11.An acid-base equilibrium system is created by dissolving 0.20mol CH3COOH in water and diluting the resulting solution to a volume of 1.0 L. What is the effect of adding 0.020 mol CH3COO (aq) to this solution? How will pH change (calculate pH before and after the addition. Ka of CH3COOH is 1.76 x10-5)? How will concentrations of CH3COOH and CH3COO at equilibrium change?

Principles of Modern Chemistry
8th Edition
ISBN:9781305079113
Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Publisher:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Chapter16: Solubility And Precipitation Equilibria
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11.An acid-base equilibrium system is created by dissolving 0.20mol CH3COOH in
water and diluting the resulting solution to a volume of 1.0 L. What is the effect of
adding 0.020 mol CH3COO (aq) to this solution? How will pH change (calculate
pH before and after the addition. Ka of CH3COOH is 1.76 x10-5)? How will
concentrations of CH3COOH and CH3COO at equilibrium change?
Transcribed Image Text:11.An acid-base equilibrium system is created by dissolving 0.20mol CH3COOH in water and diluting the resulting solution to a volume of 1.0 L. What is the effect of adding 0.020 mol CH3COO (aq) to this solution? How will pH change (calculate pH before and after the addition. Ka of CH3COOH is 1.76 x10-5)? How will concentrations of CH3COOH and CH3COO at equilibrium change?
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