Complexes For the solution containing 1 M of HCI and 0.01 M of Cd(NO_3)_2. answe

Question

Complexes For the solution containing 1 M of HCI and 0.01 M of Cd(NO_3)_2. answer the following Cd^2- + Cl^- DoubleLongLeftRightarrow CdCl^+ K_1 = 21 CdC^1+ + Cl^- DoubleLongLeftRightarrow CdCl^2 K_2 = 7.9 CdCl_2 + Cl^- DoubleLongLeftRightarrow CdCl^-_3 K_3 = 1.23 CdCl^+_3 + Cl^- DoubleLongLeftRightarrow CdCl^2-_4 K_4 = 0.35 Since the solution is strong acid, the hydrolysis of the metal ion is negligible. Assume the ionic strength efTect is also negligible. Note that Cd(NO_3)_2 is very soluble. W rite equilibrium equations for the overall formation reactions (for all relevant reactions). Write mass balance equations for Cd and Cl. Write assumptions for determining the concentrations of Cd^2+ species in the solution. Calculate the concentration(M) of Cd^2+ in the solution.

Explanation / Answer

THe reaction equilibrium equations are given by

Keq=[products]/[reactants]

so for the first reaction,

K1 = [CdCl+]/([Cd2+][Cl-])

for the second reaction,

K2 = [CdCl2]/([CdCl+][Cl-])

For the third reaction,

K3 = [CdCl3-]/([CdCl2][Cl-])

for the fourth reaction,

K4 = [CdCl4 2-]/([Cd2Cl3 -][Cl-])

2)The mass balance equations for Cd and Cl is as follows

one mole of Cl2+ uses one mole of Cl-

so,

4* mass of Cd2+ = mass of Cl- added to the solution.

4)

Therefore,

the concentration of Cd2+ in the solution be x.

total Keq=K1*K2*K3*K4

therefore, the final equation is:

Keq = [CdCl4 2-]/[Cd2+][Cl-]^4

or 21*7.9*1.23*0.35 = x/((1-4x)^4*(0.01-x))

or x=0.00986 M

so the concentration of Cd2+ in the equilibrium = 0.01-0.00986

=0.00014 M

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