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
....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
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.
And I can't read the brain waves to know what you obtained as an answer.
3.03x10^-11
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.