Problem: 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 10-5a. 4.45b. 4.81c. 9.55d. 4.64e. 9.19

FREE Expert Solution

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.


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Problem Details

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

a. 4.45

b. 4.81

c. 9.55

d. 4.64

e. 9.19

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