#1

Can anyone tell me how to solve this problem:

The concentration of H2SO4 in a solution is 3.60x10^-6 M. Assume the acid dissociates completely in solution.

What is the [H3O+] in the solution?

What is the [OH-] in the solution?

What is the pH of the solution?

What is the pOH of the solution?

The concentration of H2SO4 in a solution is 3.60x10^-6 M. Assume the acid dissociates completely in solution.

What is the [H3O+] in the solution?

What is the [OH-] in the solution?

What is the pH of the solution?

What is the pOH of the solution?

#2

your asking this question in the pit????

i laugh at you

i laugh at you

#3

1. That's just the molarity times 2.

2. The antilog of (14+log(#1)). There's an easier way, but I forget it.

3. -log(#1)

4. -log(#2)

I'm pretty sure about all of that except #2. Check your book, and look for something about the number 1 x 10^14, that's ringing a bell.

2. The antilog of (14+log(#1)). There's an easier way, but I forget it.

3. -log(#1)

4. -log(#2)

I'm pretty sure about all of that except #2. Check your book, and look for something about the number 1 x 10^14, that's ringing a bell.

#4

it is 2.

like all other math/science qustions.

like all other math/science qustions.

#5

Assume the acid dissociates completely in solution.

Trick question

*spoiler alert*

It doesn't

#6

hrdcorelaxplaya can you elaborate on # 2 please

#7

^u ar smrt

edit: was talkin to hrdcorelaxplaya

edit: was talkin to hrdcorelaxplaya

*Last edited by kenan6346 at Apr 13, 2008,*

#8

Okay

Since H2SO4 is a strong acid, it dissociates fully.

H2SO4 + H2O -> HSO4- + H3O+

So, the concentration of H3O+ is the same as the concentration of H2SO4-, which is 3.60 x 10^-6. Next, you have to do

HSO4- + H20 -> SO4-2 + H3O+

You take the given Ka, and set it equal to x^2 / 3.60 x 10^-4. Add x to the H3O+ before, and you get the [H3O+]. 1 x 10^-14 / ans = [OH-].

Then, pH = -log[H3O+], and pOH = 14 - pH.

Since H2SO4 is a strong acid, it dissociates fully.

H2SO4 + H2O -> HSO4- + H3O+

So, the concentration of H3O+ is the same as the concentration of H2SO4-, which is 3.60 x 10^-6. Next, you have to do

HSO4- + H20 -> SO4-2 + H3O+

You take the given Ka, and set it equal to x^2 / 3.60 x 10^-4. Add x to the H3O+ before, and you get the [H3O+]. 1 x 10^-14 / ans = [OH-].

Then, pH = -log[H3O+], and pOH = 14 - pH.