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

2HNO3 + Ca (OH) 2 - --> Ca (NO3) 2 + 2H2O

moles HNO3---> (0.2545 mol/L) (0.04587 L) = 0.011674 mol

use the 2 : 1 molar ratio of HNO3 to Ca (OH) 2:

2 is to 1 as 0.011674 mol is to x

x = 0.005837 mol of Ca (OH) 2

molarity of Ca (OH) 2 - --> 0.005837 mol / 0.02162 L = 0.2700 M (to four sig figs)