Approximately how much water should be added to 10.0 mL of 12.4 M HCl so that it has the same pH as 0.90 M acetic acid (Ka = 1.8 10-5)?

YOU DONOT KNOW ANY THING BROO

You're dumb, you don't know how to explain the answer.

To find out how much water should be added to the 10.0 mL of 12.4 M HCl, we need to calculate the final concentration of HCl and compare it to the concentration of acetic acid.

First, let's calculate the initial moles of HCl present in the 10.0 mL solution:

moles of HCl = concentration (M) × volume (L)
moles of HCl = 12.4 M × 0.010 L
moles of HCl = 0.124 mol

Since the concentration of acetic acid (CH3COOH) is given, we know that the concentration of H+ ions (H3O+) is equal to the concentration of CH3COO- ions.

Since acetic acid is a weak acid, it will only partially ionize in water. The ionization reaction for acetic acid is:

CH3COOH ⇌ H3O+ + CH3COO-

The equilibrium constant for this reaction, Ka, is given as 1.8 × 10^-5.

To find the concentration of H3O+ ions from acetic acid, we can use the expression for Ka:

Ka = [H3O+][CH3COO-] / [CH3COOH]

Since [H3O+] = [CH3COO-], the equation simplifies to:

[H3O+]² / [CH3COOH] = Ka

Taking the square root of both sides and rearranging the equation, we get:

[H3O+] = √(Ka × [CH3COOH])

[H3O+] = √(1.8 × 10^-5 × 0.90 M)

[H3O+] = 6.74 × 10^-3 M

So, in order to achieve the same pH as the 0.90 M acetic acid, we need the final concentration of HCl to be 6.74 × 10^-3 M.

Now we can calculate the new volume of the HCl solution by rearranging the concentration equation:

concentration (M) = moles / volume (L)

volume (L) = moles / concentration (M)

volume (L) = 0.124 mol / 6.74 × 10^-3 M

volume (L) = 18.38 L

We started with a 10.0 mL (0.010 L) solution of HCl, so the amount of water we need to add is:

volume of water = final volume - initial volume
volume of water = 18.38 L - 0.010 L
volume of water = 18.37 L

Therefore, approximately 18.37 liters of water should be added to 10.0 mL of 12.4 M HCl to achieve the same pH as 0.90 M acetic acid.

First you must calculate (H^+) for 0.90M acetic acid (HAc).

.........HAc ==> H^+ + Ac^-
I.......0.90.....0......0
C........-x......x......x
E.....0.90-x.....x......x

K = 1.8E-5 = (H^+)(Ac^-)/(HAc)
Plug in the E line and solve for x = (H^+).

So you want (H^+) in HCl to be the same.
10 mL x 12.4M = mL x MHAc from above.
mL then will be the total; subtract 10 from that to find how much must be added to 10. The reason the problem says "approximately" is because the total is not the sum of the two. Volumes are not additive although they would be so close in this instance that I expect we couldn't measure the difference.