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Suppose 0.400 g of barium acetate is dissolved in 50. mL. of a 26.0 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 Be sure your answer has the correct number of significant digits change when the barium acetate is dissolved irn it

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Answer:

See explanation below

Explanation:

First, we need to write the chemical equation that is taking place here:

Ba(CH3CO2)2 + Na2CrO4 → BaCrO4 + 2NaCH3CO2

According to the balance reaction, we have a mole ratio between barium acetate and sodium chromate of 1:1. Therefore, all we need to do is calculate the limiting reactant, and in this way we can know the moles produced in the reaction.

First, let's calculate the moles of each compound. You should use the molar mass of the barium acetate which is 245.3 g/mol.

moles = m/MM

If we have volume and concentration:

moles = M * V

moles barium acetate = 0.4/255.43 = 0.0016 moles

moles sodium chromate = 0.05 * 0.026 = 0.0013 moles

According to this, the limiting reactant is the sodium chromate, therefore, it will be produced 0.0013 moles of BaCrO4 and the remaining moles of barium acetate will be:

remaining moles Ba acetate = 0.0016 - 0.0013 = 0.0003 moles

These are the moles in solution of Ba2+. This is because, BaCrO4 is a solid compound, so, it does not provide Ba2+ in solution. Instead it will be the initial reactant which is in excess.

So, the concentration will be:

[Ba] = 0.0003 / 0.05 = 0.006 M

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