`pH=-log[H^(+)]=-log (2 xx 10^(-3))=2.7`