Multiple Choice
A 30.0-mL volume of 0.50 M CH3COOH (Ka=1.8×10⁻⁵) was titrated with 0.50 M NaOH. Calculate the pH after the addition of 30.0 mL of NaOH at 25 °C.
18. Aqueous Equilibrium
Titrations: Weak Acid-Strong Base
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A 30.00-mL sample of 0.125 M HCOOH is being titrated with 0.175 M NaOH. What is the pH after 23.0 mL of NaOH has been added? Ka of HCOOH = 1.8 × 10⁻⁴
A
5.00
B
4.25
C
4.50
D
3.75
Verified step by step guidance1
Calculate the initial moles of HCOOH using the formula: \( \text{moles} = \text{volume (L)} \times \text{molarity (M)} \). For HCOOH, use 30.00 mL and 0.125 M.
Calculate the moles of NaOH added using the formula: \( \text{moles} = \text{volume (L)} \times \text{molarity (M)} \). For NaOH, use 23.0 mL and 0.175 M.
Determine the moles of HCOOH that react with NaOH. Since NaOH is a strong base, it will react completely with HCOOH in a 1:1 ratio. Subtract the moles of NaOH from the initial moles of HCOOH to find the moles of HCOOH remaining.
Calculate the moles of the conjugate base, HCOO⁻, formed. This will be equal to the moles of NaOH added, as each mole of NaOH converts one mole of HCOOH to HCOO⁻.
Use the Henderson-Hasselbalch equation to find the pH: \( \text{pH} = \text{pK}_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \), where \( \text{pK}_a = -\log(\text{K}_a) \), \([\text{A}^-]\) is the concentration of HCOO⁻, and \([\text{HA}]\) is the concentration of HCOOH remaining.
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