Question

1. It requires 10.9 mL of 0.50 M NaOH to neutralize 15.0 mL of an unknown H3PO4 solution. a. Balance the reaction equation. H3PO4 + NaOH Na3PO4 + H2O b. Calculate the initial H3PO4 concentration. 2. It requires 28.1 mL of 16.0 mM H2SO4 to neutralize 11.0 mL of an unknown Ca(OH)2 solution. a. Balance the reaction equation. H2SO4 + Ca(OH)2 ⇋ CaSO4 + H2O b. Calculate the initial Ca(OH)2 concentration.

Answer 1

Q. 1

Given:

Volume of NaOH= 10.9 mL = 0.0109 L , [NaOH] = 0.50 M

Volume of H­­₃PO₄ = 15.0 mL = 0.015 L , [H₃PO₄] = Unknown

a) Reaction:

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

b) In order to calculate concentration of phosphoric acid we must determine the mole ratio between sodium hydroxide and phosphoric acid.

Mol ratio between phosphoric acid and sodium hydroxide is 1 : 3 . This indicates that 1 mol of phosphoric acid needs 3 moles of sodium hydroxide.

Lets calculate moles of sodium hydroxide from its volume and molarity.

n NaOH = Volume in L x molarity = 0.0109 L x 0.50 = 0.00545 mol NaOH

n H₃PO₄ = n NaOH x 1 mol H₃PO₄ / 3 mol NaOH = 0.01635 mol H₃PO₄

[H_3­PO₄] = n H₃PO₄ / Volume in L = 0.01635 mol H₃PO­₄ / 0.015 L = 1.1 M

Thefore initial concentration of phosphoric acid is 1.1 M

Q. 2

Given:

Volume of Ca(OH)₂ = 11.0 mL = 0.011 L , concentration of calcium hydroxide is unknown

Volume of H­­₂SO₄ = 28.1 mL = 0.0281 L , [H₂SO₄ ] = 16.0 mM = 0.016 M

a) Reaction:

H₂SO₄ (aq) + Ca(OH)₂ (aq) --- > CaSO₄(aq) + 2H2O (l)

b)

Calculation of moles of sulfuric acid

n sulfuric acid = 0.0281 L x 0.016 M = 0.00045 mol

Calculation of moles of calcium hydroxide

Mol calcium hydroxide

= 0.00045 mol sulfuric acid x 1 mol calcium hydroxide / 1 mol sulfuric acid

=0.00045 mol calcium hydroxide.

[Ca(OH)₂] = 0.00045 mol / 0.011 L = 0.041 M