Hmm... This does not make sense. I learned that anything over 5% error, we cannot accept so we have to use the quadratic formula.

Here is a question I got 5.831%. It didn't use the quadratic formula and it still said to set x=0.

Ka = 6.8x10^-4
HF = 0.2-x
H+ = x
F- = x

x= 0.011662

I take 0.011662 and compare it to the original one which is 0.2.

0.011662/0.2 x100 = 5.83 %

I don't understand your problem. It's over 5% so use the quadratic formula. I obtained 0.0113 which isn't that far from 0.01166 but 0.01133 satisfies the equation better than 0.01166. Technically, we're splitting hairs because you can use only 2 significant figures anyway since Ka isn't known any better than that. That being said, there isn't that much difference between 0.012 and 0.011.

Are we allowed to set x = 0 if it is like 5.83%?

Which x are you talking about?

The equation, I assume is
x^2/(0.2 - x)] = 6.8 x 10^-4.
You may NOT set the x of x^2 to zero else you can't solve the equation. Setting the x in the denominator to zero is allowed if the 5% rule (or whatever your prof sets), is not exceeded. In this case, it is exceeded so it is not allowed. [By the way, a better way of saying it is "assume 0.2-x = 0.2" and that way it isn't confusing which x we are talking about.] And for whatever it's worth, I always told my students to use 10%; however, Ks can be determined with better accuracy now so the 5% rule is a little more strict. Technically, it doesn't matter much except where x is large compared to the molarity or where Ka is large such as 10^-1 or 10^-2. Another thing to look at is that often teachers want their students to get practice at (a) solving the quadratic and (b)determining when the quadratic is appropriate so setting a smaller percent gets those two purposes accomplished. You can see how much difference those two numbers make if you substitute 0.01133 for x (and use 0.2-x) and see if the left side = 6.8 x 10^-4, then use 0.01166 and do the same thing. When I did this I came up with about 7.22 x 10^-4 when I used 0.01166 and it was 6.80 x 10^-4 when I used 0.1133. That's a 5.8% error when using 0.01166 (but you already knew that). I hope this helps.

It appears that you have encountered a problem involving an equilibrium reaction and the calculation of percent error. Let me help explain the process and clarify your confusion.

In this specific scenario, you are dealing with the equilibrium reaction between HF (hydrofluoric acid) and H+ (hydrogen ion) and F- (fluoride ion). The equilibrium constant (Ka) for this reaction is given as 6.8x10^-4.

To solve for the concentration of the hydrogen ion (H+) and the fluoride ion (F-), you assume that the initial concentration of HF (hydrofluoric acid) is 0.2 M, and both H+ and F- have an unknown concentration, which is represented by x.

The expression for the concentration of HF after the reaction has occurred can be written as 0.2 - x, since the concentration of HF decreases by x during the reaction. Likewise, the concentration of H+ and F- is x, as they are formed during the reaction.

To determine the value of x, you need to set up an equation using the equilibrium constant expression:

Ka = [H+][F-] / [HF]

Substituting the given equilibrium constant value and the corresponding concentrations, the equation becomes:

6.8x10^-4 = (x)(x) / (0.2 - x)

Now you can proceed to solve this equation. One method is using the quadratic formula since the equation is quadratic in nature. However, it seems you found another method that led to the same result.

By solving the equation using any method, you obtained the value x = 0.011662.

To calculate the percent error, you compare the calculated value with the original value of HF, which is 0.2 M. The equation for percent error is given by:

( | calculated value - original value | / original value ) x 100%

In this case, the calculated value is 0.011662 and the original value is 0.2. Plugging these values into the equation, we have:

( | 0.011662 - 0.2 | / 0.2 ) x 100% = 5.83%

Therefore, the percent error between the calculated value of x and the original value of HF is 5.83%.

In summary, it seems that you correctly followed the steps to solve the equilibrium problem and obtained a value for x. The method you used to calculate the percent error is also correct. The quadratic formula could have been used, but your approach is an alternative method that yielded the same result.