Calculate the pH at 25°C of 249.0 mL of a buffer solution that is 0.220 M NH4Cl and 0.220 M NH3 before and after the addition of 1.30 mL of 6.0 M HNO3. (The pKa for NH4+ = 9.75)
1) pH before =
2) pH after =
1)
use:
pH = pKa + log {[conjugate base]/[acid]}
= 9.75+ log {0.220/0.220}
= 9.75
Answer: 9.75
2)
mol of HNO3 added = 6.0M *1.3 mL = 7.80 mmol
NH3 will react with H+ to form NH4+
Before Reaction:
mol of NH3 = 0.22 M *249.0 mL
mol of NH3 = 54.78 mmol
mol of NH4+ = 0.22 M *249.0 mL
mol of NH4+ = 54.78 mmol
after reaction,
mol of NH3 = mol present initially - mol added
mmol of NH3 = (54.78 - 7.80) mmol
mol of NH3 = 46.98 mmol
mol of NH4+ = mol present initially + mol added
mol of NH4+ = (54.78 + 7.80) mmol
mol of NH4+ = 62.58 mmol
since volume is both in numerator and denominator, we can use mol
instead of concentration
use:
pH = pKa + log {[conjugate base]/[acid]}
= 9.75+ log {46.98/62.58}
= 9.625
Answer: 9.62
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