Question

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?

Answer 1

7.623 x 10⁻³ mol OH⁻

Step-by-step 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 *
  `(2mmolOH^-)/(1mmolH_2SO_4)` = 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⁻