Multiple Choice
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.
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18. Aqueous Equilibrium
Titrations: Weak Acid-Strong Base
Multiple Choice
A 100.0 mL sample of 0.20 mol L⁻¹ HF is titrated with 0.10 mol L⁻¹ KOH. Determine the pH of the solution after the addition of 200.0 mL of KOH. The Kₐ of HF is 3.5 × 10⁻⁴.
A
pH = 7.00
B
pH = 8.72
C
pH = 3.14
D
pH = 5.28
Verified step by step guidance1
Calculate the initial moles of HF using the formula: \( \text{moles of HF} = \text{concentration of HF} \times \text{volume of HF} \). Convert the volume from mL to L before calculation.
Calculate the moles of KOH added using the formula: \( \text{moles of KOH} = \text{concentration of KOH} \times \text{volume of KOH} \). Again, convert the volume from mL to L before calculation.
Determine the limiting reactant by comparing the initial moles of HF and the moles of KOH added. Since HF is a weak acid and KOH is a strong base, they will react in a 1:1 ratio to form water and the conjugate base, F⁻.
Calculate the moles of HF remaining and the moles of F⁻ formed after the reaction. If KOH is the limiting reactant, all of it will react, and the remaining moles of HF will be \( \text{initial moles of HF} - \text{moles of KOH} \). The moles of F⁻ formed will be equal to the moles of KOH added.
Use the Henderson-Hasselbalch equation to find the pH of the solution: \( \text{pH} = \text{pK}_a + \log \left( \frac{[\text{F}^-]}{[\text{HF}]} \right) \). Calculate \( \text{pK}_a \) from \( K_a \) using \( \text{pK}_a = -\log(K_a) \). Substitute the concentrations of F⁻ and HF, which are the moles divided by the total volume of the solution (sum of the initial HF and added KOH volumes in L).
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