Question

Calculate the pH of 100.0 mL of a buffer that is 0.0700 M  NH4Cl and 0.115 M NH3 before and after the addition of 1.00 mL of 5.10 M  HNO3.

Before=

After=

Answer 1

1)

Kb of NH3 = 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 {7*10^-2/0.115}

= 4.529

we have below equation to be used:

PH = 14 - pOH

= 14 - 4.5291

= 9.47

Answer: 9.47

2)

mol of HNO3 added = 5.1M *1.0 mL = 5.1 mmol

NH3 will react with H+ to form NH4+

Before Reaction:
mol of NH3 = 0.115 M *100.0 mL
mol of NH3 = 11.5 mmol

mol of NH4+ = 0.07 M *100.0 mL
mol of NH4+ = 7 mmol

after reaction,
mol of NH3 = mol present initially - mol added
mmol of NH3 = (11.5 - 5.1) mmol
mol of NH3 = 6.4 mmol

mol of NH4+ = mol present initially + mol added
mol of NH4+ = (7 + 5.1) mmol
mol of NH4+ = 12.1 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 {12.1/6.4}
= 5.021

we have below equation to be used:
PH = 14 - pOH
= 14 - 5.0213
= 8.98
Answer: 8.98

Feel free to comment below if you have any doubts or if this answer do not work. I will correct it and submit again if you let me know