Question

Suppose 2.19g of barium acetate is dissolved of 15oml barium of a 0.10M acetate is dissolved in of a aqueous solution of sodium chromate. Calculate the final molarity of acetate anion in the solution. You can assume the volume of the solution doesn't change when the barium acetate is dissolved in it. Be sure your answer has the correct number of significant digits.

Answer 1

The question is incomplete, here is the complete question:

Suppose 2.19 g of barium acetate is dissolved in 150 mL of a 0.10M of aqueous solution of sodium chromate. Calculate the final molarity of acetate anion in the solution. You can assume the volume of the solution doesn't change when the barium acetate is dissolved in it. Be sure your answer has the correct number of significant digits.

Answer: The final molarity of acetate ion in the solution is 0.12 M

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}$     .....(1)

• For Sodium chromate:

Molarity of sodium chromate solution = 0.10 M

Volume of solution = 150 mL

Putting values in equation 1, we get:

$0.10M=\frac{\text{Moles of sodium chromate}\times 1000}{150}\\\\\text{Moles of sodium chromate}=\frac{(0.10\times 150)}{1000}=0.015mol$

• For barium acetate:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of barium acetate = 2.19 g

Molar mass of barium acetate = 255.43 g/mol

Putting values in above equation, we get:

$\text{Moles of barium acetate}=\frac{2.19g}{255.43g/mol}=0.0086mol$

The chemical equation for the reaction of barium acetate and sodium chromate follows:

$Ba(CH_3CO_2)_2+Na_2CrO_4\rightarrow BaCrO_4(s)+2Na^+(aq.)+2CH_3CO_2^-(aq.)$

By stoichiometry of the reaction:

1 mole of barium acetate reacts with 1 mole of sodium chromate

So, 0.0086 moles of barium acetate will react with = $\frac{1}{1}\times 0.0086mol$ of sodium chromate

As, given amount of sodium chromate is more than the required amount. So, it is considered as an excess reagent.

Thus, barium acetate is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of barium acetate produces 2 moles of acetate ions

So, 0.0086 moles of barium acetate will produce = $\frac{2}{1}\times 0.0086=0.0172mol$ of acetate ion

Now, calculating the molarity of acetate ions in the solution by using equation 1:

Moles of acetate ion = 0.0172 moles

Volume of solution = 150 mL

Putting values in equation 1, we get:

$\text{Molarity of acetate ions}=\frac{0.0172\times 1000}{150}\\\\\text{Molarity of acetate ions}=0.12M$

Hence, the final molarity of acetate ion in the solution is 0.12 M