A 30.00-mL sample of 0.125 M HCOOH is being titrated with 0.175 M NaOH. What is the pH after 11.2 mL of NaOH has been added? Ka of HCOOH = 1.8 × 10⁻⁴

• A. pH = 4.25
• B. pH = 3.75
• C. pH = 5.00
• D. pH = 6.50

Answer: (A)

Verified step by step guidance

1. Step 1: Calculate the initial moles of HCOOH in the solution. Use the formula \( \text{moles} = \text{volume} \times \text{molarity} \). For HCOOH, \( \text{moles} = 30.00 \text{ mL} \times 0.125 \text{ M} \). Convert the volume from mL to L by dividing by 1000.
2. Step 2: Calculate the moles of NaOH added using the same formula \( \text{moles} = \text{volume} \times \text{molarity} \). For NaOH, \( \text{moles} = 11.2 \text{ mL} \times 0.175 \text{ M} \). Again, convert the volume from mL to L.
3. Step 3: Determine the moles of HCOOH that react with NaOH. Since NaOH is a strong base, it will react completely with HCOOH. Subtract the moles of NaOH from the initial moles of HCOOH to find the moles of HCOOH remaining.
4. Step 4: Calculate the moles of the conjugate base, HCOO⁻, formed. The moles of HCOO⁻ will be equal to the moles of NaOH added, as each mole of NaOH reacts to form one mole of HCOO⁻.
5. Step 5: Use the Henderson-Hasselbalch equation to find the pH. The equation is \( \text{pH} = \text{pKa} + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \), where \( \text{pKa} = -\log(\text{Ka}) \). Calculate \( \text{pKa} \) using the given \( \text{Ka} \) value, and then plug in the concentrations of HCOO⁻ and HCOOH to find the pH.