The pH of a solution containing 0.1 mol of `CH_(3)COOH, 0.05` mol of `NaOH, and 0.2 mol ofCH_(3)COONa,` in 1.L. `(pK_(a) of CH_(3)COOH=4.74)` is :

Options

A. `5.44`
B. `5.20`
C. 5.04
D. 4.74

Answer

A

Explanation

`{:(,CH_(3)COOH+,NaOH,rarr,CH_(3)COONa+,H_(2)O),("Initial mol",0.1,0.05,,0,0),("Final mol",(0.1-0.05),(0.05-0.05),,0.05,-),(,=0.05,=0,,,):}`
Total moles of `CH_(3)COONa = 0.2 + 0.05 = 0.25 therefore [CH_(3)COONa] = (0.25 mol)/(1L) = 0.25 M`
Moles of `CH_(3)COOH` left `= (0.1 - 0.05) = 0.05 therefore [CH_(3)COOH] = ("0.05 mol")/(1L) = 0.05 M`
Thus, acidic buffer is formed
`pH = pK_(a) + log""(["Salt"])/(["Acid"]) = 4.74 + log((0.25)/(0.05)) = 4.74 + log 5 = 4.74 + 0.7 = 5.44`