A25.0 ml sample of 0.723 m hclo4 is titrated with a 0.27303 m koh solution. the h3o+ concentration after the addition of 66.2 ml of koh is m.

This doesn't need an ICE chart. Both will fully dissociate in water.

Assume HClO4 and KOH reacts with one another. All you need to do is determine how much HClO4 will remain after the reaction. Calculate pH.

Step 1:

write out balanced equation for the reaction

HClO4+KOH ⇔ KClO4 + H2O

the ratio of HClO4 to KOH is going to be 1:1. Each mole of KOH we add will fully react with 1 mole of HClO4

Step 2:

Determining the number of moles present in HClO4 and KOH

Use the molar concentration and the volume for each:
25 mL of 0.723 M HClO4

Covert volume from mL into L:
25 mL * 1L/1000mL = 0.025 L

Remember:

M = moles/L so we have 0.025 L of 0.723 moles/L HClO4

Multiply the volume in L by the molar concentration to get:

0.025L x 0.723mol/L = 0.0181 moles HClO4.

Add 66.2 mL KOH with conc.=0.273M
66.2mL*1L/1000mL = .0662 L
.0662L x 0.273mol/L = 0.0181 moles KOH

Step 3:

Determine how much HClO4 remains after reacting with the KOH.

Since both reactants fully dissociate and are used in a 1:1 ratio, we just subtract the number of moles of KOH from the number of moles of HClO4:

moles HClO4 = 0.0181; moles KOH = 0.0181, so 0.0181-0.0181 = 0

This means all of the HClO4 is used up in the reaction.

If all of the acid is fully reacted with the base, the pH will be neutral = 7.

Determine the H3O+ concentration:

pH = -log[H3O+]; [H3O+] = 10-pH = 10-7

The correct answer is 1.0x10-7.

1.0x10⁻⁷ M

Explanation:

1) Balanced chemical equation:


HClO₄ + KOH --> KClO₄ + H₂O


2) Theoretical mole ratios:


1 mol HClO₄ : 1 mol KOH : 1 mol KClO₄ + 1 mol H₂O

3) Calcualte the number of moles of HClO₄ in the sample:

M = n / V => n = M x V = 0.723 M x 0.025 l = 0.018075 mol


4) Calculate the number of moles of KOH in the solution:


M = n / V => n = M x V = 0.27303 M x 0.0662 l = 0.018075 mol


5) Then, the acid and the base are in the ratio 1:1 which is the same theoretical ratio, wich means that the final solution is fully neutralized.

That means that all the H₃O⁺ from the acid has reacted with the OH- from the base, forming neutral water.

Neutral water ionizes into 1x10⁻⁷ M H₃O+ ions and 1x10⁻⁷ M OH⁻ ions, meaning that the final solution has a concentraion of 1x10⁻⁷ M of H3O+.