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how much Sr2+ will precipitate if 10^-3 mole Na2SO4 added per litter of solution. How much...

how much Sr2+ will precipitate if 10^-3 mole Na2SO4 added per litter of solution. How much SO42- and Sr2+ remain after precipitation. Given that 10-4 mole/L of Sr2+ and 10-4 of SO42- are originally present and that pKso = 7.8 for SrSO4 (s)

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Answer #1

given after addition of Na2SO4

total [SO4^2-] in solution = 10^-4 M x + 10^-3 M = 1.1 x 10^-3 M

pKso = -log[Kso] = 7.8

Kso = 1.6 x 10^-8

Kso = [Sr2+][SO4^2-]

at the total [SO4^2-] concentration, calculate [Sr2+] in solution,

[Sr2+] = Kso/[SO4^2-]

          = 1.6 x 10^-8/1.1 x 10^-3

          = 1.45 x 10^-5 M

So,

[Sr2+] precipitated = 10^-4 - 1.45 x 10^-5 = 8.55 x 10^-5 M

and,

[Sr2+] remained after precipitation = 1.45 x 10^-5 M

[SO4^2-] remained after presipitation = 1.1 x 10^-3 -1.45 x 10^-5 = 1.0855 x 10^-3 M

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