1. A diprotic acid is being titrated with a strong base. A volume of 0.0500 L of
ID: 1035361 • Letter: 1
Question
1.
A diprotic acid is being titrated with a strong base. A volume of 0.0500 L of 0.0750 M diprotic acid (H2A+) is combined with 0.0150 L of 0.250 M NaOH.
Ka1 for the diprotic acid is 8.7 x 10-3.
Ka2 for the diprotic acid is 6.5 x 10-7.
Calculate the pH of the solution.
2. Using the Henderson-Hasselbalch equation, calculate the pH of a buffer solution that contains 0.72 M NH4Cl and 0.48 M NH3.
3. A 1.00 L aqueous solution contains 0.10 mol acetic acid and 0.10 mol sodium acetate. The initial pH is 4.74. Calculate the pH after the addition of 0.027 mol of OH-.
(please watch for all questions sig figs, i have the most problems with sig figs. THANK YOU.)
Explanation / Answer
1. moles H2A+ = 0.075 M x 0.05 L = 0.00375 mmol
moles NaOH added = 0.250 M x 0.015 L = 0.00375 mmol
All of acid is neutralized to reach first equivalence point
pKa1 = -log(8.7 x 10^-3) = 2.06
pKa2 = -log(6.5 x 10^-7) = 6.19
pH of solution = 1/2(pKa1 + pKa2)
= 1/2(2.06 + 6.19)
= 4.125
2. [NH3] = 0.48 M
[NH4Cl] = 0.72 M
pKa = 9.25
Hendersen-Haseselbalck equation,
pH = pKa + log(NH3/NH4Cl)
= 9.25 + log(0.48/0.72)
= 9.074
3. initial moles CH3COOH = 0.1 mol
initial moles CH3COO- = 0.1 mol
added OH- = 0.027 mol
final moles CH3COOH = 0.1 - 0.027 = 0.073 mol
final moles CH3COO- = 0.1 + 0.027 = 0.127 mol
pKa = 4.74
Hendersen-Haseselbalck equation,
pH = pKa + log(CH3COO-/CH3COOH)
= 4.74 + log(0.127/0.073)
= 4.98
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