A 25.0 mL sample of 0.723 M HClO₄ is titrated with a 0.273 M KOH solution. The H₃O⁺ concentration after the addition of 60.0 mL of KOH is __________ M. A)0.0181 B)0.430 C)0.0200 D)0.273 E)none of the above

Question

Quizplus · Accepted Answer

The balanced chemical equation for the reaction between HClO and KOH is:

HClO + KOH → KClO + H2O

From the equation, we can see that the stoichiometry of the reaction is 1:1 between HClO and KOH. This means that for every mole of KOH added, one mole of HClO is consumed.

To determine the moles of HClO in the 25.0 mL sample, we use the following formula:

moles HClO = Molarity × Volume (in liters)

moles HClO = 0.723 mol/L × 0.0250 L = 0.0181 moles

Since the stoichiometry of the reaction is 1:1, the moles of KOH added are also 0.0181 moles.

The total volume after adding 60.0 mL of the 0.273 M KOH solution to the HClO solution is:

Vtotal = Vinitial + Vadded = 25.0 mL + 60.0 mL = 85.0 mL = 0.0850 L

The concentration of the resulting solution can be calculated using the following formula:

Molarity = moles of solute ÷ volume of solution (in liters)

The moles of HClO remaining after the titration can be calculated as follows:

moles HClO = moles initial - moles KOH added
moles HClO = 0.0181 moles - 0.0181 moles = 0 moles

This means that all of the HClO in the initial solution is consumed by the KOH. The moles of KClO formed can be calculated as follows:

moles KClO = moles KOH added = 0.0181 moles

The concentration of KClO in the resulting solution is:

Molarity = 0.0181 moles ÷ 0.0850 L = 0.213 M

However, we are asked to find the concentration of HSO4+ in the resulting solution. Since HSO4+ is a spectator ion in the reaction, its concentration remains the same before and after the titration. Therefore, the concentration of HSO4+ in the resulting solution is equal to the concentration of the initial HClO solution:

HSO4+ concentration = 0.723 M

Therefore, the answer is C) 0.0200 M (rounded to three significant figures).

A 250 ML Sample of 0