Strontium, Sr^2+, is a heavy metal that frequently causes concern when it is pre

Question

Strontium, Sr^2+, is a heavy metal that frequently causes concern when it is present in water. Some preliminary calculations arc necessary to determine whether it is feasible to precipitate Sr^2+ as the sulfate. Given that 10^-4 mole/L of Sr^2+ and 10^-4 mole/L of SO^2-_4 are originally present and that p_K_sp = 7.8 for SrSO_4(s), answer the following: a) How much Sr^2+ will precipitate (moles/L) if 5 times 10^-3 mole Na_2SO_4 is added per liter of solution? b) How much SO^2-^4 and Sr^2+ remain after precipitation? a) Sketch the p_C_-pCO_^2-_3 diagram for CaCO_3 and MgCO_3 at 25 degree C. b) Using this diagram, determine the [CO^2-_3] in equilibrium with 5 times 10^-5 M Ca^2+. c) What concentration of CO^2-_3 is necessary to begin precipitating Mg^2+ as the carbonate in a solution containing 3 times 10^-4 M Mg^2+?

Explanation / Answer

SrSO4 <==> Sr^2+ + SO4^2-

Ksp = [Sr^2+][SO4^2-]

pKsp = 7.8

or, Ksp = 10^-7.8 = 1.58*10^-8 M

SrSO4 <==> Sr^2+      +         SO4^2-

(a) total moles of So4^2- present in the solution = 0.0001 + 0.005 moles= 0.0051 moles

[Sr^2+] = Ksp/0.0051

           = 1.58*10^-8/0.0051

            = 3.09*10^-6 Mol/L

amount of Sr^2+ that will precipitate = 1*10^-4 mol/L -3.09*10^-6 Mol/L = 9.69*10^-5 mol/L

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amount of Sr^2+ remain = 3.09*10^-6 Mol/L

amount of SO4 ^2- that will precipitate = 9.69*10^-5 mol/L

amount of SO4^2- remains = 0.0051-9.69*10^-5 mol/L = 5*10^-3 mol/L

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