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 later posts.

To find the concentration of the unknown HClO4 solution, you can use the concept of stoichiometry and the balanced equation of the neutralization reaction provided. Here's how you can set it up:

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

2. Use the given information to create a stoichiometric ratio:
From the equation, you can see that the ratio between HClO4 and NaOH is 1:1. In other words, for every 1 mole of HClO4, 1 mole of NaOH is needed.

3. Convert the volume and concentration of NaOH to moles:
Given that the volume of NaOH used is 50.3 ml and the concentration of NaOH is 0.101 M, you can convert this to moles using the formula:

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

Convert the volume of NaOH to liters by dividing it by 1000:
50.3 ml ÷ 1000 = 0.0503 L

Now multiply the converted volume by the concentration to find moles:
moles NaOH = 0.0503 L × 0.101 M = 0.0050833 moles NaOH

4. Use the stoichiometric ratio to find moles of HClO4:
Since the ratio between HClO4 and NaOH is 1:1, the moles of HClO4 will also be 0.0050833 moles.

5. Convert the volume of HClO4 to liters:
Given that the volume of HClO4 used is 15.0 ml, convert it to liters by dividing it by 1000:
15.0 ml ÷ 1000 = 0.015 L

6. Calculate the concentration of HClO4:
To find the concentration, divide the moles of HClO4 by the volume in liters:

concentration HClO4 = moles HClO4 ÷ volume HClO4 (L)

concentration HClO4 = 0.0050833 moles ÷ 0.015 L = 0.3389 M

Therefore, the concentration of the unknown HClO4 solution is 0.3389 M.