moles of acid HA = Molarity X volume
moles of acid=0.1 X 75 mmoles = 7.5 mmoles
Moles of base added=Molarity X volume
Moles of base added =0.1 X 30= 3 mmoles
tb
he strong base added will react with weak acid to form a salt.
ANa
the moles of salt formed is = moles of base added
= 3 mmoles
moles of acid left = 7.5-3= 4.5
this will result in formation of a buffer
the pH of buffer is given by
pH = pKa + log(salt/ acid)
given pH is 5.5
therefore
5.5= pKa + log(3/4.5)
5 .5= pKa + (-0.176)
pKa = 5.68