A 50.0 mL sample of 1.50×10−2 M Na2SO4(aq) is added to 50.0 mL of 1.23×10−2 M Ca(NO3)2(aq).
What percentage of the Ca2+ remains unprecipitated?
What I did
n=CV
Na2SO4 = (1.50×10−2 M)(0.05L)
= 0.00075
Ca(NO3)2 = (1.23×10−2 M)(0.05L)
= 0.000615
0.00075 - 0.000615 = 0.000135
0.000135 / 0.000615 x100% = 21.95%
The answer is incorrect. Thanks
On the surface you pptd ALL of the Ca^2+ didn't you? You had 0.000615 mols to begin and you used 0.000615 so you used all of it and left an excess of 0.000135 mols [SO4]^2-. So the answer is you pptd 100% and 0% remains in solution UNLESS you want to consider the Ksp of CaSO4 in which case you have sulfate as a common ion which decreases the solubility of CaSO4 somewhat.. I don't know what your prof wants but if I gave a problem like this I would expect Ksp to play a role.
There isn't Ksp given in the question. What to do?
Personally, I would look up Ksp for CaSO4, use the excess sulfate as a common ion, calculate the amount of Ca^2+ in solution, then mols Ca^2+/0.000615)100 = % Ca remaining. Remember Ksp uses concentrations so you will need to convert from M to mols before that final calculations is done. Of course, I'm assuming you have had some discussions about Ksp. If not then I would go the simple route and report 0%
Okay I'll try. Thanks for your time
To calculate the percentage of Ca2+ remaining unprecipitated, you need to consider the stoichiometry of the reaction between Na2SO4 and Ca(NO3)2. The balanced equation for the reaction is:
Na2SO4(aq) + Ca(NO3)2(aq) → 2 NaNO3(aq) + CaSO4(s)
From the reaction equation, you can see that 1 mole of Ca(NO3)2 reacts with 1 mole of Na2SO4 to form 1 mole of CaSO4. Therefore, the number of moles of Ca(NO3)2 that will react is equal to the number of moles of Na2SO4 added.
Step 1: Calculate the moles of Na2SO4 and Ca(NO3)2 added.
Moles of Na2SO4 = concentration of Na2SO4 × volume of Na2SO4 = (1.50×10−2 M) × 0.050 L = 7.5×10−4 moles
Moles of Ca(NO3)2 = concentration of Ca(NO3)2 × volume of Ca(NO3)2 = (1.23×10−2 M) × 0.050 L = 6.15×10−4 moles
Step 2: Determine the limiting reactant.
To find the limiting reactant, compare the moles of Na2SO4 and Ca(NO3)2. The reactant with fewer moles is the limiting reactant.
In this case, Ca(NO3)2 is the limiting reactant because it has fewer moles (6.15×10−4 moles) compared to Na2SO4 (7.5×10−4 moles).
Step 3: Calculate the moles of CaSO4 formed.
Since the balanced equation shows a 1:1 mole ratio between Ca(NO3)2 and CaSO4, the moles of CaSO4 formed will also be 6.15×10−4 moles.
Step 4: Calculate the moles of Ca2+ that reacted to form CaSO4.
In the reaction, one mole of Ca(NO3)2 releases two moles of Ca2+. Therefore, the moles of Ca2+ that reacted will be twice the moles of Ca(NO3)2.
Moles of Ca2+ reacted = 2 × moles of Ca(NO3)2 = 2 × (6.15×10−4 moles) = 1.23×10−3 moles
Step 5: Calculate the moles of Ca2+ remaining unprecipitated.
Moles of Ca2+ remaining unprecipitated = Total moles of Ca2+ - Moles of Ca2+ reacted
= (Total moles of Ca(NO3)2 × 2) - Moles of Ca2+ reacted
= (2 × (6.15×10−4 moles)) - (1.23×10−3 moles)
= -6.15×10−4 moles
It seems you made a calculation error in your previous working. The correct value for the moles of Ca2+ remaining unprecipitated is -6.15×10−4 moles.
However, the question asks for the percentage of Ca2+ remaining unprecipitated, so you need to convert the moles to a percentage. Use the following formula:
Percentage = (Moles of Ca2+ remaining unprecipitated / Initial moles of Ca2+) × 100
Since the initial moles of Ca(NO3)2 = 6.15×10−4 moles and each mole of Ca(NO3)2 releases 2 moles of Ca2+, the initial moles of Ca2+ = 2 × 6.15×10−4 moles = 1.23×10−3 moles.
Now substitute the values into the percentage formula:
Percentage = ((-6.15×10−4 moles) / (1.23×10−3 moles)) × 100
= (-0.5) × 100
= -50%
The negative sign indicates that all of the Ca2+ reacted with Na2SO4 to form CaSO4 and there is no Ca2+ remaining unprecipitated.
Therefore, the correct answer is that 0% of the Ca2+ remains unprecipitated.