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-BARSIC- [3]
3 years ago
11

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 o

f OH- are unreacted in the solution after the neutralization is complete?
Chemistry
1 answer:
Tom [10]3 years ago
5 0

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:

<em>Moles H₂SO₄:</em>

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

<em>Moles KOH:</em>

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 =

<h3>0.01228 moles OH⁻</h3>
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