Question

Calculate the pH of 100.0 mL of a buffer that is 0.0850 M NH4Cl and 0.180 M NH3 before and after the addition of 1.00 mL of 5.50 M HNO3.

Can you please do a step-by-step of how to solve this, paying special attention to how to get the Ka in the Henderson Hasselbalch Equation? That's the part I'm not understanding.

Answer 1

1) Before addition of HNO3

Kb of NH3 = 1.8*10^-5

pKb = - log (Kb)

= - log(1.8*10^-5)

= 4.745

use:

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

= 4.745+ log {8.5*10^-2/0.18}

= 4.419

use:

PH = 14 - pOH

= 14 - 4.4189

= 9.5811

Answer: 9.58

2)

mol of HNO3 added = 5.5M *1.0 mL = 5.5 mmol

NH3 will react with H+ to form NH4+

Before Reaction:

mol of NH3 = 0.18 M *100.0 mL

mol of NH3 = 18 mmol

mol of NH4+ = 0.085 M *100.0 mL

mol of NH4+ = 8.5 mmol

after reaction,

mol of NH3 = mol present initially - mol added

mmol of NH3 = (18 - 5.5) mmol

mol of NH3 = 12.5 mmol

mol of NH4+ = mol present initially + mol added

mol of NH4+ = (8.5 + 5.5) mmol

mol of NH4+ = 14 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

use:

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

= 4.745+ log {14/12.5}

= 4.794

use:

PH = 14 - pOH

= 14 - 4.7939

= 9.2061

Answer: 9.21