## The Quadratic Equation

Annalyn Diaz 1J
Posts: 61
Joined: Fri Sep 28, 2018 12:15 am

### The Quadratic Equation

Is there a reason why we can “bypass” the quadratic equation when calculating for x with weak acids?

Samantha Kwock 1D
Posts: 61
Joined: Fri Sep 28, 2018 12:24 am

### Re: The Quadratic Equation

If the K constant for the weak acid is less than 10^-3, the quadratic equation can be bypassed because the change in concentration is so small compared to the actual initial concentration that it is negligible. After you solve for the equilibrium concentration, if the answer is less than 5% of the initial concentration, then this shortcut is valid.

Julia Lee
Posts: 80
Joined: Fri Sep 28, 2018 12:29 am

### Re: The Quadratic Equation

This is what Professor Lavelle posted previously on this topic:

"When K is smaller than 10-3 then X is much smaller than the initial concentration, [I], and therefore: [I] - X is approximately equal to [I].

Note: X is not zero.

But X is much smaller than [I] and therefore the difference is approximately the initial value.

For example, someone with \$1 million gives \$1,000 to someone with no money. The \$1,000 is meaningful (not zero) to the person receiving it.
But initial - change is essentially the initial (\$1 million - \$1,000 is approximately \$1 million)."

Philipp_V_Dis1K
Posts: 32
Joined: Fri Sep 28, 2018 12:20 am

### Re: The Quadratic Equation

Bypassing the quadratic equation because when on the bottom of the (x^2)/(concentration-x), the x, or change in concentration, is so small that the concentration is basically untouched and it is needless to use it. Therefore the equation becomes (x^2)/concentration. Then it's very easy to solve for the x.

Return to “Equilibrium Constants & Calculating Concentrations”

### Who is online

Users browsing this forum: No registered users and 1 guest