pH Acid-Bases Chem, Buffers & Titrations

Question

Consider a buffer made by adding 132.8 g of NaC₇H₅O₂ to 300.0 mL of 2.27 M HC₇H₅O₂ (Ka = 6.3 x 10⁻⁵)

a) What is the pH of this buffer?

b) What is the pH of the buffer after 0.750 mol of H⁺ have been added?

c) What is the pH of the buffer after 0.250 mol of OH⁻ have been added?

Answer 1

molar mass NaC₇H₅O₂ = 144 g / mol

[NaC₇H₅O₂ = 132.8 g x 1 mol / 144 g = 0.922 mols / 0.300 L = 3.07 M

[HC₇H₅O₂] = 2.27 M

pKa = -log Ka = -log 6.3x10^-5 = 4.20

ALL OF THE FOLLOWING CALCULATIONS ASSUME NO CHANGE IN VOLUME

Henderson Hasselbalch equation

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

(a) pH = 4.20 + log (2.27 / 3.07) = 4.20 + (-0.13) = 4.07

(b) 0.750 mol H⁺ / 0.300 L = 2.50 M H⁺ which will react with the NaC₇H₅O₂ to produce HC₇H₅O₂

[NaC₇H₅O₂] = 3.07 - 2.50 = 0.57 M

[HC₇H₅O₂] = 2.27 + 2.50 = 4.77

pH = pKa + log (0.57 / 4.77) = 4.20 + (-0.92) = 3.28

(c) 0.250 mol OH^- / 0.300 L = 0.833 M OH^- which will react with HC₇H₅O₂ to produce NaC₇H₅O₂

[NaC₇H₅O₂] = 3.07 + 0.83 = 3.90 M

[HC₇H₅O₂] = 2.27 - 0.83 = 1.44 M

pH = pKa + log (3.90 / 1.44) = 4.20 + 0.43 = 4.63