Question

Calculate the pH of 100.0 mL of a buffer that is 0.0800 M NH4Cl and 0.100 M NH 3 before and after the addition of 1.00 mL of 5.25 M HNO3- 1st attempt Part 1 (0.5 point) M See Periodic Table See = pH before Part 2 (0.5 point) = pH after

Answer 1

1)
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 {8*10^-2/0.1}
= 4.648

use:
PH = 14 - pOH
= 14 - 4.6478
= 9.3522
Answer: 9.35

2)

mol of HNO3 added = 5.25M *1.0 mL = 5.25 mmol

NH3 will react with H+ to form NH4+

Before Reaction:
mol of NH3 = 0.1 M *100.0 mL
mol of NH3 = 10 mmol

mol of NH4+ = 0.08 M *100.0 mL
mol of NH4+ = 8 mmol

after reaction,
mol of NH3 = mol present initially - mol added
mmol of NH3 = (10 - 5.25) mmol
mol of NH3 = 4.75 mmol

mol of NH4+ = mol present initially + mol added
mol of NH4+ = (8 + 5.25) mmol
mol of NH4+ = 13.25 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 {13.25/4.75}
= 5.19

use:
PH = 14 - pOH
= 14 - 5.1902
= 8.8098
Answer: 8.81