Question

`20 mol` of `M//10 CH_(3)COOH` solution is titrated with `M//10 NaOH` solution. After addition of `16 mL` solution of `NaOH`. What is the `pH` of the solution `(pK_(a) = 4.74)`
A. `5.05`
B. `4.15`
C. `4.75`
D. `5.35`

Answer 1

Correct Answer - D
`mEq of CH_(3)COOH = 20 xx (1)/(10) = 2`
mEq of `NaOH = 16 xx (1)/(10) = 1.6`
So `1.6 mEq` of `NaOH` reacts with `1.6 mEq` of `CH_(3)COOH` to form `1.6 mEq` of salts `CH_(3)COONA` and `(2-1.6 = 0.40 mEq)` of `W_(A) (CH_(3)COOH)` is left and an acidic buffer is formed. So to calculate `pH`, buffer equation is used.
`pH = pK_(a) + log (("Salt")/("Acid")) = 4.74 + log ((1.6)/(0.4))`
`= 4.74 +2 log 2`
`= 4.74 +2 xx 0.3010 = 5.35`