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Multiple Choice
A 30.00-mL sample of 0.125 M HCOOH is being titrated with 0.175 M NaOH. What is the pH after 15.0 mL of NaOH has been added? Ka of HCOOH = 1.8 × 10⁻⁴
A
4.50
B
3.75
C
5.00
D
4.25
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Verified step by step guidance
1
Calculate the initial moles of HCOOH using the formula: \( \text{moles} = \text{volume} \times \text{molarity} \). For HCOOH, use \( 30.00 \text{ mL} \times 0.125 \text{ M} \).
Calculate the moles of NaOH added using the formula: \( \text{moles} = \text{volume} \times \text{molarity} \). For NaOH, use \( 15.0 \text{ mL} \times 0.175 \text{ M} \).
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: \( \text{pH} = \text{pKa} + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \). Calculate \( \text{pKa} \) from \( \text{Ka} \) using \( \text{pKa} = -\log(\text{Ka}) \), and substitute the concentrations of HCOO⁻ and HCOOH into the equation.