50 mL of 0 . 02 M Na OH solution is mixed 50 mL of 0 . 6 M acetic acid solution, the pH of resulting solution is
(`pK_(a)` pf acetic acid is 4 . 76, log 5 = 0 . 70 )

Options

A. 5 . 06
B. 4 . 06
C. 5 . 46
D. 4 . 46

Answer

D

Explanation

Given,
Volume of NAOH solution = 50 mL
Molarity of NaOH solution = 0 . 02 M
Meq. Of Na OH ` = 50 xx 0 . 2 = 1 meq`
Volume of `CH_(3) COOH ` solution = 50 mL
Molarity of `CH_(3) COOH = 0 . 0 6 M`
Meq. Of `CH_(3) COOH = 50 xx 0 . 0 6 ` = 3 meq .
1 meq , of NaOH compbines with 3 meq. of
`CH_(3) COOH ` and forms 1 meq. of `CH_(3) COONa` .
`:.` 2 meq . of `CH_(3) COOH` is left . So it forms an acidic buffer .
Now, for an acidic buffer
`pH = ph_(a) + log [("Salt")/("Acid")]`
`pH = pH_(a) + log ([CH_(3) COOH Na)]/([CH_(3) COOH])`
`:. pH = 4 76 + log [(1)/(2)]`
`pH = 4.76 + log (5 xx 10^(-1))`
pH = 4.76 + log 5 - log 10
= 4 . 76 + 0 . 70 - 1
= 4 . 46