In decinormal solution, `CH_(3)COOH` acid is ionised to the extent of `1.3%`. If `log 1.3 = 0.11,` what is the `pH` of the solution?

Options

A. `3.89`
B. `2.89`
C. `4.89`
D. Unpredictable

Answer

B

Explanation

`{:(,CH_(3)COOH+,H_(2)OhArr,CH_(3)COO^(Theta)+,H_(3)O^(o+)),("Initial",1,,0,0),("Final",C(1-alpha),,Calpha,Calpha):}`
C = Decinormal `= 0.1N`
`alpha = 1.3% = (1.3)/(100) = 1.3 xx 10^(-2)`
`[H_(3)O^(o+)] =C alpha = 0.1 xx 1.3 xx 10^(-2) = 1.3 xx 10^(-3)`
`pH =- log (1.3 xx 10^(-3)) =- log 1.3 - log 10^(-3)`
`=- 0.11 +3 = 2.89`