# Equilibrium Equation?

• Dec 6th 2009, 10:11 AM
Ponderous
Equilibrium Equation?
A saturated solution of hydrogen sulfide (0,1 M concentration) dissociates into cation [H+] and anion [HS-], where [H+] = [HS-]. When this solution dissociates, the following equilibrium equation is established: http://www.webassign.net/cgi-bin/sym...%2A%2A%28-7%29. Find [H+]. (If [H+] = x, then [HHS] = 0.1 - x.)

I found this problem in my algebra book and I'm not entirely sure how to do it. I've took chemistry before but it's been quite a while, so..

I was wondering if i can seek some assistance, please?

To: x^2 = 1.0 X 10^-7/0.1-x

Then: x^2 = 0.1-x/1.0 X 10^7

I'm unsure what to do from there.
• Dec 6th 2009, 10:15 AM
e^(i*pi)
Quote:

Originally Posted by Ponderous
A saturated solution of hydrogen sulfide (0,1 M concentration) dissociates into cation [H+] and anion [HS-], where [H+] = [HS-]. When this solution dissociates, the following equilibrium equation is established: http://www.webassign.net/cgi-bin/sym...%2A%2A%28-7%29. Find [H+]. (If [H+] = x, then [HHS] = 0.1 - x.)

I found this problem in my algebra book and I'm not entirely sure how to do it. I've took chemistry before but it's been quite a while, so..

I was wondering if i can seek some assistance, please?

To: x^2 = 1.0 X 10^-7/0.1-x

Then: x^2 = 0.1-x/1.0 X 10^7

I'm unsure what to do from there.

$\displaystyle \frac{x^2}{0.1-x} = 10^{-7}$

$\displaystyle x^2 = 10^{-8} - x \times 10^{-7}$

$\displaystyle x^2+x \times 10^{-7} - 10^{-8} = 0$

You can solve this just like a normal quadratic

Just to point out that since $\displaystyle (10^{-7})^2 << 4 \times 1 \times 10^{-8}$ you may take $\displaystyle b^2-4ac = -4ac$