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 M HF is titrated with 0.10 M KOH. Determine the pH of the solution after the addition of 100.0 mL of KOH. The Ka of HF is 3.5 × 10^-4.
A
pH = 7.00
B
pH = 3.14
C
pH = 8.00
D
pH = 5.14
Verified step by step guidance1
Calculate the initial moles of HF using the formula: \( \text{moles of HF} = \text{volume (L)} \times \text{molarity (M)} \). For 100.0 mL of 0.20 M HF, convert the volume to liters and multiply by the molarity.
Calculate the moles of KOH added using the formula: \( \text{moles of KOH} = \text{volume (L)} \times \text{molarity (M)} \). For 100.0 mL of 0.10 M KOH, convert the volume to liters and multiply by the molarity.
Determine the limiting reactant by comparing the moles of HF and KOH. 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. Subtract the moles of KOH from the initial moles of HF to find the moles of HF remaining. 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: \( \text{pH} = \text{pK}_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \), where \( \text{pK}_a = -\log(\text{K}_a) \), \([\text{A}^-]\) is the concentration of F\(^-\), and \([\text{HA}]\) is the concentration of HF remaining. Calculate \( \text{pK}_a \) and substitute the concentrations into the equation to find the pH.
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