How much of a 1.50 M sodium sulfate solution in milliliters is required to completely precipitate all of the barium in 150.0 mL of a 0.250 M barium nitrate solution?

Problem 118P :

Step 1:

Given :

Molarity of Sodium sulfate(Na2SO4) solution = 1.50 M

Molarity of barium nitrate(Ba(NO3)2) solution = 0.250 M

Volume of silver nitrate(AgNO3) solution = 150.0 mL

Volume of solution is in mL, so let’s convert to L :

1 L = 1000 mL

Therefore, 150.0 mL in L will be :

= 150 mL

= 0.150 L.

First, Let’s write the chemical reaction that’s happening :

Na2SO4 (aq) + Ba(NO3)2 (aq) → 2NaNO3 (aq) + BaSO4 (s)

Now, let’s find the number of moles of barium nitrate( Ba(NO3)2) using the molarity formula as barium nitrate is the solute :

Molarity =

Therefore, the number of moles(m) can be derived from the above formula as :

Moles of solute = Molarity Volume of Solution (in L)

n = 0.250 M0.150 L

n = 0.0375 moles