Calculate the change in `pH` if `0.02 mol CH_(3)COONa` is added to `1.0L` of this solution. ltbRgt `K_(a)` of `CH_(3)COOH = 1.8 xx 10^(-5)`.

Explanation

`CH_(3)COONa` contains `CH_(3)COO^(Θ)` ions which react with `H_(3)O^(o+)` ion from `HCI`. Assume complete reactions of `H_(3)O^(o+)` with `CH_(3)COO^(Θ)` ion.
`{:(,CH_(3)COO^(Theta)+,H_(3)O^(oplus),rarr,CH_(3)COOH +,H_(2)O),("Initial concentration",rArr 0.02 M,0.01 M,,-,-),("Concentration reacted",rArr 0.02 -0.01,0.01,,-,-),("Final concentration",rArr 0.01("excess"),0,,0.01,0.01):}`
`CH_(3)COOH +H_(2)O hArr CH_(3)COO^(Theta) +H_(3)O^(oplus)`
`K_(a) = ([CH_(3)COOH^(Θ)][H_(3)O^(o+)])/([CH_(3)COOH] ) = ((0.01)[H_(3)O^(o+)])/((0.01)) = 1.8 xx 10^(-5)`
`:. [H_(3)O^(o+)] = 1.8 xx 10^(-5), so pH = 4.74`
Change in `pH (DeltapH) = 4.74 - 2.00 = 2.74`