Respuesta :
The question is incomplete, here is the complete question:
Suppose 2.19 g of barium acetate is dissolved in 150. mL of a 0.10 M 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 ions in the solution is 0.115 M
Explanation:
- For barium acetate:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of barium acetate = 2.19 g
Molar mass of barium acetate = 255.43 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of barium acetate}=\frac{2.19g}{255.43g/mol}=0.0086mol[/tex]
- For sodium chromate:
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex] .....(1)
Molarity of sodium chromate solution = 0.10 M
Volume of solution = 150 mL = 0.150 L (Conversion factor: 1 L = 1000 mL)
Putting values in equation 1, we get:
[tex]0.10M=\frac{\text{Moles of sodium chromate}}{0.150L}\\\\\text{Moles of sodium chromate}=(0.10mol/L\times 0.150L)=0.015mol[/tex]
The chemical equation for the reaction of barium acetate and sodium chromate follows:
[tex]Ba(CH_3COO)_2+Na_2CrO_4\rightarrow BaCrO_4+2CH_3COONa[/tex]
By Stoichiometry of the reaction:
1 moles of barium acetate reacts with 1 mole of sodium chromate
So, 0.0086 moles of barium acetate will react with = [tex]\frac{1}{1}\times 0.0086=0.0086mol[/tex] 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 moles of barium acetate produces 2 moles of sodium acetate
So, 0.0086 moles of barium acetate will produce = [tex]\frac{2}{1}\times 0.0086=0.0172moles[/tex] of sodium acetate
1 mole of sodium acetate produces 1 mole of sodium ions and 1 mole of acetate ions
Now, calculating the molarity of acetate ions by using equation 1:
Moles of acetate ions = 0.0172 moles
Volume of solution = 0.150 L
Putting values in equation 1, we get:
[tex]\text{Molarity of }CH_3COO^-\text{ ions}=\frac{0.0172mol}{0.150L}\\\\\text{Molarity of }CH_3COO^-\text{ ions}=0.115M[/tex]
Hence, the final molarity of acetate ions in the solution is 0.115 M
