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?

Question

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?

cuanchi · Answer

25mL of NaOH is the equivalence point, then 30mL of NaOH will be 5mL past the equivalence point