A sample of tap water is found to be 0.025 M in Ca2+. If 105 mg of Na2SO4 is added to 100.0 mL of the tap water, will any CaSO4 precipitate?

Question

Given:

My Attempt:

Step 1 - figure out molar solubility of CaSO4

√7.10*10^-5 = 8.43*10^-3M

Step 2 - convert Na2SO4 from mass to moles

105mg = 0.105g 0.105/142.04 = 7.39*10^-4 mol

Step 3 - figure out moles of CaSO4 produced

Since we need 1 mol of Ca2+ & 1 mol of Na2SO4 to form 1 mol of CaSO4 And we have less Na2SO4 than Ca2+

CaSO4 produced = 7.39*10^-4 mol

Step 4 - figure out molarity of CaSO4

7.39*10^-4mol / 100ml = 7.39*10^-3 / L = 7.39*10^-3M

Step 5 - figure out if any CaSO4 precipitates

8.43*10^-3M > 7.39*10^-3M Therefore NO CaSO4 precipitates

The answer in the book is a Yes, can anyone please explain to me where I have gone wrong? Thanks!

DavePhD · Accepted Answer · 2014-05-28 14:51:48Z

$K_{sp} =[Ca^{2+}][SO_4^{2-}] = 7.10 \times 10^{-5} M^2$

$0.025M[SO_4^{2-}] = 7.10 \times 10^{-5} M^2$

$[SO_4^{2-}] = 0.00284M$ (this is the maximum soluble concentration of $SO_4^{2-}$ in the presence of $0.025M Ca^{2+}$)

However, 0.000739 moles of $SO_4^{2-}$ are added to 100 mL of water, corresponding to a concentration of $0.00739M SO_4^{2-}$.

Precipitation occurs because 0.00739M is greater than 0.00284M.

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