Question

A solution is prepared by diluting \(225 \mathrm{~mL}\) of \(0.1885 \mathrm{M}\) aluminum sulfate solution with water to a final volume of 1.450 L. Calculate (a) the number of moles of aluminum sulfate before dilution. (b) the molarities of the aluminum sulfate, aluminum ions, and sulfate ions in the diluted solution.

Short answer

Question: Calculate (a) the number of moles of aluminum sulfate before dilution and (b) the molarities of aluminum sulfate, aluminum ions, and sulfate ions in a diluted solution having an initial molarity of 0.1885 M, a volume of 225 mL before dilution, and a final volume of 1.450 L after dilution. Answer: (a) The number of moles of aluminum sulfate before dilution is 0.0424 mol. (b) The molarities of aluminum sulfate, aluminum ions, and sulfate ions in the diluted solution are 0.0291 M.

Step-by-step solution

Calculate the number of moles of aluminum sulfate before dilution

To find the number of moles of aluminum sulfate before dilution, we can use the formula: moles = molarity × volume The molarity of the aluminum sulfate solution before dilution is 0.1885 M, and its volume is 225 mL. We need to convert the volume from mL to L by dividing it by 1000: Volume in L = \(\frac{225}{1000}\) = 0.225 L Now we can calculate the number of moles of aluminum sulfate: moles = 0.1885 M × 0.225 L moles = 0.0424 mol

Calculate the molarity of the diluted aluminum sulfate solution

After dilution, the final volume of the solution is 1.450 L. To calculate the molarity of the diluted aluminum sulfate solution, we can use the dilution formula: M1 × V1 = M2 × V2 Where M1 is the initial molarity, V1 is the initial volume, M2 is the final molarity, and V2 is the final volume. We know M1 (0.1885 M), V1 (0.225 L), and V2 (1.450 L). We need to find M2. 0.1885 M × 0.225 L = M2 × 1.450 L Now solve for M2: M2 = \(\frac{0.1885 \times 0.225}{1.450}\) M2 ≈ 0.0291 M

Calculate the molarities of aluminum ions and sulfate ions in the diluted solution

The molarity of the aluminum sulfate solution is the same as the molarities of the aluminum ions and sulfate ions because each aluminum sulfate molecule dissociates into one aluminum ion and one sulfate ion when it dissolves in water. So, the molarity of the aluminum ions and sulfate ions in the diluted solution is the same as the molarity of the diluted aluminum sulfate solution, which is 0.0291 M. So, the final answers are: (a) the number of moles of aluminum sulfate before dilution: 0.0424 mol (b) the molarities of the aluminum sulfate, aluminum ions, and sulfate ions in the diluted solution: 0.0291 M