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While performing a neutralization reaction, Jonna added 22.63 mL of 0.142 M H2SO4 to 46.21 mL of 0.304 M KOH. How many moles of OH- are unreacted in the solution after the neutralization is complete?

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Answer:

7.623 x 10⁻³ mol OH⁻

Explanation:

The reaction that takes place is:

  • H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O

Now we calculate how many moles of each reagent were added:

  • H₂SO₄ ⇒ 22.63 mL * 0.142 M = 3.213 mmol H₂SO₄
  • KOH ⇒ 46.21 mL * 0.304 M = 14.05 mmol KOH

We calculate how many OH⁻ moles reacted with H₂SO₄:

  • 3.213 mmol H₂SO₄ * [tex]\frac{2mmolOH^-}{1mmolH_2SO_4}[/tex] = 6.427 mmol OH⁻

Finally we substract the OH⁻ moles that reacted from the added ammount of OH⁻ moles:

  • 14.05 mmol KOH - 6.427 mmol OH⁻ = 7.623 mmol OH⁻
  • 7.623 mmol /  1000 = 7.623 x 10⁻³ mol OH⁻

The number of moles of OH⁻ that are unreacted in the solution after the neutralization reaction is complete is 0.00770 moles OR 7.70×10⁻³ moles

First, we will write a balanced chemical equation for the neutralization reaction

The balanced chemical equation for the neutralization reaction is

H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O

This means

1 mole of H₂SO₄ reacts with 2 moles of KOH to give 1 mole of K₂SO₄ and 2 moles of H₂O

To determine the number of moles of OH⁻ that are unreacted after the neutralization is complete,

We will determine the number of moles of KOH remaining in the solution after the neutralization reaction is complete

First, we will determine the number of moles of each reactant present

  • For H₂SO₄

Volume = 22.36 mL = 0.02236 L

Concentration = 0.142 M

From the formula,

Number of moles = Concentration × Volume

∴ Number of moles of H₂SO₄ = 0.142 × 0.02236

Number of moles of H₂SO₄ = 0.00317512 moles

  • For KOH

Volume = 46.21 mL = 0.04621 L

Concentration = 0.304 M

∴ Number of moles of KOH = 0.304 × 0.04621

Number of moles of KOH = 0.01404784 moles

(NOTE: This is equal to the number of moles of OH⁻ at the beginning of the reaction)

From the equation of reaction, we have that

1 mole of H₂SO₄ will neutralize 2 moles KOH

Therefore,

0.00317512 moles of H₂SO₄ will neutralize 2×0.00317512 moles KOH

2×0.00317512 = 0.00635024 moles

This means only 0.00635024 moles of KOH reacted

(NOTE: This is equal to the number of moles of OH⁻ that reacted)

Now, for the number of moles of unreacted OH⁻

Number of moles of unreacted OH⁻ = Total number of moles OH⁻ at the beginning of the reaction - Number of moles of OH⁻ that reacted

∴ Number of moles of unreacted OH⁻ = 0.01404784 moles - 0.00635024 moles

Number of moles of unreacted OH⁻ = 0.0076976 moles

Number of moles of unreacted OH⁻ ≅ 0.00770 moles OR 7.70×10⁻³ moles

Hence, the number of moles of OH⁻ that are unreacted in the solution after the neutralization is complete is 0.00770 moles OR 7.70×10⁻³ moles

Learn more here: https://brainly.com/question/13190988