Asked by Elise

At 25 degrees Celsius Ksp for CaSO4 is 2.5 x 10^-5 suppose 100mL of 0.5M Na2SO4 is mixed with 100mL of 1.65 M Ca(NO3)2 find the initial concentrations of Ca^2+ and SO4^2-
Answered by DrBob222
What do you mean by initial concentration? Is that BEFORE you calculate to see if a ppt of CaSO4 will form?
That would be Na2SO4 = 0.5M x (100/200) = ?
Ca(NO3)2 = 1.65M x (100/200) = ?
Therefore, initial (Ca^2+) = [Ca(NO3)2] from above and
(SO4^2-) = (Na2SO4) from above.
Answered by Doc48
Just in case you need the ion concentrations after mixing, here's my solution...
Na₂SO₄ + Ca(NO₃)₂ => 2NaNO₃ + CaSO₄(s)
1:1 Rxn Ratio between Na₂SO₄ (limiting reactant) in an excess of Ca(NO₃)₂
100ml(0.50M Na₂SO₄) + 100ml(1.65M Ca(NO₃)₂) …
 0.10(0.50) mole Na₂SO₄ + 0.10(1.65) mole Ca(NO₃)₂
 0.050 mole Na₂SO₄ + 0.165 mole Ca(NO₃)₂ …
 (0.165 – 0.05) mole Ca(NO₃)₂ excess => 0.115 mole Ca⁺² unreacted …
 (0.115/0.20)M Ca⁺² = 0.575M Ca⁺² (unreacted)
 From Solubility of CaSO₄ =√Ksp = √2.5 x 10¯⁵ M = 0.005M CaSO₄(ppt ionized back into solution)…
 0.005M Ca⁺² + 0.005M SO₄¯² from ionization of CaSO₄(s) formed in reaction.
 [Ca⁺²] = 0.575M (fm unreacted Ca(NO₃)₂) + 0.005M (ionized fm CaSO₄ formed in Rxn)
 0.580M[Ca⁺²]Total + 0.005M[SO₄¯²]fm ionization of ppt formed in rxn
Spec Ions => 0.50M Na⁺ & 0.165M NO₃¯ .
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