By how much will the ph change if 0.025 mol of hcl is added to 1.00 l of the buffer that contains 0.15 m hc2h3o2 and 0.25 m c2h3o2–? use ka = 1.8 × 10–5 for hc2h3o2.

According to the reaction equation:

              CH3COO-  + H+   → CH3COOH 

initial         0.25                             0.15

change     - 0.025                          + 0.025

Equ         (0.25-0.025)                 (0.15 + 0.025)

first, we have to get moles acetate and moles acetic acid:

moles of acetate = 0.25 - 0.025 = 0.225 moles

∴ [CH3COO-] = 0.225 mol / 1 L = 0.225 M

moles of acetic acid = 0.15 + 0.025 = 0.175 moles

∴ [ CH3COOH] = 0.175 mol / 1L = 0.175 M 

Pka = -㏒ Ka 

       = -㏒ 1.8 x 10^-5
       
       = 4.74

from H-H equation we can get the PH value:

PH = Pka + ㏒ [acetate / acetic acid]

PH = 4.74 + ㏒[0.225/0.175]

∴ PH = 4.8