1. In examining another candidate drug, CAND, for use in removing Cd2 ions from
ID: 984064 • Letter: 1
Question
1. In examining another candidate drug, CAND, for use in removing Cd2 ions from blood, an effective equilibrium constant for the reaction was found to be 5.0 x 1020 CAND Cd2+ Cd (CANDO? A dose of CAND was added at a concentration of 0.0010 M to a sample of blood containing 1.0 x 10s M Cd2 ions. What would you predict as the resulting equilibrium concentration of Cd2 ions? 2. What is the equilibrium concentration of Cd2 ions when cadmium oxalate, Cd(ox, is dissolved in water to make an initial concentration of 0.0100 M? All of the cadmium oxalate dissolves but only a fraction of it dissociates as indicated below. What are the equilibrium concentrations of the three species shown in the reaction equation below? Cd(ox) Cd2 ox K 2.0 x 104Explanation / Answer
Solution :-
When the CAND and Cd2+ is mixed then they will form complex Cd(CAND)^2+
So initially all the Cd^2+ is coinverted to complex
So
Lets write the reverse equation equation
Cd(CAND)^2+ ------- > Cd^2+ + CAND
1.0*10^-8 0 0.001 M
-x +x +x
1.0*10^-8 –x x 0.001+x
1/K=[x][0.001+x]/[1.0*10^-8 –x]
1/5.0*10^20 = [x][0.001+x]/[1.0*10^-8 –x]
Solving for the x we get
X= 2*10^-26
So the equilibrium concetration of the Cd2+ is 2.0*10^-26 M
Q2)
Cd(Ox) -------- > Cd2+ + OX^2- K = 2.0*10^-4
0.001 M 0 0
-x +x +x
0.001-x x x
K = [Cd2+][OX^2-]/[Cd(OX)]
2.0*10^-4 = [x][x]/[0.001-x]
Solving for the x we get
X=3.58*10^-4 M
So the equilibrium concetration of the three species are as follows
[Cd(OX)] = 0.001-x = 0.001 – 3.58*10^-4 = 6.42*10^-6 M
[Cd2+] = 3.58*10^-4 M
[OX2-] = 3.58*10^-4 M
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