Asked by Neeta

What concentration of Ag is in equilibrium with 1.4 × 10^-6 M Co(CN)63– and Ag3Co(CN)6(s)? The solubility constant of Ag3Co(CN)6 is 3.9 x 10^-26
Answered by DrBob222
....Ag3Co(CN)6 ==> 3Ag^3+ + [Co(CN)6]^-3
I..solid............0.........0
C.....solid.........3x........x
E.....solid.........3x........x

Ksp = 3.9E-26 x (3x)^3(x)
Solve for x. (Ag^+) = 3x
Answered by Neeta
I'm still not getting it right.. But I don't understand what I did wrong.

I did 3.9x10^-26 = [3x]^3 [1.4x10^-6]

then isolated [3x]^3, then took the cube root of both sides.
Answered by DrBob222
And I can't read the brain waves to know what you obtained as an answer.
Answered by Neeta
3.03x10^-11
Answered by DrBob222
How about 1.01E-7?
You should have 3.9E-26/1.4E-6 and that = 3(X)^3
I suspect your error is in the next step. You should divide by 27 and take the cube root of the what is left.

But note that if you had shown what you did at the beginning I would have already seen the found the error and we could have gone on to other things much sooner. And all of this typing would have been unnecessary.
Answered by DrBob222
Note that 1.01E-7 is too many significant figures.
Answered by Neeta
Oh, that is what I did wrong. It was actually 1.01E-11, not 1.01E-7. Sorry to waste your valuable time.
Answered by Neeta
My bad, it was 1.01E-7. I made a typing error on the calculator.
Answered by DrBob222
You didn't waste my valuable time as much as you wasted your valuable time. But I would round my answer to 1.0E-7M.
Answered by DrBob222
I'm glad you corrected that. I was scratching my head trying to figure out your last post. Good luck in your studies.