We are asked to calculate the pH of the solution after 0.08 mol of HCl is added.
CH3COONa will break up in the solution:
CH3COONa(aq) → Na+(aq) + CH3COO-(aq)
▪ K+ and Na+ → group 1A → (+1) charge
Main group metals should have a +3 charge or higher to be considered acidic
Na+and K+ neutral and will not contribute to pH
CH3COOH is a weak acid and based on the Bronsted-Lowry definition, an acid is a proton (H+). Once CH3COOH loses its proton, its conjugate base is formed:
CH3COOH(aq) + H2O(l) → CH3COO-(aq) + H3O+(aq)
(weak acid) (base) (conjugate base) (conjugate acid)
Whenever we have a conjugate base and a weak acid, we have a buffer.
A solution with a volume of 1.00 L is 0.450 M in CH3COOH(aq) and 0.350 M in CH3COONa(aq). What will the pH be after 0.0800 mol of HCl is added to the solution? Ka for CH3COOH(aq) = 1.80 x 10-5
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