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-
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.
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¯ .
To find the initial concentrations of Ca^2+ and SO4^2-, we need to consider the stoichiometry of the reaction and use the principles of solubility product (Ksp) and molarity.
First, let's write the balanced equation for the reaction between Ca(NO3)2 and Na2SO4:
Ca(NO3)2 (aq) + Na2SO4 (aq) -> CaSO4 (s) + 2NaNO3 (aq)
From the balanced equation, we can see that for every one mole of Ca(NO3)2, we get one mole of CaSO4. Also, since Na2SO4 has two sodium ions (Na^+) and one sulfate ion (SO4^2-), it would require two moles of Na2SO4 to react with one mole of Ca(NO3)2.
Given:
- Volume of Na2SO4 solution (V1) = 100 mL = 0.1 L
- Molarity of Na2SO4 solution (M1) = 0.5 M
- Volume of Ca(NO3)2 solution (V2) = 100 mL = 0.1 L
- Molarity of Ca(NO3)2 solution (M2) = 1.65 M
To find the initial concentrations, we'll start by converting the volumes and molarities of the solutions into moles:
Moles of Na2SO4 (n1) = Molarity (M1) x Volume (V1) = 0.5 M x 0.1 L = 0.05 moles
Moles of Ca(NO3)2 (n2) = Molarity (M2) x Volume (V2) = 1.65 M x 0.1 L = 0.165 moles
Since we have a 1:1 stoichiometry between Ca(NO3)2 and CaSO4, the moles of Ca^2+ (n_Ca^2+) will be equal to n2 = 0.165 moles.
Since each mole of Ca(NO3)2 produces one mole of CaSO4 and each mole of CaSO4 produces one mole of SO4^2-, the moles of SO4^2- (n_SO4^2-) will also be equal to n2 = 0.165 moles.
Finally, to find the initial concentrations, we divide the moles by the total volume of the mixture:
Initial concentration of Ca^2+ = n_Ca^2+ / Total volume = 0.165 moles / (0.1 L + 0.1 L) = 0.165 moles / 0.2 L = 0.825 M
Initial concentration of SO4^2- = n_SO4^2- / Total volume = 0.165 moles / (0.1 L + 0.1 L) = 0.165 moles / 0.2 L = 0.825 M
Therefore, the initial concentrations of Ca^2+ and SO4^2- in the mixture are both 0.825 M.