molar concentration [ENDORSED]

[Maribel 1E]

When solving for x using the quadratic formula when you are trying to find the equilibrium molar concentrations, which value do you use if you get two positive x values? And why?

[Emma MacLachlan 1D]

Sometimes one out of the two positive x values that is given when using the quadratic formula cannot be used because it is higher than the initial concentration. This means that you would get a negative equilibrium molarity value when using that value, in which case you would use the other positive number.

[davidzhang2K]

Out of two positive x-value solutions that can come from the quadratic equation, there is often only one answer that fits within the question. There is often larger number that cannot be used in the ICE table. In case you're unsure, you can always plug the value in and double check it!

[davidzhang2K]

Out of two positive x-value solutions that can come from the quadratic equation, there is often only one answer that fits within the question. There is often larger number that cannot be used in the ICE table. In case you're unsure, you can always plug the value in and double check it!

[Sophi_Whitman_1I]

I was wondering the same thing! I found plugging it in as x and solving for the concentrations usually tells you which number to use. You don't want it to be larger than an initial concentration because that would mean the final concentration would be negative.

[Adelyne_Tran_1J]

I had trouble with this concept at first as well! However, I found that the two calculated x values usually ended a larger number and a smaller one. Oftentimes, plugging the larger x value in the equation would result in a negative value(ex. if x=5.2 and 0.023 and the equation to calculate the equilibrium concentration is 0.278-x, subtracting 5.2 from the value would equal a negative number). Therefore, that x value would be crossed out since it is not possible to have a negative equilibrium concentration.

[Payton Zarceno 1E]

When I was doing the achieve HW I came across this same problem, but in most cases, one of the positive values will be higher than the initial concentration so when you subtract it you'll get a negative value at equilibrium! One of your answers, for most cases, will be higher than the initial value so you can rule it out.

[Sophie Vansomphone 1J]

If the quadratic gives you two positive x values, typically you use the smaller one. However, that is not the case for all answers so you will just have to test it out. When you test them out and get a negative number at equilibrium, you know that is the one not to use.

[Alysha Dijamco 3E]

Hi! I was also initially confused about what to do if I the quadratic formula gave me two values for x. However, after doing a couple practice problems where this was the case, I noticed that there would only be one value of x that would yield the correct answer. One of the values of x would give you a negative equilibrium concentration, and concentrations must be positive values. Hope this helps!

[Andres Ibarra 2F]

I was also confused about this. I believe you just use the positive x value that does not give you a negative equilibrium concentration. I usually just plug in the numbers to make sure.

[Yasmine_B_1H]

Hello,

When using the quadratic formula to solve for x, you always use the positive x value because there cannot be a negative concentration. However, when both values are positive you will take x plug it back into the equilibrium and use the result that is not higher than the initial or a negative equilibrium molarity value. In these cases you will use the other positive value.

[Levon_Avedian_2H]

When you use the quadratic equation, only one answer works given the ICE table. Use process of elimination to find the correct value!

[Eden K 1B]

If you get two positive values for x after using the quadratic formula to solve compare these two values to the initial molar concentration you started with. One of the x values is usually bigger than the initial you started with.