A 15.0 ml sample of an unknown HClO4 solution requires 50.3 ml if 0.101 M NaOH for complete neutralization.

What was the concentration of the unknown HClO4 solution? The neutralization reaction is:

HClO4(aq)+ NaOh(aq)yields H2O(l)+ NaClO4 (aq)

If anyone can help me to set this up, thank you

See your post above.

To find the concentration of the unknown HClO4 solution, you need to use the concept of stoichiometry and the volume and concentration of the NaOH solution used for neutralization. Here's how you can set up the problem:

1. Write the balanced chemical equation for the neutralization reaction:
HClO4(aq) + NaOH(aq) → H2O(l) + NaClO4(aq)

2. Determine the moles of NaOH used in the reaction:
Moles of NaOH = Volume of NaOH solution (in liters) × Concentration of NaOH solution (in moles per liter)
Moles of NaOH = 50.3 ml × 0.101 M = 5.0973 mmol

Note: Converted the given volume of NaOH solution from milliliters (ml) to liters (L).

3. Use the stoichiometry of the balanced equation to determine the moles of HClO4 reacted:
From the balanced equation, we can see that the ratio of NaOH to HClO4 is 1:1. This means that the moles of NaOH used is equal to the moles of HClO4 reacted.

Moles of HClO4 reacted = Moles of NaOH = 5.0973 mmol

4. Calculate the concentration of the HClO4 solution:
Concentration of HClO4 solution = Moles of HClO4 (in moles) / Volume of HClO4 solution (in liters)

Given that the volume of the HClO4 solution is 15.0 ml, convert it to liters:
Volume of HClO4 solution = 15.0 ml = 0.015 L

Concentration of HClO4 solution = (5.0973 mmol) / (0.015 L)
Concentration of HClO4 solution = 339.82 mmol/L
Concentration of HClO4 solution ≈ 0.340 M

Therefore, the concentration of the unknown HClO4 solution is approximately 0.340 M.