A solution of NaOH with pH 13.68 requires 35.00 mL of 0.128 M HClO_4_ to reach the equivalence point. How do you calculate [Na^+^] and [ClO_4_^-^] at the aquicalence point? (volumes are additive)

We've been here before. What's the problem? You have M NaOH (from the pH = 13.68), you have mL HClO4 and you have M HClO4. There is only one unknown in

mL NaOH x M NaOH = mL HClO4 x M HClO4.

To calculate the concentration of Na^+ and ClO_4^- at the equivalence point, we need to use the concept of stoichiometry and the information given in the question.

First, let's break down the steps to approach this problem:

Step 1: Find the moles of HClO_4 used in the reaction.
Step 2: Determine the stoichiometry between HClO_4 and NaOH to find the moles of NaOH.
Step 3: Calculate the concentration of Na^+ and ClO_4^- at the equivalence point using the moles of NaOH and the volume of the NaOH solution.

Now let's calculate step-by-step:

Step 1: Moles of HClO_4 used
Moles = Molarity x Volume (in L)
Moles of HClO_4 = 0.128 mol/L x 0.03500 L
Moles of HClO_4 = 0.00448 mol

Step 2: Stoichiometry between HClO_4 and NaOH
From the balanced equation, we know that 1 mol of HClO_4 reacts with 1 mol of NaOH.
Therefore, the moles of NaOH = moles of HClO_4 = 0.00448 mol

Step 3: Concentration of Na^+ and ClO_4^- at the equivalence point
To calculate the concentration, divide the moles of NaOH by the total volume at the equivalence point.
We know that the total volume is 35.00 mL (given in the question).

Concentration = Moles / Volume
For Na^+:
[Na^+] = (0.00448 mol) / (0.03500 L)
[Na^+] = 0.1279 M

For ClO_4^-:
[ClO_4^-] = (0.00448 mol) / (0.03500 L)
[ClO_4^-] = 0.1279 M

Therefore, the concentration of Na^+ and ClO_4^- at the equivalence point is 0.1279 M.