pH = pKa + log (base)/(acid)
base is 0.050M Na2HPO4
acid is 0.10 M KH2PO4
pKa = 7.199
pH = 7.199 + log (0.05)/(0.1)
pH = 6.89
100.0 mL x 0.050 = 5 mmols base
Na2HPO4
100.0 mL x 0.10 = 10 mmols acid KH2PO4
added 5 mL x 0.20 M HCl = 1.0 mmols
. HPO4-
  + H+ ==>
H2PO4-
                            Â
Initial                                 Â
5Â Â Â Â Â Â Â Â Â Â Â Â
0Â Â Â Â Â Â Â Â Â Â Â Â
10
                            Â
Add      Â
                                           Â
1.0
                            Â
Change
-1Â Â Â Â Â Â Â Â Â Â Â
-1Â Â Â Â Â Â Â Â Â Â Â
+1
             Â
             Â
Equilibrium                      Â
4Â Â Â Â Â Â Â Â Â Â Â Â
0Â Â Â Â Â Â Â Â Â Â Â Â
11
pH = pKa2 + log (base)/(acid)
base is 4 mmols Na2HPO4
acid is 11 mmol KH2PO4
pH = 7.199 + log 4/11
pH = 6.76