. A 13 mL sample of a solution of AlBr3 was diluted with water to 56 mL. A 25 mL sample of the dilute solution was found to contain 2.21 M Br-. What was the concentration of AlBr3 in the original undiluted solution?
13 mL is diluted to 56 mL so everything in the original sample is now in 56 mL. If we take 25 mL of that and the M is 2.21M, the the amount in the 56 mL is 2.21M x (56 mL/25 mL) = ?
That's for Br^-. Divide by 3 for AlBr3.
To solve this problem, we need to use the principles of dilution and molarity.
First, let's calculate the moles of Br- in the 25 mL sample:
Moles of Br- = Molarity × Volume
Moles of Br- = 2.21 M × 0.025 L
Moles of Br- = 0.05525 moles
Next, we need to calculate the dilution factor, which is the ratio of the final volume to the initial volume:
Dilution factor = Final volume / Initial volume
Dilution factor = 56 mL / 13 mL
Dilution factor = 4.3077 (approximately)
Since the dilution factor is equal to the ratio of the final concentration to the initial concentration, we can use this information to find the initial concentration of Br-:
Final concentration of Br- = Initial concentration of Br- / Dilution factor
2.21 M = Initial concentration of Br- / 4.3077
Solving for the initial concentration of Br-:
Initial concentration of Br- = 2.21 M × 4.3077
Initial concentration of Br- = 9.508 M (approximately)
Now, since AlBr3 is a 1:3 electrolyte and the concentration of Br- is three times the concentration of AlBr3, the concentration of AlBr3 will be one-third of the initial concentration of Br-:
Concentration of AlBr3 = (1/3) × Initial concentration of Br-
Concentration of AlBr3 = (1/3) × 9.508 M
Concentration of AlBr3 = 3.169 M (approximately)
Therefore, the concentration of AlBr3 in the original undiluted solution is 3.169 M.