A 25.0 mL sample of 0.723 M HClO₄is titrated with a 0.273 M KOH solution. The H₃O⁺ concentration after the addition of 10.0 mL of KOH is M. A) 1.00 × 10^-⁷ B) 0.273 C) 0.723 D) 0.440 E) 2.81 × 10^-¹³

Question

Quizplus · Accepted Answer

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

HClO + KOH → KClO + H2O

This equation shows that for every mole of KOH used, one mole of HClO is consumed. Therefore, the number of moles of HClO initially present in the 25.0 mL sample is:

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

When 10.0 mL of 0.273 M KOH is added, the number of moles of KOH added is:

moles KOH = M × V = 0.273 mol/L × 0.0100 L = 0.00273 mol

Since KOH and HClO react in a 1:1 ratio, the number of moles of HClO remaining after the addition of KOH is:

moles HClO remaining = moles HClO initially present - moles KOH added
moles HClO remaining = 0.0181 mol - 0.00273 mol = 0.0154 mol

The volume of the solution after the addition of KOH is:

final volume = initial volume + volume of KOH added
final volume = 25.0 mL + 10.0 mL = 35.0 mL = 0.0350 L

Therefore, the concentration of HClO remaining in the solution is:

[HClO] = moles HClO remaining / final volume
[HClO] = 0.0154 mol / 0.0350 L = 0.440 mol/L

So the answer is D: 0.440.

A 250 ML Sample of 0