Posted by **Xi** on Tuesday, November 25, 2008 at 3:49pm.

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 %

- Chemistry -
**DrBob222**, Tuesday, November 25, 2008 at 4:32pm
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.

- Chemistry -
**Xi**, Tuesday, November 25, 2008 at 6:11pm
Are we allowed to set x = 0 if it is like 5.83%?

- Chemistry -
**DrBob222**, Tuesday, November 25, 2008 at 9:11pm
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.

## Answer this Question

## Related Questions

- Chemistry - Okay so I'm having a brainfart here. I'm doing a titration problem ...
- algebra,math - My qustion is for the following the problem reads as follows (x-1...
- Math - Use the quadratic formula to solve each of the following quadratic ...
- chemistry - Complete the table below: Note: Make simplifying assumptions, do not...
- math - (4.35E-2 - x)(7.56E-2 -x ) ============================ (0.324 + 2x)2 ...
- math,algebra - can someone help me out factor out this 0=16t^2-122t+180 i have ...
- Pre Calculus - ok I appologize for my stupidity... I'll answer your question... ...
- physics - Two identical small spherical conductors (point charges), separated by...
- Chemistry - Question: Suppose you have 557 ml of 0.0300M HCL and you want to ...
- Math ,help - how can i simplify this more its for the following problem: Problem...

More Related Questions