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?
Calculate the moles of Ca(NO3)2 you have (concentration * volume in L). Moles of Ca(NO3)2 = moles of Ca+2 per the chemical formula.
What i did but incorrect please help thanks;)
The calculate the moles of Na2SO4 you have. That will tell you how many moles of Ca+2 can be reacted based on the equation.
Remaining moles of Ca+2 = original moles of Ca+2 - moles of Ca+2 that reacted
remaining moles of Ca+2 / original moles of Ca+2 * 100 is the percentage remaining.
Na2SO4 = (1.50×10−2 M)(0.05L) = 0.00075mol
Ca(NO3)2 = ( 1.23×10−2 M)(0.05L) = 0.000615mol
0.00075-0.000615=1.35x10^-4
(1.35x10^-4 / 0.000615 mol ) x2 = 44%
44% not the right answer
Your math os OK but your reasoning is faulty. According to yuour numbers, you have an excess of Na2SO4 and ALL of your Ca(NO3)2 has been used so you might think 100% of the Ca^2+ has been pptd. After the reaction this is what you have.
You have 0.000615 mols CaSO4 solid pptd in a 100 mL solution of 0.000135 Na2SO4. However, the CaSO4 has some solubility as determined by the Ksp for CaSO4. Essentially you have a saturated solution of CaSO4 with an excess of 0.000135 mols [SO4]^2- in 100 mL solution. That maks the sulfate a common ion. Look up Ksp for CaSO4, solve for (Ca^2+) in that solution and calculate % from that. Remember that Ksp calculations must be done with concn and not mols.
To find the percentage of Ca2+ that remains unprecipitated, you need to calculate the moles of Ca2+ originally present in the solution and compare it to the moles of Ca2+ that reacted with the Na2SO4.
First, calculate the moles of Ca(NO3)2:
Moles of Ca(NO3)2 = concentration of Ca(NO3)2 × volume of Ca(NO3)2 in liters
= (1.23×10−2 M)(0.050 L)
= 0.000615 mol
Next, calculate the moles of Na2SO4:
Moles of Na2SO4 = concentration of Na2SO4 × volume of Na2SO4 in liters
= (1.50×10−2 M)(0.050 L)
= 0.00075 mol
Now, we need to determine the mole ratio between Ca(NO3)2 and Na2SO4. Looking at the balanced equation:
Ca(NO3)2(aq) + Na2SO4(aq) → CaSO4(s) + 2NaNO3(aq)
We can see that for every 1 mole of Ca(NO3)2, 1 mole of Ca2+ is produced. Therefore, the number of moles of Ca2+ that reacted with the Na2SO4 is also 0.000615 mol.
To find the moles of Ca2+ remaining unreacted, subtract the moles of Ca2+ that reacted with Na2SO4 from the original moles of Ca2+:
Moles of Ca2+ remaining = original moles of Ca2+ - moles of Ca2+ that reacted
= 0.000615 mol - 0.000615 mol
= 0 mol
Since no moles of Ca2+ remain, the percentage of Ca2+ remaining unprecipitated is 0%.
Therefore, the correct answer is 0%.