Question

A beaker with 180 mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M . A student adds 6.20 mL of a 0.440 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.

Answer 1

Let 'x' be the concentration of acid
Let 'y' be the concentration of conjugate base
The total molarity of acid and conjugate base in this buffer is 0.100 M .
Therefore, x + y = 0.100 M

According to Henderson-Hasselbach equation,

pH = pKa + log [conjugate base] [acid]

5.00 = 4.760 + log (y x)

Solving the above equation:

5.00 - 4.760 = log (y x)
0.24 = log (y x)
10^0.24 =1.74 = y x
1.74 x = y
x + 1.74 x = 0.100 M
2.74 x = 0.100 M
x =0.0365 M = concentration of acid
0.100 - 0.0365 = 0.0635 M = concentration of conjugate base

Number of moles of acetic acid = 0.180 L 0.0365 M= 0.00657
Number of moles of conjugate base = 0.180 L 0.0635 M = 0.01143

Number of moles of HCl = 6.20 10^-3 L x 0.440 M=0.00273

A^- + H⁺  HA

Number of moles of conjugate base = 0.01143 - 0.00273 = 0.0087
Number of moles of acid = 0.00657 + 0.00273 = 0.0093

Total volume = 180 + 6.20 = 186.2 mL = 0.1862 L

Therefore, concentration of acid = 0.0093 0.1862 = 0.05 M
Concentration of conjugate base = 0.0087 0.1862 = 0.047 M

pKa of acetic acid is 4.760.

Applying Henderson-Hasselbach equation, we get

pH = 4.760 + log (0.047 0.050) = 4.760 - 0.027 = 4.73

Therefore, the change in pH = 5.00 - 4.73 = 0.27