Eryn M. answered • 05/27/19

Experienced and Effective STEM Tutor

Hello Mikaela,

This is an RICE table problem!

First lets start with our **R**eaction: __HA → H__^{+}__ + A__^{-}

Plug in the** I**nitial Concentrations: 0.10M 0M 0M

*If you are not explicitly given an initial concentration you can assume it's just going to be 0.

Now plug in the **C**hange -x +x +x

*Because the leading coefficient of every compound is just one, we can assume we're losing and gaining one x.

Find the **E**quilibrium concentrations:** 0.10-x x x **

Now we know that K_{a} = [products]/[reactants] at equilibrium so we have

K_{a} = __x • x __ = 4.0 x 10^{-9} (given to you)

0.10-x

Now we can solve for x!

__x__^{2}__ __ = 4.0 x 10^{-9} => x^{2} = 4.0 x 10^{-9} (0.10-x) => x^{2} = 4.0 x 10^{-11} - 4.0 x 10^{-9} x

0.10-x

This is a quadratic so let's get everything to one side so we can set it equal to zero. I am going to subtract 4.0 x 10^{-11} - 4.0 x 10^{-9} x from both sides:

x^{2} - 4.0 x 10^{-11} + 4.0 x 10^{-9} x

and then rewrite this in standard form (ax^{2} + bx +c±):

x^{2} + 4.0 x 10^{-9} x - 4.0 x 10^{-11}

We can now use the quadratic formula: -b ± (√b^{2} - 4ac) ÷ 2a

So we have -4.0 x 10^{-9} ± [√(4.0 x 10^{-9})^{2} - (4 • 1 • - 4.0 x 10^{-11})] /2

You're only looking for the positive value for x since a concentration cannot be negative.

Therefore x, which is the value for H^{+} = **6.32 x 10**^{-6}** M**

Ellen E.

Careful. There's an issue with the quadratic solution here. I will post answer in a moment. Notice that : a) x2 = 4.0 x 10-11 - 4.0 x 10-9 x should be x2 = 4.0 x 10-10 - 4.0 x 10-9 x and b) x2 - 4.0 x 10-11 + 4.0 x 10-9 x should be x2 + 4.0 x 10-10 - 4.0 x 10-9 x AND anyway, you should use an approximation here. I will explain when I post my answer in a moment!05/27/19

Mikaela B.

Thank you! This was really helpful.05/27/19