<Dl> <Dd> (H) 2 + K a (H) − K a C a = 0 (\ displaystyle ((\ ce (H))) ^ (2) + K_ (a) ((\ ce (H))) - K_ (a) C_ (a) = 0) </Dd> </Dl> <Dd> (H) 2 + K a (H) − K a C a = 0 (\ displaystyle ((\ ce (H))) ^ (2) + K_ (a) ((\ ce (H))) - K_ (a) C_ (a) = 0) </Dd> <P> Solution of this quadratic equation gives the hydrogen ion concentration and hence p (H) or, more loosely, pH . This procedure is illustrated in an ICE table which can also be used to calculate the pH when some additional (strong) acid or alkaline has been added to the system, that is, when C ≠ C . </P> <P> For example, what is the pH of a 0.01 M solution of benzoic acid, pK = 4.19? </P>

(a) what is the ph of 3.3× 10–3 m hcl