Suppose 5.87 g of barium acetate is dissolved in 300. mL of a 71.0 m M aqueous solution of sodium chromate.

calculate the final molarity of barium cation in the solution. You can assume the volume of the solution doesn't change when the barium acetate is dissolved in it.

Round your answer to 2 significant digits.

@ DrBob222

I know my ions are the same as my moles but I am having trouble understanding what I do next. Can you explain how you broke down this problem? What steps am I missing? Thank you!

To calculate the final molarity of the barium cation in the solution, we first need to determine the number of moles of barium acetate.

Step 1: Calculate the number of moles of barium acetate.
Using the given mass of barium acetate (5.87 g) and the molar mass of barium acetate (255.43 g/mol), we can calculate the number of moles:

moles of barium acetate = mass of barium acetate / molar mass of barium acetate
= 5.87 g / 255.43 g/mol

Step 2: Convert the given volume of the solution from mL to L.
The volume given in the problem is 300 mL. To convert it to liters, divide by 1000:

volume of solution = 300 mL / 1000
= 0.3 L

Step 3: Calculate the final molarity of the barium cation.
Divide the moles of barium acetate by the volume of the solution in liters:

final molarity of barium cation = moles of barium acetate / volume of solution
= (5.87 g / 255.43 g/mol) / 0.3 L

Step 4: Simplify the expression and round the answer to two significant digits.
Perform the calculation to find the final molarity of the barium cation and round it to two significant digits.

final molarity of barium cation = (5.87 g / 255.43 g/mol) / 0.3 L
≈ 0.0769 mol/L
≈ 0.077 mol/L

Therefore, the final molarity of the barium cation in the solution is approximately 0.077 mol/L.

To calculate the final molarity of the barium cation in the solution, we need to determine the number of moles of barium acetate and the final volume of the solution.

First, let's find the number of moles of barium acetate:
Given mass of barium acetate = 5.87 g

The molar mass of barium acetate (Ba(CH3COO)2) can be calculated as follows:
Ba = 1 * 137.33 g/mol = 137.33 g/mol
C = 2 * 12.01 g/mol = 24.02 g/mol
H = 6 * 1.01 g/mol = 6.06 g/mol
O = 4 * 16.00 g/mol = 64.00 g/mol

Now, add up the individual molar masses:
Molar mass of barium acetate = 137.33 + 24.02 + 6.06 + 64.00 = 231.41 g/mol

Next, calculate the number of moles of barium acetate:
Number of moles = mass / molar mass = 5.87 g / 231.41 g/mol ≈ 0.0254 mol

Since the volume of the solution doesn't change when barium acetate is dissolved in it, the final volume remains 300. mL or 0.3 L.

Now, we can calculate the final molarity of the barium cation:
Molarity (M) = moles of solute / volume of solution in liters
Molarity of barium cation = 0.0254 mol / 0.3 L ≈ 0.085 M

Therefore, the final molarity of the barium cation in the solution is approximately 0.085 M, rounded to 2 significant digits.

Barium acetate = Ba(C2H3O2) = Ba(Ac)2

Ba(Ac)2 + Na2CrO4 ==> BaCrO4 + 2NaAc
mols Ba(Ac)2 initially = grams/molar mass
mols Na2CrO4 initially = 0.2L x 0.071 M = ?(I assume 71.0 m M is 71 millimolar.) and that makes all of the Na2CrO4 used.
mols Ba(Ac)2 remaining = inital Ba(Ac)2 - mols Na2C4O4 = ?
Then M (Ba^2+) = mols/L soln. L soln = 0.300 + 0.071 = ?
Then remember 2 s.f.
Again, you can add what little Ba^2+ comes from the BaCrO4 ppt (use Ksp and the common ion effect for this) but I think this is insignificant and can be neglected.