A solution containing 1000 ml of 0135 m ch3cooh ka18e-5 is


A solution containing 100.0 mL of 0.135 M CH3COOH (ka=1.8E-5) is being titrated with 0.54 M NaOH. calculate the ph:  at the equivalence point and 5 mL past the equivalence point.

how much is 5 mL past? i do not know how to do it?

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Chemistry: A solution containing 1000 ml of 0135 m ch3cooh ka18e-5 is
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