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?

A 75.0mL sample of 1.48×10−2M Na2SO4(aq) is added to 75.0mL of 1.28×10−2M of Ca(NO3)2(aq).


What percentage of the remains unprecipitated?
Express your answer using two significant figures.

To determine the percentage of Ca+ that remains unprecipitated, we need to first understand the reaction that occurs between Na2SO4 and Ca(NO3)2. These two ionic compounds will form a precipitate, which is a solid that is insoluble in water.

The balanced chemical equation for the reaction is:

Na2SO4(aq) + Ca(NO3)2(aq) → CaSO4(s) + 2NaNO3(aq)

In this reaction, the Ca2+ ions from Ca(NO3)2 combine with the SO42- ions from Na2SO4 to form the insoluble precipitate CaSO4.

Since the volumes of both solutions are equal (75.0 mL), we can assume they react completely according to the stoichiometry of the balanced equation. This means that all the Ca2+ ions from Ca(NO3)2 will react with the SO42- ions from Na2SO4.

To calculate the amount of Ca2+ ions that react, we need to use the concentration (Molarity) and volume of the Ca(NO3)2 solution.

Given:
Volume of Ca(NO3)2 solution = 75.0 mL
Molarity of Ca(NO3)2 solution = 1.22×10−2 M

First, we need to convert the volume of the solution from mL to L (liters):
75.0 mL = 75.0 / 1000 = 0.075 L

Next, we can calculate the number of moles of Ca(NO3)2 using the formula:

moles = Molarity × volume (in liters)
moles of Ca(NO3)2 = 1.22×10−2 M × 0.075 L

Now, we can determine the number of moles of Ca2+ ions:
Since the balanced equation states that one mole of Ca(NO3)2 forms one mole of Ca2+ ions, the number of moles of Ca2+ ions is equal to the number of moles of Ca(NO3)2.

To calculate the number of moles of unreacted Ca2+ ions, we subtract the number of moles of Ca2+ ions that reacted from the total number of moles of Ca2+ ions available in the Ca(NO3)2 solution.

Now, let's plug in the values to calculate the number of moles of unreacted Ca2+ ions:

moles of unreacted Ca2+ = moles of Ca(NO3)2 − moles of reacted Ca2+
moles of unreacted Ca2+ = (1.22×10−2 M × 0.075 L) − (1.22×10−2 M × 0.075 L)

Since both solutions have the same volume and concentration, the number of moles of unreacted Ca2+ ions will be zero.

Therefore, the percentage of Ca+ ions that remain unprecipitated is 100%.