Answer:
5.916x10⁻³ mol OH⁻
Explanation:
The reaction that takes place is:
- H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O
First we calculate the added moles of each reagent, using the given volumes and concentrations:
- H₂SO₄ ⇒ 0.144 M * 27.55 mL = 3.967 mmol H₂SO₄
- KOH ⇒ 0.316 M * 43.84 mL = 13.85 mmol KOH
Now we calculate how many KOH moles reacted with 3.967 mmol H₂SO₄:
- 3.967 mmol H₂SO₄ * [tex]\frac{2mmolKOH}{1mmolH_2SO_4}[/tex] = 7.934 mmol KOH
Finally we calculate how many OH⁻ moles remained after the reaction
- 13.85 mmol - 7.934 mmol = 5.916 mmol OH⁻
- 5.916 mmol / 1000 = 5.916x10⁻³ mol OH⁻
