Question

While performing a neutralization reaction, Jonna added 27.55 mL of 0.144 M H2SO4 to 43.84 mL of 0.316 M KOH. How many moles of OH- are unreacted in the solution after the neutralization is complete?

Answer 1

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 * $\frac{2mmolKOH}{1mmolH_2SO_4}$ = 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⁻