[NaH2M] = molarity x volume in Litres = 0.2 M X 0.03
L = 0.006 mol
[NaOH] = molarity x volume in Litres = 0.1 M X 0.03 L = 0.003
mol
NaH2M
         + NaOH
---------------> Na2HM +
H2OÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
pKa2 = 6
0.006
mol          Â
0.003
mol                        Â
0
----------------------------------------------------------------------------
0.006 -
0.003Â Â Â Â Â Â Â Â Â Â
0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
0.003 mol
= 0.003 mol
Hence,
[NaH2M ] = 0.003 mol
[Na2HM] = 0.003 mol
pKa = pKa2 = 6
From Henderson-Hasselbalch equation,
pH = pKa + log [ conjugate base]/ [acid]
    = pKa + log [Na2HM] /
[NaH2M]
     = 6 + log ( 0.003/0.003)
     = 6 + 0
     = 6
Therefore,
pH = 6
