Acetic Acid, NaOH titration

[WhiskeyTangoFoxtrot]

Need help understanding orange text below. I can understand conceptually why it becomes more basic, but when the using HH equation why is the salt concentration put in the [A-] spot and acetic acid in the [HA]? Thanks!
Question came from here.

What is the pH when 25.0 mL of 0.200 M of CH3COOH has been titrated with 35.0 mL of 0.100 M NaOH? Acetic acid pka is 4.752.

Solution:

1) Determine moles of acetic acid and NaOH before mixing:

CH3COOH: (0.200 mol/L) (0.0250 L) = 0.00500 mol
NaOH: (0.100 mol/L) (0.0350 L) = 0.0035 mol

2) Determine moles of acetic acid and sodium acetate after mixing:

CH3COOH: 0.00500 mol - 0.00350 mol = 0.00150 mol
CH3COONa: 0.0035 mol

The reaction:

CH3COOH + NaOH <===> CH3COONa + H2O

3) Use the Henderson-Hasselbalch Equation:

pH = 4.752 + log [(0.00350 mol/0.060 L) / (0.0015 mol/0.060 L)]
pH = 4.752 + log 2.333

pH = 4.752 + 0.368 = 5.120

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[Dr. Doctor]

Because the [A] represents a base(the A has no hydrogen and represents that it can accept the H from the acid) and the [HA] represents an acid. The HH is based this idea of ratios between the acid and conj base so if you have 10x more acid than the conj base you would have a lower pH or log(1/10)=-1 to the pKa of the acid; if you had equal acid to conj base your ratio would be one to one and you would get log(1/1)=0 so the pH of the solution would be the pKa; if you had 10x the amount of base you would have the ratio log(10/1)=+1 which you would add to the pKa increasing the alkalinity of the solution.

 

[WhiskeyTangoFoxtrot]

Thanks!