Consider a buffer made by adding 132.8 g of NaC₇H₅O₂ to 300.0 mL of 2.31 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

pKa for HC₇H₅O₂ = 4.20

Henderson Hasselbalch equation: pH = pKa + log [salt]/[acid]

Initial concentrations:

[NaC₇H₅O₂] = 132.8 g x 1 mol / 144.8 g = 0.9171 mols / 0.300 L = 3.057 M

[HC₇H₅O₂] = 2.31 M

HC₇H₅O₂ + OH^- ==> C₇H₅O₂^- + H2O

mols OH- added = 0.250 mols in 0.300 L = 8.833 M

Final concentrations after addition of 0.250 mols OH- are as follows:

[HC₇H₅O₂] = 2.31 M - 0.833 = 1.477 M

[C₇H₅O₂^-] = 3.057 M + 0.833 M = 3.89 M

pH = pKa + log [salt]/[acid]

pH = 4.20 + log (3.89/1.477) = 4.20 + log 2.633

pH = 4.20 + 0.42

pH = 4.62