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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a solution containing 1000 ml of 0135 m ch3cooh ka18e-5 is being titrated with 054 m naoh calculate the phnbsp at the
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