Question

Write the balanced neutralization reaction between H2SO4 and KOH in aqueous solution. Phases are optional.

0.450 L of 0.490 M H2SO4 is mixed with 0.400 L of 0.210 M KOH. What concentration of sulfuric acid remains after neutralization?

Answer 1

Answer – Given, [H₂SO₄] = 0.490M , volume = 0.450 L

[KOH] = 0.210 M , volume 0.400 L

Step 1) Balanced neutralization reaction between H₂SO₄ and KOH in aqueous solution-

H₂SO₄ + 2 KOH -------> K₂SO₄ + 2H₂O

Step 2) calculate the moles of each

Moles of H₂SO₄ = 0.490 M * 0.450 L = 0.2205 moles

Moles of KOH = 0.210 M * 0.400 L = 0.0840 moles

Step 3) Calculating the limiting reactant

Moles of KOH is less than moles of H₂SO₄ and there is also mole ratio 2:1

So limiting reactant is KOH

So moles of H₂SO₄ reacted –

2 moles of KOH = 1 moles of H₂SO₄

So, 0.0840 moles KOH = ?

= 0.0840 moles KOH * 1 moles of H₂SO₄ / 2 moles of KOH

= 0.0420 moles o H₂SO₄

Step 4) Calculating the concentration of sulfuric acid remains after neutralization

We calculate the moles of H₂SO₄ reacted, so remaining moles of H₂SO₄ is

remaining moles of H₂SO₄ = 0.2205 mole – 0.0420 moles

                                          = 0.1785 moles

Total volume after neutralization = 0.450 L + 0.400 L

                                               = 0.850 L

So, concentration of sulfuric acid remains after neutralization

[H₂SO₄] = 0.1785 moles / 0.850 L

              = 0.210 M

So, concentration of sulfuric acid remains after neutralization is 0.210 M