You are performing a titration of a triprotic acid, when you spill water on your lab notebook. you can read that: pka 1 = 1.40, pka 3 = 9.80. you have determined experimentally that the ph at the first equivalence point is 3.35, and the ph at the second equivalence point is 7.55. what is pka 2 for this acid?

According to the PH formula:
PH= Pka +㏒ [strong base/weak acid]
when we have PH at the first equivalence =3.35 and the Pka1 = 1.4
So, by substitution, we can get the value of ㏒[strong base / weak acid]
3.35 = 1.4 + ㏒[strong base/ weak acid]
∴㏒[strong base/weak acid] = 3.35-1.4 = 1.95 
to get the Pka2 we will substitute with the value of ㏒[strong base/ weak acid] and the value of PH of the second equivalence point
∴Pk2 = PH2 - ㏒[strong base/ weak acid]
          = 7.55 - 1.95 = 5.6 

Neither model is able to show the reactivity of potassium.