Consider a buffer made by adding 132.8 g of NaC₇H₅O₂ to 300.0 mL of 1.43 M HC₇H₅O₂ (Ka = 6.3 x 10⁻⁵) What is the pH of the buffer after 0.250 mol of OH⁻ have been added?

Answer 1

You will need to use Henderson Hasselbach equation which is:

pH = pKa + log(A-/HA)

NaC7H5O2 will be your conjucate base and HC7H5O2 will be the acid.

First find the pKa which can be found by taking the -log(6.3x10^-5). This is equal to 4.2

Next find the concentration of your conjugate base. To do so you will need to know how many mols of it you have. The molar mass can be found by adding the atomic weights which comes out to 144g/mol. Dividing this by the 133 g givs 0.92 mols. The concentration can be found by dividing by the volume which is 0.3L which comes out to 3.08 M.

Next find the concentration of Hydoxide. Do this by dividing the number of mols by the total volume. 0.25 mols/0.3L = 0.83 M.

The reaction is as follows

HA + -OH --> H2O + A-

1.43. -0.83 3.08

0.6. 0 3.91

The first row of numbers shows the initial concentrations of the buffer and how the hydroxide will react. The second row shows the final concentration after the reaction.

We can now plug all our numbers into the HH equation.

pH = 4.2 + log(3.91/0.6)

pH = 4.2 + 0.81

pH = 5.01