Question

1.What volume (in L) of 0.500 M sodium sulfate will react with 275 mL of 0.250 M barium chloride to completely precipitate all Ba2+ in the solution?

Answer 1

Let's start by writing a balanced equation for the reaction:
BaCl2 + Na2SO4 -> BaSO4 + 2NaCl

We can see that 1 mole of BaCl2 reacts with 1 mole of Na2SO4 to produce 1 mole of BaSO4.

First, we need to determine the number of moles of BaCl2 present in 275 mL of 0.250 M solution:
0.250 moles/L x 0.275 L = 0.06875 moles of BaCl2

Since 1 mole of BaCl2 reacts with 1 mole of Na2SO4, we need 0.06875 moles of Na2SO4 to react completely with the BaCl2.

Next, we can use the concentration and volume of the Na2SO4 solution to determine the amount of Na2SO4 needed:
0.500 moles/L x V = 0.06875 moles
V = 0.06875 moles / 0.500 moles/L
V = 0.1375 L or 137.5 mL

Therefore, we need 137.5 mL of 0.500 M sodium sulfate to react completely with 275 mL of 0.250 M barium chloride.