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pH of complex

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

K2(Cd(CN)4) :rarrow: 2K+ + Cd(CN)42-
Cd(CN)4 ::equil:: 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 ::equil:: 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 pKHCN = 10, write the equation for which K = 10-10.

AWK:

--- Quote from: mjc123 on January 21, 2021, 06:34:06 AM ---If pKHCN = 10, write the equation for which K = 10-10.

--- End quote ---

--- Quote from: kamilast on January 20, 2021, 03:37:21 PM ---CN- +H2O ::equil:: HCN+OH-    Ka=10^-10
I did another table from where Ka=x^2/9,16x10^-5
--- End quote ---
Rather, the problem lies in the wrong concentration of CN- ions.

mjc123:
It's both. And also the assumption that x << [CN-].

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