Question

g While performing a neutralization reaction, Jonna added 21.73 mL of 0.142 M H2SO4 to 48.93 mL of 0.377 M KOH. How many moles of OH- are unreacted in the solution after the neutralization is complete?

Answer 1

Answer:

0.01228 moles OH⁻

Explanation:

The reaction of H₂SO₄ with KOH is:

H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O

Thus, we need to determine the moles of H2SO4 and KOH that reacts:

Moles H₂SO₄:

0.02173L * (0.142mol/L) = 3.086x10⁻³mol

Moles KOH:

0.04893L * (0.377mol/L) = 0.01845 mol KOH

The moles of KOH that reacts are:

3.086x10⁻³mol H₂SO₄ * (2 mol KOH / 1 mol H₂SO₄) = 6.172x10⁻³ moles KOH

And will remain in solution:

0.01845 mol KOH - 6.172x10⁻³ moles KOH = 0.01228 moles of KOH =

0.01228 moles OH⁻