What is the pH of acetic acid at equilibrium, given that acetic acid concentration is `0.1` M and it is `30%` dissociated at equilibrium ?
(log `3=0.47)`

Options

A. `2.00`
B. `1.53`
C. `3.53`
D. `3.00`

Answer

B

Explanation

The required realtin is
`CH_(3) COOH hArr CH_(3) COO ^(-) + H^(+)`
`because` Dissociation occurs `=30%` means 100 moles of
`CH_(3) COOH,` gives =30 moles of `H ^(+)-` ions.
Thus, `0.1` mole of `CH_(3) COOH` gives
`= (30xx0.1)/(100) = 0.03` moles of `H ^(+)` ions.
`(because` Concentrationg of `CH_(3) COOH=0.1M)`
Also, `becausepH =-log`[Concentration]
`pH=- log [H^(+)]`
` pH =- log [0.03]`
`pH =-log [(3xx 10^(-2)) ]`
`pH =-0.47 +2`
`pH =1.53`