<Dd> K a 1 ⋅ K a 2 = c (H 3 O +) 2 (\ displaystyle K_ (a_ (1)) \ cdot K_ (a_ (2)) = c ((\ ce (H3O+))) ^ (2)) </Dd> <P> Taking the square root and logarithm on both sides yields: </P> <Dl> <Dd> pH = p K a 1 + p K a 2 2 (\ displaystyle (\ ce (pH)) = (\ frac ((\ ce (p)) K_ (a_ (1)) + (\ ce (p)) K_ (a_ (2))) (2))) </Dd> </Dl> <Dd> pH = p K a 1 + p K a 2 2 (\ displaystyle (\ ce (pH)) = (\ frac ((\ ce (p)) K_ (a_ (1)) + (\ ce (p)) K_ (a_ (2))) (2))) </Dd>

The ph at which the amino acid is in the dipolar form is known as