equlibrium buffer solution

Question

4-8. Which of the following solutions show buffer properties? Calculate the pH of each solution that is buffered. pK of NH3 = 4.75.

(a) 0.10 L of 0.25 M NH4Cl + 0.15 L 0.25 M NaOH

(b) 0.10 L of 0.25 M NH4Cl + 0.05 L 0.25 M NaOH

(c) 0.10 L of 0.25 M NH4Cl + 0.05 L 0.25 M HCI

(d) 0.10 L of 0.25 M NH3 + 0.05 L 0.25 M HCI

Answer 1

A buffer consists of a weak acid or a weak base PLUS the conjugate acid or conjugate base.

(a). NH4Cl + NaOH is NOT a buffer. Has the conjugate acid (NH4+), but no weak base

(b). NH4Cl + NaOH is NOT a buffer. Same explanation as for (a)

(c). NH4Cl + HCl is NOT a buffer. Same explanation as for (a)

(d). NH3 + HCl WILL form a BUFFER containing NH3 (weak base) and NH4Cl (conjugate acid).

So working with solution (d), we have the following situation:

0.1 L of 0.25 M NH3 + 0.05 L 0.25 M HCl

moles NH3 = 0.1 L x 0.25 mols / L = 0.025 mols NH3

moles HCl = 0.05 L x 0.25 mol / L = 0.0125 mols HCl

NH3 + HCl ==> NH4Cl

0.025.......0.0125.........0.......Initial

-0.0125...-0.0125...+0.0125..Change

0.0125........0...........0.0125...Equilibrium

So, at equilibrium we have 0.0125 mols NH3 and 0.0125 mols NH4+. Whenever the mols weak base = the moles of conjugate acid (as is the case here), the pOH will equal the pKb. You can see this from the Henderson Hasselbalch equation (see below).

So pOH = pKb

pOH = 4.75

pH = 14 - 4.75

pH = 9.25

Using the Henderson Hasselbalch equation for an alkaline buffer and finding pH:

pOH = pKb + log [conjugate acid] / [weak base]

pOH = 4.75 + log (0.0125) / (0.0125)

pOH = 4.75 + log 1

pOH = 4.75

pH = 14 - 475

pH = 9.25