Question

A student is asked to standardize a solution of potassium hydroxide. He weighs out 1.08 g potassium hydrogen phthalate (KHC8H4O4, treat this as a monoprotic acid). It requires 36.8 mL of potassium hydroxide to reach the endpoint. A. What is the molarity of the potassium hydroxide solution? M This potassium hydroxide solution is then used to titrate an unknown solution of perchloric acid. B. If 10.1 mL of the potassium hydroxide solution is required to neutralize 27.6 mL of perchloric acid, what is the molarity of the perchloric acid solution? M

Answer 1

Answer:

A. 0.143 M

B. 0.0523 M

Explanation:

A.

Let's consider the neutralization reaction between potassium hydroxide and potassium hydrogen phthalate (KHP).

KOH + KHC₈H₄O₄ → H₂O + K₂C₈H₄O₄

The molar mass of KHP is 204.22 g/mol. The moles corresponding to 1.08 g are:

1.08 g × (1 mol/204.22 g) = 5.28 × 10⁻³ mol

The molar ratio of KOH to KHC₈H₄O₄ is 1:1. The reacting moles of KOH are 5.28 × 10⁻³ moles.

5.28 × 10⁻³ moles of KOH occupy a volume of 36.8 mL. The molarity of the KOH solution is:

M = 5.28 × 10⁻³ mol / 0.0368 L = 0.143 M

B.

Let's consider the neutralization of potassium hydroxide and perchloric acid.

KOH + HClO₄ → KClO₄ + H₂O

When the molar ratio of acid (A) to base (B) is 1:1, we can use the following expression.

$M_{A} \times V_{A} = M_{B} \times V_{B}\\M_{A} = \frac{M_{B} \times V_{B}}{V_{A}} \\M_{A} = \frac{0.143 M \times 10.1mL}{27.6mL}\\M_{A} =0.0523 M$