What is the pH of a buffer system?

Question

What is the pH of a buffer system made by dissolving 10.70 grams of NH Cl and 20.00 mL of 12.0 M

NH in enough water to make 1.000 L of solution? K_b = 1.8 × 10 for NH₃

The correct answer is 9.34

Answer 1

I'll assume the buffer is NH₄Cl and NH₃. And I'll assume Kb NH3 = 1.8x10^-5. Let's hope so.

10.79 g NH4Cl x 1 mol NH4Cl/53.5 g = 0.202 moles/L = 0.202 M

(20 ml)(12 M) = (1000 ml )(x M) and x = 0.24 M NH₃

Using a form of the Henderson Hasselbalch equation for basic buffers...

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

pKb = -log Kb = -log 1.8x10^-5 = 4.74

pOH = 4.74 + log (0.202/0.240)

pOH = 4.74 + log 0.842

pOH = 4.74 + (-0.075)

pOH = 4.67

pH = 14 - 4.67 = 9.33 (differs due to rounding)