For the titration of 50.0 ml of 0.020 m aqueous salicylic acid with 0.020 m koh (aq), calculate the ph after the addition of 55.0 ml of the base. for salycylic acid, pka = 2.97.'

First by getting moles of salicylic acid:

moles of salicylic acid = molarity * volume

= 0.02 * 0.05 L

= 0.001 mol

then moles of base KOH = molarity * volume

= 0.02 * 0.055L

= 0.0011

when total volume = 0.05 + 0.055 = 0.105 L

[salisalic acid] = moles / total volume

= 0.001 / 0.105

= 0.0095

[KOH] = moles / total volume

= 0.0011 / 0.105

= 0.01

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

PH = Pka + ㏒[salt/acid]

by substitution:

PH = 2.97 + ㏒[0.01 / 0.0095]

= 2.99