A 27.4 mL sample of an unknown HClO4 solution requires 45.4 mL of 0.101 M NaOH for complete neutralization. What was the concentration of the unknown HClO4 solution? The neutralization reaction is shown below. HClO4(aq) + NaOH(aq) H2O(l) + NaClO4(aq)

To determine the concentration of the unknown HClO4 solution, you need to use the stoichiometry of the neutralization reaction and apply the concept of molarity.

First, we need to write and balance the equation for the neutralization reaction:

HClO4(aq) + NaOH(aq) → H2O(l) + NaClO4(aq)

From the balanced equation, we can see that the mole ratio between HClO4 and NaOH is 1:1. This means that 1 mole of HClO4 reacts with 1 mole of NaOH.

Next, let's calculate the number of moles of NaOH used in the neutralization:

moles of NaOH = volume of NaOH (L) × concentration of NaOH (M)

Given:
Volume of NaOH = 45.4 mL = 0.0454 L
Concentration of NaOH = 0.101 M

moles of NaOH = 0.0454 L × 0.101 M
moles of NaOH = 0.0045954 mol

Since the mole ratio between HClO4 and NaOH is 1:1, we can conclude that 0.0045954 mol of HClO4 was present in the 27.4 mL sample.

Now, we can calculate the concentration of HClO4:

concentration of HClO4 = moles of HClO4 / volume of HClO4 (L)

Given:
Volume of HClO4 = 27.4 mL = 0.0274 L

concentration of HClO4 = 0.0045954 mol / 0.0274 L
concentration of HClO4 ≈ 0.1675 M

Therefore, the concentration of the unknown HClO4 solution is approximately 0.1675 M.