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

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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

Postby Matt_Fontila_2L_Chem14B » Fri Dec 08, 2017 4:19 pm

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

Postby lauren chung 2f » Fri Dec 08, 2017 6:06 pm

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

Postby Kaelie Blanes-Ronda 2L » Fri Dec 08, 2017 10:24 pm

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


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