Equilibrium Equation?

**Ponderous** · Dec 6th 2009, 10:11 AM

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:

. 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?

I've had already tried: x^2/0.1-x = 1.0 X 10^-7

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.

**e^(i*pi)** · Dec 6th 2009, 10:15 AM

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:

. 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?

I've had already tried: x^2/0.1-x = 1.0 X 10^-7

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.

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

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

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

You can solve this just like a normal quadratic

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

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