The solubility-product constants, K sp, at 25 C for two compounds [iron(II) carb

Question

The solubility-product constants, Ksp, at 25 C for two compounds [iron(II) carbonate, FeCO3, and cadmium(II) carbonate, CdCO3] are given by the table

Part A

A solution of Na2CO3 is added dropwise to a solution that contains 1.04×102M Fe2+ and 1.50×102M Cd2+. What concentration of CO32 is need to initiate precipitation? Neglect any volume changes during the addition.

Express the molar concentration numerically.

Part B

In the solution from Part A, what will the concentration of CO32 be when Fe2+ begins to precipitate?

Express the molar concentration numerically.

1.80×1014

1.80×1014

Explanation / Answer

Part A

Dissolution equillibrium of FeCO3 is

FeCO3(s) <------> Fe2+(aq) + CO32-(aq)

Ksp = [Fe2+][CO32-]= 2.10*10-11

given concentration of Fe2+ = 0.0104M

So,

0.0104M * [CO32-] = 2.10*10-11M2

[CO32-]=2.10*10-11M2/0.0104M

= 2.02*10-9

So, when [CO32-] reaches 2.02*10-9M , FeCO3 start to precipitate

dissolution equillibrium of CdCO3 is

CdCO3(s) <-----> Cd2+(aq) + CO32-(aq)

Ksp = [ Cd2+][CO32-]=1.80*10-14

given concentration of Cd2+ = 0.015M

So,

0.015*[CO32-] = 1.80*10-14M2

[CO32-] = 1.20*10-12M

So, when [CO32-] reaches 1.20*10-12M , CdCO3 start to precipitate

therfore,

CdCO3 will precipitate first

Part B

When [ CO32-] reaches 2.02*10-9M , FeCO3 start to precipitate

[Cd2+][CO32-] = 1.80*10-14M2

When , [CO32-] = 2.02*10-9M

[Cd2+]*2.02*10-9M = 1.80*10-14M2

[Cd2+] = 1.22*10-5M

So, when FeCO3 start to precipitate

[Cd2+] = 1.22*10-5M

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