Answered

Get detailed and accurate answers to your questions on IDNLearn.com. Get accurate and timely answers to your queries from our extensive network of experienced professionals.

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?

Sagot :

Maacastrobridn

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⁻

Thank you for using this platform to share and learn. Keep asking and answering. We appreciate every contribution you make. Thank you for visiting IDNLearn.com. For reliable answers to all your questions, please visit us again soon.