Buffer/Titration Chemistry Problem??! PLEASE HELP - I AM LOST!!!

Question

A 25.0 mL sample of 0.723 M HClO4 is titrated with a 0.273 M KOH solution. The H3O+ concentration after the addition of 66.2 mL of KOH is __________ M.

0.723
1.00 × 10-7
2.81 × 10-13
0.439
0.273

 

I think well with ICE charts, but I don't even know how to set one up for this. I am so so so lost and would love if you could show each step along the way!!! I am so stressed out that I have no clue what I am doing!!!!!

Answer 1

Hi Angelica,

HClO₄ is a strong acid and KOH is a strong base, so you don't need to set up a ICE chart because both will fully dissociate in water.

Basically, you can assume that the HClO₄ and KOH will react fully with one another, so you just need to determine how much HClO₄ will remain after the reaction and calculate pH based on that.

 

Step 1: Write out a balanced equation for the reaction:

HClO₄+KOH ⇔ KClO4 + H₂O

Both sides balance without needing to add any reaction coefficients, so 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: Determine the number of moles of HClO4 and KOH present.

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.

 

We are adding 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

So the correct answer is 1.0x10^-7.

Message me if you have questions!