60.0 mL of 0.100 M Ca(OH)2 solution is mixed with 50.0 mL of 0.100 M HNO3. Calculate the volume of 0.100 M H2SO4 needed to neutralize the resulting mixture.

The volume of 0.100 M H₂SO₄ needed to neutralize the resulting mixture containing an excess of 10 ml of 0.100 M Ca(OH)₂  is 10 ml of 0.100 M H₂SO₄ .

Explanation:

Ca(OH)₂ + HNO₃ = Ca(NO₃)₂ + H₂O

From the above reaction;

1 mole of Ca(OH)₂ is required to neutralize 1 mole of HNO₃ also

0.1 mole of Ca(OH)₂ will be required to neutralize 0.1 mole of HNO₃

Since there is 60 ml of 0.100 M Ca(OH)₂ and  50.0 mL of 0.100 M HNO₃, then on completion of neutralization, there will be 10 ml of 0.100 M Ca(OH)₂ left to be consumed

When H₂SO₄ is added to the solution, the reaction will be as follows;

Ca(OH)₂ + H₂SO₄ → CaSO₄ + H₂O

Therefore we have;

1 mole of  H₂SO₄ is required to completely react with 1 mole of Ca(OH)₂

Hence,

0.1 mole of  H₂SO₄ will completely react with 0.1 mole of Ca(OH)₂

Therefore the volume of 0.100 M H₂SO₄  required to neutralize  the resulting mixture containing an excess of 10 ml of 0.100 M Ca(OH)₂  is 10 ml

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