In examining another candidate drug, CAND, for use in removing Cd2+ ions from bl
ID: 983812 • Letter: I
Question
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 times 1020. CAND + Cd2+ leftrightarrow Cd(CAND)2+ A dose of CAND was added at a concentration of 0.0010 M to a sample of blood containing 1.0 times 10-8 MCD2+ 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) leftrightarrow Cd2+ + ox2- K = 2.0 times 10-4Explanation / 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