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

To standardize the KOH solution, first calculate the moles of KHP and then use the titration volume to find the molarity of KOH. To find the molarity of the perchloric acid solution, use the volume of KOH required for neutralization and the standardized molarity of KOH.

Step-by-step explanation:

To find the molarity of the potassium hydroxide solution, use the molar mass of potassium hydrogen phthalate (204.22 g/mol) and the equation of the reaction, which is a simple neutralization where KHC₈H₄O₄reacts with KOH in a 1:1 ratio. Calculate the number of moles of KHC₈H₄O₄as follows:

1. Moles of KHC₈H₄O₄ = mass (g) / molar mass (g/mol) = 1.08 g / 204.22 g/mol = 0.00529 mol.
2. Since the molar ratio is 1:1, moles of KOH = moles of KHC8H4O4 = 0.00529 mol.
3. To find molarity (M) of KOH: M = moles of solute / volume of solution (L). Therefore, M KOH = 0.00529 mol / 0.0368 L = 0.1437 M.

Titration of Perchloric Acid with Standardized KOH

Using the standardized KOH solution to titrate perchloric acid (HClO₄), we follow a similar process as above. The reaction between HClO₄ and KOH is also a 1:1 stoichiometry. Thus:

1. Moles of KOH = volume (L) × molarity (M) = 0.0101 L × 0.1437 M = 0.0014537 mol.
2. Since the reaction is 1:1, moles of HClO₄ = moles of KOH = 0.0014537 mol.
3. Molarity of HClO₄ = moles / volume (L) = 0.0014537 mol / 0.0276 L = 0.0526739 M.

Answer 2

A. 0.143 M

B. 0.0523 M

Step-by-step 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) * V_(A) = M_(B) * V_(B)\\M_(A) = (M_(B) * V_(B))/(V_(A)) \\M_(A) = (0.143 M * 10.1mL)/(27.6mL)\\M_(A) =0.0523 M`