How do you use approximation instead if the quadratic formula?

Question

In this exmaple: x^2 / (0.0550-x) = 2.13 x 10^-11

How would I use approximation method instead of using the quadratic formula?? I know that we assume that x is small and we set up...

x^2 / (0.0550)= 2.13 x 10^-11

Solving for X= 1.08 x 10^-6 whenI plug this back in it works BUT if it didn't how would I continue... I don't get that part =(

Answer 1

You're on the right track. You've assumed that x is a small enough quantity such that:

x^2/(0.0550-x)

is approximately equal to

x^2/(0.0550)

All you need to do to use successive approximation is take your answer (2.13x10^-11), treat it as x, and plug it back into your original formula for equilibrium product concentration such that you have:

x^2/(0.0550 - 2.13x10^-11)

Solve for x again, and you should come up with a smaller quantity. Continue to take each new x and plug it back into the equation.

Once you receive the same answer for x twice in a row, you'll know to stop your successive approximations. At that point, you will have a reasonable approximation of what x is.

You should be able to solve your problem from there.

I hope that helps!

Answer 2

The previous answer gives you a way to find the true value of x in several steps. This you would apply if you used a program since calculating manually x in such a multi-step manner is much more tedious than solving the quadratic.

In chemistry we usually do this assumption in equilibrium problems in order to avoid solving the quadratic.
When x< The more x approximates C the bigger the deviation of the K value you calculate by plugging-in the approximate x value compared to the actual K. When this is deviation is big you can no longer use the assumprion but you have to solve the quadratic.

Usually you just have to look at x and C and if x<0.001C (or mayebe even x<0.005C) the approximation is OK

Answer 3

whilst the consistent is under one, including yours, you may anticipate that x is going to account for under a 5% distinction interior the reaction. it somewhat is primary simply by fact the "5% Rule" you probably did it properly