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A 25.00 mL sample of 1.03 × 10⁻³ M Ba(OH)₂ was titrated with 1.88 × 10⁻³ M HClO₄. What is the pH of the solution after the addition of 38.1 mL of titrant?

Answer :

3.50 is the pH of the solution after the addition of 38.1 mL of titrant.

  • Calculate the moles of Ba(OH)₂ initially present

Ba(OH)₂ dissociates completely in water to produce Ba²⁺ and 2 OH⁻ ions. The number of moles of Ba(OH)₂ is given by:

moles of Ba(OH)₂ = Molarity × Volume
moles of Ba(OH)₂ = (1.03 × 10⁻³ M) × (25.00 mL × 10⁻³ L/mL)
moles of Ba(OH)₂ = 2.575 × 10⁻⁵ moles

  • Calculate the moles of OH⁻ produced by Ba(OH)₂

Since each mole of Ba(OH)₂ produces 2 moles of OH⁻, the moles of OH⁻ are:
moles of OH⁻ = 2 × moles of Ba(OH)₂
moles of OH⁻ = 2 × 2.575 × 10⁻⁵
moles of OH⁻ = 5.15 × 10⁻⁵ moles

  • Calculate the moles of HClO₄ added

moles of HClO₄ = Molarity × Volume
moles of HClO₄ = (1.88 × 10⁻³ M) × (38.1 mL × 10⁻³ L/mL)
moles of HClO₄ = 7.1628 × 10⁻⁵ moles

  • Determine the reaction between OH⁻ and HClO₄

HClO₄ is a strong acid and will react completely with OH⁻:
HClO₄ + OH⁻ → H₂O + ClO₄⁻

  • Since there are fewer moles of OH⁻ than HClO₄, OH⁻ is the limiting reactant. All of the OH⁻ will react, and some HClO₄ will remain unreacted.

moles of OH⁻ remaining = 0 (since all OH⁻ has reacted)
moles of HClO₄ remaining = moles of HClO₄ - moles of OH⁻
moles of HClO₄ remaining = 7.1628 × 10⁻⁵ - 5.15 × 10⁻⁵
moles of HClO₄ remaining = 2.0128 × 10⁻⁵ moles

  • Calculate the concentration of H⁺ in the solution

The remaining moles of HClO₄ contribute to the concentration of H⁺ in the solution.
Total volume of the solution = volume of Ba(OH)₂ + volume of HClO₄
Total volume of the solution = 25.00 mL + 38.1 mL = 0.0631 L
Concentration of H⁺ = moles of HClO₄ remaining / Total volume
Concentration of H⁺ = 2.0128 × 10⁻⁵ moles / 0.0631 L
Concentration of H⁺ = 3.19 × 10⁻⁴ M

  • Calculate the pH of the solution

pH is given by:
pH = -log[H⁺]
pH = -log(3.19 × 10⁻⁴)
pH = 3.50

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