25.0 mL of a 0.21 M formic acid, HCOOH (Ka = 1.8 x 10-4) is titrated with a 0.15M KOH solution. Calculate the pH when 10.0 mL of KOH has been added to the formic acid solution.
a) Write the correct equation
b) Write the balanced equation and show the stoichiometric relationship in the space provided in the next question
25.0 mL of a 0.21 M formic acid, HCOOH (Ka = 1.8 x 10-4) is titrated with a 0.15M KOH solution. Calculate the pH when 10.0 mL of KOH has been added to the formic acid solution.
The correct answer and explanation is :
a) Correct Chemical Equation
This is an acid-base neutralization reaction between formic acid (HCOOH), a weak acid, and potassium hydroxide (KOH), a strong base:
HCOOH (aq) + OH⁻ (aq) → HCOO⁻ (aq) + H₂O (l)
KOH completely dissociates in water, so the hydroxide ion (OH⁻) is the actual reacting species.
b) Balanced Chemical Equation & Stoichiometric Relationship
HCOOH (aq) + OH⁻ (aq) → HCOO⁻ (aq) + H₂O (l)
Stoichiometric relationship: 1 mole of formic acid reacts with 1 mole of hydroxide ion to produce 1 mole of formate ion and 1 mole of water.
pH Calculation – Step-by-Step Explanation
Step 1: Moles of HCOOH and OH⁻
- Volume HCOOH = 25.0 mL = 0.0250 L
- Molarity HCOOH = 0.21 M
⇒ Moles HCOOH = 0.21 × 0.0250 = 0.00525 mol - Volume KOH added = 10.0 mL = 0.0100 L
- Molarity KOH = 0.15 M
⇒ Moles OH⁻ = 0.15 × 0.0100 = 0.00150 mol
Step 2: Limiting Reactant and Reaction Progress
Since OH⁻ is less than HCOOH, it’s the limiting reactant. It reacts completely with an equal amount of HCOOH:
- Moles of HCOOH remaining = 0.00525 – 0.00150 = 0.00375 mol
- Moles of HCOO⁻ produced = 0.00150 mol
Step 3: Use Henderson-Hasselbalch Equation
We now have a buffer (HCOOH/HCOO⁻). Use the Henderson-Hasselbalch equation:
[
\text{pH} = \text{p}K_a + \log \left(\frac{[\text{A}^-]}{[\text{HA}]}\right)
]
- Ka = 1.8 × 10⁻⁴ ⇒ pKa = –log(1.8 × 10⁻⁴) ≈ 3.74
- Concentrations (in total volume 25.0 + 10.0 = 35.0 mL = 0.035 L):
- [HCOOH] = 0.00375 / 0.035 ≈ 0.107 M
- [HCOO⁻] = 0.00150 / 0.035 ≈ 0.0429 M
Now plug into the equation:
[
\text{pH} = 3.74 + \log\left(\frac{0.0429}{0.107}\right) = 3.74 + \log(0.401) \approx 3.74 – 0.396 = \boxed{3.34}
]
Final Answer:
- pH ≈ 3.34
- At this point, the solution is a buffer containing unreacted formic acid and its conjugate base, formate ion. The buffer resists pH change due to the presence of both acidic and basic components in a defined ratio, calculated via Henderson-Hasselbalch.
