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Multiple Choice
What is the pH at the equivalence point for the titration of 20.0 mL of 0.500 M acetic acid (HC2H3O2) with 0.350 M potassium hydroxide (KOH), given that the Ka for acetic acid is 1.8 × 10^-5?
A
8.72
B
3.45
C
5.28
D
7.00
Verified step by step guidance
1
Identify the type of titration: This is a weak acid (acetic acid) being titrated with a strong base (potassium hydroxide). At the equivalence point, the solution will contain the conjugate base of acetic acid, which is acetate ion (C2H3O2^-).
Calculate the initial moles of acetic acid: Use the formula \( \text{moles} = \text{volume} \times \text{concentration} \). For acetic acid, \( \text{moles} = 20.0 \text{ mL} \times 0.500 \text{ M} \). Convert mL to L by dividing by 1000.
Determine the volume of KOH needed to reach the equivalence point: Since the reaction is 1:1, the moles of KOH needed will be equal to the moles of acetic acid. Use the formula \( \text{volume} = \frac{\text{moles}}{\text{concentration}} \) to find the volume of KOH.
Calculate the concentration of acetate ion at the equivalence point: The total volume at equivalence point is the sum of the volumes of acetic acid and KOH. Use \( \text{concentration} = \frac{\text{moles}}{\text{total volume}} \) to find the concentration of acetate ion.
Determine the pH using the hydrolysis of acetate ion: Acetate ion will hydrolyze in water to form acetic acid and hydroxide ion. Use the formula \( \text{pH} = 14 - \text{pOH} \), where \( \text{pOH} = -\log[\text{OH}^-] \). Calculate \([\text{OH}^-]\) using the equilibrium expression for the hydrolysis reaction and the given \( K_a \) value.