Suppose 0.490g of lead(II) acetate is dissolved in 50 mL of a 57.0 m M aqueous solution of ammonium sulfate. Calculate the final molarity of acetate anion in the solution. You can assume the volume of the solution doesn’t change when the lead(II) acetate is dissolved in it. Round your answer to 2 significant digits
To calculate the final molarity of acetate anion in the solution, we need to determine the number of moles of acetate anion and the final volume of the solution.
First, let's calculate the number of moles of lead(II) acetate:
moles = mass / molar mass
moles = 0.490g / (207.2 g/mol + 2 * 12.01 g/mol + 4 * 16.00 g/mol)
moles = 0.490g / 227.23 g/mol
moles = 0.00215 mol
Next, let's determine the final volume of the solution. We are given that the initial volume is 50 mL and the volume doesn't change when the lead(II) acetate is dissolved, so the final volume is also 50 mL or 0.050 L.
Now, let's calculate the final molarity of acetate anion:
molarity = moles / volume
molarity = 0.00215 mol / 0.050 L
molarity = 0.0430 M
Therefore, the final molarity of acetate anion in the solution is 0.0430 M.