Question

How do I solve this for c?

Ka = (0.02c)(0.02c)/(c-0.02)

1.7x10^-5=(0.02c)(0.02c)/(c-0.02)

Thanks.

Answer 1

Use algebra to solve the quadratic equation
1.7*10^-5 *(c - 0.02) = 4*10^-4 c^2
which can be rewritten
4*10^-4 c^2- 1.7*10^-5 c +3.4*10^-7 = 0
c^2 -4.25*10-2 c + 8.5*10^-4 = 0
It appears there is no real-number solution to this equation, because B^2 - 4AC is negative. (A, B and C refer to the polynomical coefficients: 1, -4.25*10-2, and 8.5*10^-4.

Answer 2

I made a typo: it should be 1.8 x10^-5.

So in the quadratic it should be: 0=4.0x10^-4-1.8x10^-5c + 3.6x10^-8.

When I solve I get 2 positive answers, so I am not sure how to figure out which one is inadmissible. Thanks.

Answer 3

The quadratic should then be:
0 = 4.0x10^-4 c^2 -1.8x10^-5 c + 3.6x10^-7. (The last term is not 3.6*10^-8)
c^2 - 4.5*10^-2 c + 9.00*10^-4 = 0
B^2 - 4AC is still negative.

Answer 4

hmhm that is odd. The question is supposed to have an answer of 0.044M, so that is weird. Thanks very much for trying to help me with this.

-Sadie

Answer 5

If 0.044 is supposed to be what c is,
(0.02c)(0.02c)/(c-0.02) = 3.2*10^-5

That does not agree with your starting equation.

Are you sure you are starting with the correct equation? If you are dealing with a reaction A -> B + C, 2% of A dissociates, and the starting concentration is c, then at equilibrium
(.02 c)(0.02c)/0.98 c = Kc = 1.8*10^-5
(0.02)^2 c = 4*10^-4 c = = 1.8*10^-5
c = 0.045
That is close to the book's answer.
You apparently started with the wrong equation, with the wrong term in the denominator