For GC problems, will Ka, Ksp always be negligibly small?

[Teeth12345]

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Hi,
I did a search on this and couldn't find anything.
For most of the GC problems I've come across in Destroyer, I've been able to ignore the minus X part during solubility/acid, base equilibrium problems. This is nice because I don't have to deal with crazy quadratic equations.
Is this how it will be on the exam? All really small Ka and Ksp?

Thanks!!

Sorry if I start posting a lot of new threads - my test is coming up and I'm going crazy!!! 😱

 

[KillaCam]

In all of the problems that I have encountered in study prep and the real thing, Yes, the [-X] is always so small that it is negligable. It surely makes things easier and faster!!!

 

[wired202808]

Teeth you are correct, as mention in Chad's videos because the values tend to be so insignificant like 1.5 x 10^-10, you can simply ignore the X and do it the way you were. The answer would be approximate and you should be able to determine it.

 

[jeo]

Generally, yes, the Ka and Ksp will be negligibly small. Thus, when setting up your "ICE" table, disregard the [-X]. Remember, a quick solution to check if the [X] is inconsequential is to take the X value and divide it by the initial concentration and times by 100. if the answer results in a change less than 5% of the initial concentration, the error is not to be considered and you can avoid using the quadratic formula.

Think of it like weighing yourself on a scale with a wristband. Initially you weigh 200 lb. When you take the wristband off, you still weigh 200 lb. The wristband is so negligible compared to your body that you can neglect it.
Initial body weight - weight of wristband = final body weight ≈ initial body weight

Example:
A(g) <-----> B(g) + C(g) |||||Kc = 5.0 X10E-4; initial [A] = .500 M||||| Find the value of

Set up your ICE table

Concentration (M)|| A(g) <------> B(g) |||+||| C(g)
-------------------------------------------------------------
Initial ||||||||||||||| .500 |||||||| 0 |||||||||| 0
Change|||||||||||||| -X |||||||||| +X ||||||||| +X
--------------------------------------------------------------
Equilibrium|||||||||||| .500 - X ||||+ X|||||||||||| + X

Kc = [C] / [A] = X2/ .500 - X = 5.0 X10E-4

Assume X is so small that it is negligible and omit it. Therefore, we can avoid the quadratic formula.

X2 / .500 = 5.0 X10E-4 Times .500 in the form of 1 on both sides to get rid of the denominator on the left side

X2 = (5.0 X10E-4)(.500) Times out the right side

X2 = 2.5 X10E-4 Take the square root of both sides to find X

X = 1.6 X10E-2

Now check the 5% rule by taking X and dividing it by our initial concentration times 100
(X / [A]inital)100 = (1.6 X10E-2 X / .500)100 = 3.2%

It's <5%. Thus, we can keep our X value, and we successfully avoided the evil quadratic equation!

Therefore = X = 1.6 X10E-2 M

Hope this helps!

 

[Teeth12345]

Thanks for the helpful input guys!

And thank you Jeo for that nice problem and explanation.