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25. 50.00 mL of 0.10 M CH3COOH (Ka = 1.8 x 10) is titrated with a 0.10 M KOH solution. After 75.00 mL of the KOH solution is

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Answer #1

CH coolt molarity = o.lm | KoH molarity= o LM volume=50m1=0.05 I volume 75ml=0.0752 M = 2 = 0.0015 no of moles = 0.24 0.05 I

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25. 50.00 mL of 0.10 M CH3COOH (Ka = 1.8 x 10") is titrated with a...
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