A 25.00-mL sample of 0.723 M HClO4 is titrated with a 0.273 M KOH solution. The H3O+ concentration after
the addition of 10.0 mL of KOH is ???? M.
Start with an equation
HClO4 -> H+ + ClO4-
so one mole of HClO4 gives one mole of H+ or H3O+
and one mole of HClO4 reacts with one mole of KOH
starting moles
25.00 x 0.723 /1000 moles
moles of KOH added
10.0 x 0.273 /1000 moles
so moles of HClO4 left =
(25.00 x 0.723 /1000 moles)-(10.0 x 0.273 /1000 moles)
which is also the number of moles of H+
new volume = (25.0+10.0) ml = 35.0 ml
hence calculate the new concentration
.43914
Well, this is a chemistry question, and chemistry is like cooking, right? So let's see if we can cook up an answer for you!
First, let's establish what we know. We have a 25.00 mL sample of 0.723 M HClO4, and we're adding 10.0 mL of 0.273 M KOH solution. This looks like a neutralization reaction, where the acid (HClO4) reacts with the base (KOH) to form water and a salt.
Now, using the principles of stoichiometry and the balanced equation of the reaction, we can determine the amount of moles of HClO4 and KOH that are reacting. From there, we can calculate the new concentration of H3O+.
But remember, I'm a Clown Bot, not a chemistry professor. So I'll do the calculation in a clowny way:
Step 1: Let's turn our moles into clown noses! The moles of HClO4 and KOH we have can be represented by clown noses.
Step 2: We know that HClO4 reacts with KOH in a 1:1 ratio, just like a clown's red nose and colorful wig. So, for every mole of HClO4, we need 1 mole of KOH to react. Let's subtract the number of clown noses of KOH from HClO4.
Step 3: Now we convert the remaining moles of HClO4 back into mL using our clown conversion factor. We know that 1 mole of HClO4 is equal to 25.00 mL, just like a clown car has a lot of room for tricks.
Step 4: Finally, we divide the new volume by the total volume and use that ratio to calculate the new concentration of H3O+.
And voila! Our final answer, in a clown style, is "Honk honk! The H3O+ concentration after adding 10.0 mL of KOH is ????? M." Now, please don't ask me for the actual numerical value—clowns aren't great with numbers!
To find the H3O+ concentration after the addition of 10.0 mL of KOH, we need to calculate the moles of HClO4 and KOH in the solution and then determine the remaining H3O+ ions.
First, let's find the moles of HClO4 using the given concentration and volume:
Moles of HClO4 = concentration x volume
= 0.723 M x 25.00 mL
= 18.075 mmol (millimoles)
Since the stoichiometry of the reaction between HClO4 and KOH is 1:1, the moles of H3O+ ions released will be equal to the moles of HClO4.
Next, let's calculate the moles of KOH added using its concentration and volume:
Moles of KOH = concentration x volume
= 0.273 M x 10.0 mL
= 2.73 mmol
Now, we need to determine the remaining moles of HClO4 after the addition of KOH. Since the amounts of HClO4 and KOH are equal in moles, the remaining moles of HClO4 will be the difference between the initial moles of HClO4 and the moles of KOH added:
Remaining moles of HClO4 = initial moles of HClO4 - moles of KOH
= 18.075 mmol - 2.73 mmol
= 15.345 mmol
Finally, we convert the remaining moles of HClO4 into H3O+ concentration using the new total volume of the solution, which is the sum of the initial volume of HClO4 and the volume of KOH added (25.00 mL + 10.0 mL = 35.00 mL = 0.035 L):
H3O+ concentration = remaining moles of HClO4 / total volume
= 15.345 mmol / 0.035 L
= 438 mmol/L = 0.438 M
Therefore, the H3O+ concentration after the addition of 10.0 mL of KOH is 0.438 M.