Chemistry Buffer Problems

Question

1. Calc‎ulate the p‎H of a buf‎fer conta‎ining 0.10 M NH‎₃ and 0.20 M NH₄‎Cl . K_b NH₃ = 1.8 x 10^-5
2. A che‎mist ‎needs a bu‎ffer with p‎H 4.‎35. How m‎any millilit‎ers of pu‎re ace‎tic a‎cid (d‎ensity = 1.049‎ g/mL ) mu‎st be a‎dded to 465‎mL of 0.0‎941 M NaOH sol‎ution to obtain such a ‎buffer?

Answer 1

The conventional form of the Henderson Hasselbalch equation is

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

An alternative form (for basic buffers) is

pOH = pKb + log [salt]/[base]

pKb = -log Kb = -log 1.8x105 = 4.74

pOH = 4.74 + log (0.20 / 0.10) = 4.74 + 0.301 = 5.04

pH = 14 - pOH = 14 - 5.04

pH = 8.96

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

For acetic acid, we'll use a pKa = 4.75

4.35 = 4.75 + log [salt]/[acid]

log [salt]/[acid] = -0.40

[salt]/[acid] = 0.398

mols NaOH = 0.465 L x 0.0941 mol/L = 0.0438 mols

CH3COOH + NaOH ==> CH3COONa + H2O

x....................0.0438............0...........

-0.0438........-0.0438..........+0.0438.....

x-0.0438.........0...................0.0438....

0.0438 / x-0.0438 = 0.398

x = 0.154 mols CH3COOH

0.154 mols CH3COOH x 60.1 g /mol = 9.26 g CH3COOH

9.26 g x 1 ml / 1.049 g = 8.82 mls CH3COOH

(this neglected the small (<2%) increase in volume from 465 ml to 473.8 ml. If you must include the change in volume, you can use the appropriate algebraic equations to include this small change in volume. I chose not to do so.