Question

A beaker with 155 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 4.10 mL of a 0.470 MHCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.

Express your answer numerically to two decimal places. Use a minus ( − ) sign if the pH has decreased.

Answer 1

pka of aceticacid = 4.74

pH Of acidic buffer = pka + log(salt(or)base/acid)

no of mole of buffer = 155*0.1 = 15.5 mmol

    pH = 5

no of mole of aceticacid = x

no of mole of acetate = 15.5-x

5 = 4.74+log((15.5-x)/x)

x = 5.5 mmol

no of mole of aceticacid = x = 5.5 mmol

no of mole of acetate = 15.5-x = 15.5 - 5.5 = 10 mmol

after addition of HCl

pH Of acidic buffer = pka + log(salt(or)base - HCl/acid+ HCl)

no of mole of HCl added = 4.1*0.47 = 1.927 mmole

pH = 4.74 + log((10-1.927)/(5.5+1.927))

     = 4.78

change in pH = 4.78-5 = -0.22