A500.0 ml buffer solution contains 0.30 m acetic acid (ch3cooh) and 0.20 m sodium acetate(nach3coo). what will the ph of this solution be after the addition of 20.0 ml of 1.00 m naohsolution? [ka(ch3cooh) = 1.8 × 10–5]

First, we need to get no.of moles of acetic acid = molarity * volume

                                                  = 0.3 m * 0.5 L

                                                  = 0.15 moles
then, we need to get no. of moles acetate = molarity *volume 

                                                                       = 0.2m * 0.5L

                                                                       = 0.1 moles

and we have also to get moles of NaOH = molarity * volume 

                                                                   = 1 m* 0.02L

                                                                   = 0.02 moles

according to the reaction equation and by using ICE table:

CH3COOH  + OH- → CH3COO- + H2O

∴ moles acetic acid = 0.15 - 0.02 = 0.13 moles 

when [acetic acid] = moles / total volume

                               = 0.13 moles / (0.5L + 0.02L)

                               = 0.25 M

and moles of acetate = 0.1 + 0.02 = 0.12 moles 

∴[acetate] = moles / total volume

                 = 0.12 moles / 0.52L

                = 0.23 M

and when we have the value of Ka so we can get the Pka from this formula:

Pka = -㏒ Ka 

       = -㏒ 1.8 x 10^-5

       = 4.7 

by using H-H equation we can get the PH:

∵PH = Pka + ㏒[acetate] /[acetic acid]

∴PH = 4.7 + ㏒(0.23 /0.25]

        = 4.66

Hcn is your answer my guy : )