Question

A beaker with 1.30×102 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 5.00 mL of a 0.370 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.

Answer 1

he has given

[CH3COOH] + [CH3COONa] = 0.1M ----1

use the handerson equation

pH = Pka + log([CH3COONa]/[CH3COOH])

5.5 = 4.74 + log([CH3COONa]/[CH3COOH])

log([CH3COONa]/[CH3COOH]) = 5.5-4.74 = 0.76

[CH3COONa]/[CH3COOH] = 10^0.76 = 5.75 ------2

from 1 and 2 we can solve th eboth concentrations

from 2

[CH3COONa] = 5.75 x [CH3COOH]

put this value in equation1

[CH3COOH] +5.75 x [CH3COOH] = 0.1M

[CH3COOH] 6.75 = 0.1M

[CH3COOH] = 0.1 / 6.75

[CH3COOH] = 0.01481 M

put this value in equation1

0.01481 + [CH3COONa] = 0.1M

[CH3COONa] = 0.1M - 0.001481 M

[CH3COONa] = 0.0985 M

moles of CH3COOH = 0.01481 M x 0.13 L = 0.001925 moles

moles of CH3COONa = 0.0985 M x 0.13L = 0.01276 moles

now no of moles of H+ added = 0.37 M x 0.005L = 0.00185 moles

so 0.00185 moles of CH3COONa will convert in to CH3COOH

after addition

no of moles of CH3COOH = 0.001925 + 0.00185 = 0.00375 moles

no of moles of CH3COONa = 0.01276-0.00185 = 0.01091

now

pH = 4.74 + log(0.01091 / 0.00375)

pH = 4.75 + 0.464

pH = 5.21