A 75.0mL sample of 1.56×10−2M Na2SO4(aq) is added to 75.0mL of 1.22×10−2M of Ca(NO3)2(aq).
What percentage of the Ca+ remains unprecipitated?
CaSO4(s) ==> Ca^+2 + SO4^=
Ksp = (Ca^+2)(SO4^=)
I would calculate moles Ca^+2 and moles SO4^=, from an ICE chart you will know which is the limiting reagent and from there which is the common ion (that will be the one in excess). Then calculate the solubility and from there how much remains. Find percent of that.
the answer is not clear
To find the percentage of Ca+ that remains unprecipitated, we first need to determine the amount of Ca+ ions that precipitate with the sulfate ions (SO4²-) from the Na2SO4.
1. Calculate the moles of Ca+ and SO4²- ions:
- For Ca+ ions:
- Molarity (Ca+): 1.22×10^(-2) M
- Volume (Ca+): 75.0 mL
- Convert mL to L: 75.0 mL ÷ 1000 = 0.075 L
- Moles (Ca+): Molarity × Volume = 1.22×10^(-2) M × 0.075 L = 9.15×10^(-4) moles
- For SO4²- ions:
- Molarity (SO4²-): 1.56×10^(-2) M
- Volume (SO4²-): 75.0 mL
- Convert mL to L: 75.0 mL ÷ 1000 = 0.075 L
- Moles (SO4²-): Molarity × Volume = 1.56×10^(-2) M × 0.075 L = 1.17×10^(-3) moles
2. The balanced chemical equation for the reaction between Ca(NO3)2 and Na2SO4 is:
Ca(NO3)2(aq) + Na2SO4(aq) ⟶ CaSO4(s) + 2NaNO3(aq)
From the equation, we can see that one mole of Ca(NO3)2 reacts with one mole of Na2SO4 to produce one mole of CaSO4. Therefore, the mole ratio between Ca+ and SO4²- is 1:1.
3. Since the mole ratio is 1:1, the limiting reactant in this reaction is the one with fewer moles. In this case, we have 9.15×10^(-4) moles of Ca+ and 1.17×10^(-3) moles of SO4²-.
4. As there are more moles of SO4²- than Ca+, the amount of Ca+ that will precipitate is equal to 9.15×10^(-4) moles.
5. To find the percentage of Ca+ that remains unprecipitated, we need to compare the amount of Ca+ that reacts with the total amount of Ca+ initially present.
- Total amount of Ca+ initially present:
- Molarity (Ca+): 1.22×10^(-2) M
- Volume (Ca+): 75.0 mL
- Convert mL to L: 75.0 mL ÷ 1000 = 0.075 L
- Moles (Ca+): Molarity × Volume = 1.22×10^(-2) M × 0.075 L = 9.15×10^(-4) moles (same as the amount that precipitates)
- Percentage of Ca+ remaining unprecipitated = (Amount of Ca+ remaining / Initial amount of Ca+) × 100
= (9.15×10^(-4) moles / 9.15×10^(-4) moles) × 100
= 100%
Therefore, 100% of the Ca+ ions remain unprecipitated.