<Dd> p H = p K a + log ⁡ ((base) (acid)) (\ displaystyle \ mathrm (pH) = \ mathrm (p) K_ (\ mathrm (a)) + \ log \ left ((\ frac (((\ mbox (base)))) (((\ mbox (acid))))) \ right)) </Dd> <Dd> p H = p K a + log ⁡ (1) (\ displaystyle \ mathrm (pH) = \ mathrm (p) K_ (\ mathrm (a)) + \ log (1) \,) </Dd> <Dd> p H = p K a (\ displaystyle \ mathrm (pH) = \ mathrm (p) K_ (\ mathrm (a)) \,) </Dd> <P> Therefore, one can easily find the pK of the monoprotic acid by finding the pH of the point halfway between the beginning of the curve and the equivalence point, and solving the simplified equation . In the case of the sample curve, the K would be approximately 1.78 × 10 from visual inspection (the actual K is 1.7 × 10) </P>

The first and last sections of any titration curve are