An aqueous solution of perchloric acid is standardized by titration with a 0.111 M solution of calcium hydroxide. If 13.7 mL of base are required to neutralize 17.5 mL of the acid, what is the molarity of the perchloric acid solution?

0.174M of  HClO₄

Explanation:

The reaction of perchloric acid (HClO₄) with calcium hydroxide (Ca(OH)₂) is:

2 HClO₄ + Ca(OH)₂ → Ca(ClO₄)₂ + 2 H₂O

Moles of 13.7mL of Ca(OH)₂ 0.111M are:

0.0137L ₓ (0.111mol / 1L) = 1.52x10⁻³ moles of Ca(OH)₂. Based on the reaction, 2 moles of acid requires 1 mole of base, thus:

1.52x10⁻³ moles of Ca(OH)₂ ₓ (2 mol HClO₄ / 1 mol Ca(OH)₂) = 3.04x10⁻³ moles of HClO₄

As you neutralize 17.5mL, molarity of perchloric acid is:

3.04x10⁻³ moles of HClO₄ / 0.0175L = 0.174M of  HClO₄

Explanation:

1000cm³ = 1mL = 1dm³

Ca(OH)₂ + 2HCLO₄ → Ca(ClO)₄ + 2H₂O

If 0.111M = 1000cm³

X M = 13.7 cm³

X = (13.7 * 0.111) / 1000 = 0.00152moles

From the stoichiometry if the reaction,

1 mole of Ca(OH)₂ is required to neutralise 2 moles of HCLO₄

1 mole Ca(OH)₂ = 2 moles of HCLO₄

0.00152 moles of Ca(OH)₂ = y

y = ( 0.00152 * 2 ) / 1 = 0.0030414 moles

If 0.0030414 moles of HCLO₄ is present in 17.5cm³,

Z moles would be present in 1000cm³

0.0030414 moles = 17.5 cm³

Z moles = 1000cm³

Z = (0.0030414 * 1000) / 17.5

Z = 0.1737M

Molarity of HCLO₄ is 0.1737M

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