In examining another candidate drug, CAND, for use in removing Cd^2+ ions from b

Question

In examining another candidate drug, CAND, for use in removing Cd^2+ ions from blood, an effective equilibrium constant for the reaction was found to be 5.0 times 10^20. CAND + Cd^2 reversible Cd(CAND)^2+ A dose of CAND was added at a concentration of 0.0010 M to a sample of blood containing 1.0 x 10^-8 M Cd^2+ ions. What would you predict as the resulting equilibrium concentration of Cd^2+ ions What is the equilibrium concentration of Cd^2+ 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) reversible Cd^2+ + ox^2- K = 2.0 x 10^-4

Explanation / Answer

I have a little doubt in question 1, so I'll answer question 2 instead:

Cd(ox) <--------> Cd2+ + ox2-   K = 2x10-4

i. 0.010 0 0

e. 0.010-x x x

2x10-4 = x2 / 0.010-x but K is low, so x will be low too, therefore 0.010-x = 0.010

2x10-4(0.010) = x2

x = (2x10-4 * 0.010)1/2 = 1.4142x10-3 M

[Cd(ox)] = 0.010 - 1.4142x10-3 = 0.0085 M

Hope this helps. Post question 1 in another question thread.

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