Asked by Xi
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 %
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 %
Answers
Answered by
DrBob222
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.
Answered by
Xi
Are we allowed to set x = 0 if it is like 5.83%?
Answered by
DrBob222
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.
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.
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