[kamilast]

The task is
calculate pH of the  0,1M solution K₂Cd(CN)₄. For complex logβ=18,9 and pK_HCN=10
This is what I tried to do:

K₂(Cd(CN)₄) 2K⁺ + Cd(CN)₄^2-
Cd(CN)₄ Cd²⁺ + 4CN^-
then i convert β to K by K=1\β  and I made "before\after" table from where I got concentrations: Cd(CN)₄=0,2-x ;Cd=x; Cn=4x
K=4x^5/0,1 and its 9,16x10^-5
CN^- +H₂O HCN+OH^-    Ka=10^-10
I did another table from where Ka=x^2/9,16x10^-5. when i put this into formula for pH it turned out that the answer is wrong. 
I would be so greatfull for any hints 

[AWK]

4⁴ = 4  ?

[mjc123]

If pK_HCN = 10, write the equation for which K = 10^-10.

[AWK]

If pK_HCN = 10, write the equation for which K = 10^-10.

CN- +H2O HCN+OH-    Ka=10^-10
I did another table from where Ka=x^2/9,16x10^-5

Rather, the problem lies in the wrong concentration of CN- ions.

[mjc123]

It's both. And also the assumption that x << [CN^-].

[kamilast]

Now i see how many mistakes I did. I have olso 1 question, is it right that I change β to K? Im completly not sure about it.

[AWK]

It does not matter.
This is just replacing the numerator with the denominator in the formula.