Question

If 100.mL of 0.035 M HCl solution is added to 220.mL of a buffer solution which is 0.20M in NH3 and 0.18M in NH4Cl, what will be the pH of the new solution? The Kb of NH3 is 1.8x10^-5.

Please explain how the answer was obtained. Thank you!

Answer 1

mol of HCl added = 100.0M *0.035 mL = 3.50 mmol

NH3 will react with H+ to form NH4+

Before Reaction:

mol of NH3 = 0.18 M *220.0 mL

mol of NH3 = 39.6 mmol

mol of NH4+ = 0.2 M *220.0 mL

mol of NH4+ = 44 mmol

after reaction,

mol of NH3 = mol present initially - mol added

mmol of NH3 = (39.6 - 3.50) mmol

mol of NH3 = 36.1 mmol

mol of NH4+ = mol present initially + mol added

mol of NH4+ = (44 + 3.50) mmol

mol of NH4+ = 47.5 mmol

since volume is both in numerator and denominator, we can use mol instead of concentration

Kb = 1.8*10^-5

pKb = - log (Kb)

= - log(1.8*10^-5)

= 4.745

we have below equation to be used:

This is Henderson–Hasselbalch equation

pOH = pKb + log {[conjugate acid]/[base]}

= 4.745+ log {47.5/36.1}

= 4.864

we have below equation to be used:

PH = 14 - pOH

= 14 - 4.8639

= 9.1361

Answer: 9.14