General Chemistry 2

Question

Consider a buffer made by adding 132.8 g of NaC₇H₅O₂ to 300.0 mL of 1.41 M HC₇H₅O₂ (Ka = 6.3 x 10⁻⁵) What is the pH of this buffer?

Answer 1

molar mass NaC₇H₅O₂ = 144 g / mol

[NaC₇H₅O₂] = 132.8 g x 1 mol / 144 g = 0.9222 mol / 0.300 L = 3.074 M

[HC₇H₅O₂] = 1.41 M

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

Henderson Hasselbalch equation:

pH = pKa + log [conj.base]/[acid]

pH = 4.20 + log (3.07/1.41)

pH = 4.20 + log 2.18

pH = 4.20 + 0.338

pH = 4.54