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 are 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_4^2- are originally present and that pK_so = 7.8 for SrSO_4(s), answer the following: How much Sr^2+ will precipitate (moles/L) if 10^-3 mole Na_2SO_4 is added per liter of solution? How much SO_4^2- and Sr^2+ remain after precipitation?

Explanation / Answer

SrSO4 <----> Sr2+ + SO42-

pKsp = 7.8, i.e. -log Ksp = 7.8. Hence, Ksp = 1.58 * 10^-8

Let's assume, s is the number of moles of Sr2+ at equlibrium, hence number of moles of SO42- at equilibrium will be (s+0.001) [considering addition of Na2SO4]

Hence, s(s+0.001) = 1.58 * 10^-8

i.e. s = 1.56 * 10^-5

So, precipitated amount of Sr2+ = (10^-4 - 1.56 * 10^-5) = 8.44 * 10^-5 mol/L

b) Sr2+ remains after precipitation = 1.56 * 10^-5 mol/L

and amount of SO42- = (10^-4 + 10^-3 - 8.44 * 10^-5) mol/L = 1.0156 * 10^-3 mol/L

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