Question

The pH of a buffer is calculated by using the Henderson-Hasselbalch equation:

pH=pKa +log[Base]/[Acid]

Part A:

What is the pH of a buffer prepared by adding 0.809mol of the weak acid HA to 0.406mol of NaA in 2.00 L of solution? The dissociation constant Ka of HA is 5.66

Answer 1

(A) pH = pKa + log([NaA]/[HA])

= -log Ka + log(moles of NaA/moles of HA)

= -log(5.66 x 10^(-7)) + log(0.406/0.809)

= 5.948

(B) NaA + HCl => NaCl + HA

Moles of HA = 0.809 + 0.150 = 0.959 mol

Moles of NaA = 0.406 - 0.150 = 0.256 mol

pH = pKa + log([NaA]/[HA])

= -log Ka + log(moles of NaA/moles of HA)

= -log(5.66 x 10^(-7)) + log(0.256/0.959)

= 5.674

(C) HA + NaOH => NaA + H2O

Moles of HA = 0.809 - 0.195 = 0.614 mol

Moles of NaA = 0.406 + 0.195 = 0.601 mol

pH = pKa + log([NaA]/[HA])

= -log Ka + log(moles of NaA/moles of HA)

= -log(5.66 x 10^(-7)) + log(0.601/0.614)

= 6.238