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.
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
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