Question 34

[nsross]

what is the purpose of x10 in the solution of this problem?
([F-] = sq root (6.5 x 10-9 x 10) = 2.5 x 10-4 M)

[admin]

Just above the line that you quoted in your question, on the left hand side of the equation, because the [Mg] was 0.1 M, you had:

0.1 [F]^2 = blah blah

So to isolate [F]^2, you would have to multiply both sides by 10.

[marlins173272]

isnt the concentration of the flourine ions 2x^2 and not just x^2

[nedaa.asba6809]

From the following equation,
MgF_(2 (s)) → 〖Mg〗_((aq))^(2+)+ 2F_((aq))^-
We have Ksp = [Mg2+] [F-]2, then let the concentration of F- be
6.5 X10-9 = [0.1] [2x]2, this would give x= 1.27 X 10-4 M, and then the concentration of [F-] is
2x = 2.5 * 10-4.

[wesley.pau5578]

So, I have looked at the question and answer in the discussion for Question 33 that explains why the concentration of F- is denoted [2F-] (which is then squared, becoming 4F-^2. However, I don't really understand why we let the concentration of F- be [2F-] for Question 33, and just [F-] for Question 34.

[jellywing_2058]

In this case the 2 would be implicit within the concentration so you are actually looking for F- only which would also be [2F-] = x and so you do not consider the 2 in this calculation. However the above explanation IS CORRECT.

[wesley.pau5578]

word.

[drgshahbaz1193]

Hey could someone please tell me how to calculate concentration using Ksp. I am really confused because sometimes we have to double the value of the one of the products and sometimes we dont and I dont see the distinction. Help.