Neutralising a puffersolution.

Question

If you add 20 mL 0.100M NaOH to 100 mL of puffersolution, which consists of NH_3 and NH_4Cl, both with the same molar concentrtion 0.050M. What is the pH change?

Answer 1

You can use a variation of the Henderson Hasselbalch equation

pOH = pK_b + [BH⁺]/[B]

After adding 20 ml (0.020 L) of 0.100 M NaOH to 100 ml of buffer:

NH₄⁺ + OH^- ==> NH₃ + HOH so [NH4⁺] decreases and [NH₃] increases

0.02 L x 0.1 mol/L = 0.002 moles NaOH added

Initial moles NH₄⁺ and NH₃ = 0.1 L x 0.050 mol/L = 0.005 moles

After NaOH addition, moles NH₄⁺ = 0.005 - 0.002 = 0.003; moles NH₃ = 0.005+0.002 = 0.007

FINAL [NH₄⁺] = 0.003 moles/0.12 L = 0.025 M

FINAL [NH₃] = 0.007 moles/0.12 L = 0.058 M

Using 4.7 as the pK_b for NH₃ we have

pOH = 4.7 + log [0.025]/[0.058] = 4.7 + (-0.37) = 4.33

pH = 9.67

Initial pH: pOH = 4.7 + log 0.05/0.05 = 4.7 and pH = 9.3

SO CHANGE IN pH = 9.7 - 9.3 = 0.4