In: Chemistry
1) A 75.0-mL volume of 0.200 M NH3 (Kb=1.8×10−5) is titrated with 0.500 M HNO3. Calculate the pH after the addition of 23.0 mL of HNO3.
2)A 52.0-mL volume of 0.35 M CH3COOH (Ka=1.8×10−5) is titrated with 0.40 M NaOH. Calculate the pH after the addition of 21.0 mL of NaOH.
Part.1
Initial moles of NH3 = 0.075 L x 0.200 M = 0.015 mol
Moles of HNO3 added = 0.023 L x 0.500 M = 0.0115 mol
NH3 + HNO3 ----------> NH4+ + NO3-
Moles of NH3 left = 0.015 - 0.0115 = 0.0035 mol
Moles of NH4+ = 0.0115 mol
Ka(NH4+) = Kw/Kb(NH3)
= 10-14/1.8 x 10-5 = 5.556 x 10-10
Henderson-Hasselbalch equation:
pH = pKa + log([NH3]/[NH4+])
= - log Ka + log 0.0035/0.0115
= -log(5.556 x 10-10) + log 0.0035/0.0115
= 9.26 + log 0.3043
= 9.26 - 0.5167
pH = 8.74
Part.2
Initial moles of CH3COOH = 0.052 L x 0.35 M = 0.0182 mol
Moles of NaOH added = 0.021 L x 0.400 M = 0.0084 mol
CH3COOH + NaOH ---------> CH3COO- + Na+
Moles of CH3COOH left = 0.0182 - 0.0084 = 0.0098 mol
Moles of CH3COO- = 0.0084 mol
Henderson-Hasselbalch equation:
pH = pKa + log([CH3COO-]/[CH3COOH])
= -log Ka + log([CH3COO-]/[CH3COOH])
= -log(1.8 x 10-5) + log(0.0084/0.0098)
= 4.74 + log 0.8571
= 4.74 - 0.06697
pH = 4.67