This question stumped me and for future reference, I want to know how to solve t

Question

This question stumped me and for future reference, I want to know how to solve this step by step. Q6) (12 points) An solution containing 0.000022 M of Fea. and 0.00065 M Mg If co? is used to precipitate out one of these ions from 2 6.82x10 solution. for Feco, 3.07x10-11, p for MgCO, a) What is the minimum [cos2 is required to precipitate each ion? (No Pop' doc 22 b) If 3.54 x 104 M solution containing co32. ions is added to the above solution, which ions will remain in the solution and which ions will precipitate out? (show your work to receive credit)

Explanation / Answer

a) i)MgCO3(s) <------> Mg2+(aq) + CO32-(aq)

Ksp = [Mg2+] [ CO32-]

Ksp = 6.82×10^-6

For precipitation

Qsp > Ksp

Where, Qsp is solubility product quotient

Qsp = [ Mg2+][CO32-]

[Mg2+]= 0.00065M

6.82×10^-6= 0.00065 × [ CO32-]

[CO32-] = 1.05 × 10^-2M

So, [ CO32-] should be > 1.05 × 10^-2M to start precipitation of MgCO3

To precipitate all Mg2+ ion

[CO32-] = 0.00065 + 1.05×10^-2=0.01115M

So, minimum [CO32-] required is 0.01115M to precipitate 0.00065M Mg2+

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

Ksp = 3.07 ×10^-11

3.07× 10^-11 = [ Fe2+][CO32-]

[Fe2+] = 0.000022M

3.07×10^-11 = 0.000022 × [CO32-]

[ CO32- ] = 1.39 × 10^-6M

So, to start precipitation of FeCO3 , Concentration of CO32- should be greater than 1.39×10^-6

[ CO32-] required to precipitate all Fe2+ ion

[ CO32- ] = 0.000022 M + 1.39 × 10^-6M = 2.33×10^-5M

So, 2.33×10^-5M of CO32- required to precipitate 0.000022 M of Fe2+ ion

b) From part a, we know that [ CO32-] required to precipitate 0.000022M Fe2+ ion is 2.33×10^-5 M but added concentration of CO32- is 3.54×10^-4M so, Fe2+ ion completely precipitate out as FeCO3.

From part a ,we know that [CO32-] required to precipitate 0.00065M of Mg2+ ion is 0.01115M but Concentration of CO32- remaining after precipitation of Fe2+ is 0.00033M which is not enough concentration to precipitate 0.00065M Mg2+ ion. So, Mg2+ remain in solution.

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