Acid base titration

Question

59.) What is the molarity of a solution of HCl if 5.00 mL of the HCI solution is titrated with 28.6 mL of 0.145 M NaOH solution?

HCl(aq) + NaOH(aq) -------> NaCl(aq) + H₂O(l)

61.) If 38.2 mL of a 0.163 M NaOH is used to neautralize 25.0 mL of a H₂SO₄ solution, what is the molarity of the HSO solution?

H₂SO₄(aq) + 2NaOH(aq) ---------> K₂SO₄(aq) + 2H₂O(l)

63.) A solution of 0.204 M NaOH is used to neutralize 50.0 mL of a 0.0224 M H₃PO₄(aq) solution. What volume, in milliliters, of the NaOH solution is required to reach the endpoint?

H₃PO₄(aq) + 3NaOH(aq) --------->Na₃PO₄(aq) + 3H₂O(l)

Answer 1

These are all problems dealing with the same concept, which is titration and stoichiometry. If you know the moles of the titrant, you can find moles of the analyte. If you know the volume of analyte, you can then determine molarity of analyte.

59). HCl(aq) + NaOH(aq) ==> NaCl(aq) + H2O(l)

moles NaOH = 28.6 ml NaOH x 1 L / 1000 ml x 0.145 mol/L = 0.004147 moles NaOH

moles HCl = 0.004147 mols NaOH x 1 mol HCl / mol NaOH = 0.004147 moles HCl

Molarity of HCl = mols HCl/L = 0.004147 mols / 5.00 ml x 1000 ml / L = 0.829 mol/L = 0.829 M

61). H₂SO₄(aq) + 2NaOH(aq) ---------> K₂SO₄(aq) + 2H₂O(l)

Do the same way as #59 but note it takes TWO moles NaOH for each 1 mol H2SO4. Use this stoichiometry when calculating moles of H2SO4 present.

63). H₃PO₄(aq) + 3NaOH(aq) --------->Na₃PO₄(aq) + 3H₂O(l)

This problem is only slightly different in that it is asking for the VOLUME of NaOH needed.

moles H3PO4 present = 50.00 ml x 1 L /1000 ml x 0.0224 mol/L = 0.00112 mols H3PO4

moles NaOH needed = 0.00112 mols H3PO4 x 3 mol NaOH/mol H3PO4 = 0.00336 mols NaOH needed

Volume of 0.204 M NaOH needed = 0.00336 mols NaOH x 1 L / 0.204 mols = 0.0165 L = 16.5 mls