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.
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I know my ions are the same as my moles but I am having trouble understanding what I do next. What steps am I missing? If someone could please break it down for me it will help for future problems. Thank you!
I posted a response yesterday.
To calculate the final molarity of the barium cation in the solution, we need to follow a few steps:
Step 1: Convert the mass of the barium acetate to moles.
To convert grams to moles, divide the mass by the molar mass of barium acetate. The molar mass of barium acetate (Ba(CH3COO)2) can be calculated by adding up the atomic masses:
Ba: 137.33 g/mol
C: 12.01 g/mol
H: 1.01 g/mol (there are 4 hydrogens in the acetate)
O: 16.00 g/mol (there are 4 oxygens in the acetate)
The molar mass of barium acetate = (137.33) + 2(12.01) + 4(1.01) + 2(16.00) = 255.42 g/mol
To calculate moles, divide the given mass (5.87 g) by the molar mass:
moles of barium acetate = 5.87 g / 255.42 g/mol
Step 2: Calculate the moles of barium cation.
Since barium acetate dissociates into barium cation (Ba2+) and acetate anions (CH3COO-), we know that 1 mole of barium acetate yields 1 mole of barium cation.
Therefore, the moles of barium cation present in the solution are equal to the moles of barium acetate.
Step 3: Convert the volume of the solution to liters.
The given volume of the solution is 300. mL. To use it in the molarity calculation, we need to convert it to liters.
1 L = 1000 mL, so 300 mL = 300/1000 = 0.3 L.
Step 4: Calculate the final molarity of the barium cation.
Molarity (M) is defined as moles of solute divided by liters of solution.
The final molarity of the barium cation can be calculated using the formula:
Molarity (M) = moles of barium cation / volume of solution (in liters)
Final Molarity of barium cation = moles of barium cation / 0.3 L
By following these steps, you should be able to calculate the final molarity of the barium cation in the solution.