## ICE Box: How to find the equation to plug into the Quadratic Formula

Matt_Fontila_2L_Chem14B
Posts: 39
Joined: Fri Sep 29, 2017 7:07 am

### ICE Box: How to find the equation to plug into the Quadratic Formula

How does one find the equation (a,b, and c) to plug into the Quadratic Formula from the ICE Box?

My TA gave us the problem: C6H5COOH <---> H+ + C6H5COO- , the Ka = 6.28*10^-5, and the initial concentration for C6H5COOH = 0.0237

I filled in the ICE box and got to Ka=(x^2)/(0.0237-x)=6.28*10^-5

but then it was assumed that we could just get to this equation:

0=x^2+6.28*10^-5 x - 1.48836*10^-6

I understand that they are a, b, and c that we could plug into the Quadratic Formula, but how do I get to that equation above?

lauren chung 2f
Posts: 50
Joined: Fri Sep 29, 2017 7:03 am

### Re: ICE Box: How to find the equation to plug into the Quadratic Formula

You get the equation by cross multiplying (x^2)/(0.0237-x)=6.28*10^-5. You solve the equation by multiplying either side with (0.0237-x) to get x^2= -6.28*10^-5 x + 1.48836*10^-6. Then you move everything onto one side to get 0=x^2+6.28*10^-5 x - 1.48836*10^-6.

Kaelie Blanes-Ronda 2L
Posts: 42
Joined: Fri Sep 29, 2017 7:07 am
Been upvoted: 1 time

### Re: ICE Box: How to find the equation to plug into the Quadratic Formula

When it comes to an equation what has the coefficient to do with the reaction? Like when figuring out when equilibrium is achieved?