Multiple Choice
What is the pH at the equivalence point for a titration involving 25 mL of 0.1 M acetic acid and 0.2 M KOH, given that the acid dissociation constant (Ka) for acetic acid is 1.8 × 10^-5?
18. Aqueous Equilibrium
Titrations: Weak Acid-Strong Base
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20.0 mL of 0.500 M HC2H3O2 is titrated with 0.350 M KOH. Determine the pH after 15.0 mL of the base has been added. Ka for HC2H3O2 = 1.8 × 10−5.
A
4.29
B
4.59
C
4.71
D
4.74
E
4.79
Verified step by step guidance1
Calculate the initial moles of acetic acid (HC2H3O2) using the formula: \( \text{moles} = \text{volume (L)} \times \text{molarity} \). Convert 20.0 mL to liters and multiply by 0.500 M.
Calculate the moles of KOH added using the formula: \( \text{moles} = \text{volume (L)} \times \text{molarity} \). Convert 15.0 mL to liters and multiply by 0.350 M.
Determine the moles of acetic acid that react with KOH. Since KOH is a strong base, it will react completely with the acetic acid in a 1:1 ratio. Subtract the moles of KOH from the initial moles of acetic acid to find the moles of acetic acid remaining.
Calculate the moles of acetate ion (C2H3O2-) formed, which will be equal to the moles of KOH added, since each mole of KOH reacts to form one mole of acetate ion.
Use the Henderson-Hasselbalch equation to find the pH: \( \text{pH} = \text{pK}_a + \log \left( \frac{[\text{C2H3O2}^-]}{[\text{HC2H3O2}]} \right) \). Calculate \( \text{pK}_a \) from \( K_a \) and use the moles of acetate and acetic acid to find their concentrations in the total volume of the solution (initial volume of acetic acid + volume of KOH added).
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