0.550 L of 0.410 M H2SO4 is mixed with 0.500 L of 0.270 M KOH. What concentration of sulfuric acid remains after neutralization?

Answer 1

First write the balanced equation:

H₂SO₄ + 2KOH ==> K₂SO₄ + 2H₂O

Next, find moles H₂SO₄ and moles of KOH present:

moles H₂SO₄ = 0.550 L x 0.410 moles/L = 0.2255 moles H₂SO₄

moles KOH = 0.500 L x 0.270 moles/L = 0.135 moles KOH

Using the stoichiometry of the balanced equation, we can determine how many moles of H₂SO₄ have been neutralized with the KOH, and then how many moles remain:

0.135 moles KOH x 1 moles H₂SO₄/2 moles KOH = 0.0675 moles H₂SO₄ neutralized

Since we started with 0.2255 moles of H₂SO₄, the amount remaining after neutralization is the difference:

0.2255 moles - 0.0675 moles = 0.158 moles H₂SO₄ remaining.

The concentration of remaining H₂SO₄ is the moles remaining/total volume in liters:

total volume = 0.550 L + 0.500 L = 1.05 L

Final concentration of H₂SO₄ = 0.158 moles/1.05 L = 0.150 M