How many milliliters of 0.580 M aluminum bromide solution are required to prepare 89.3 mL of a solution that is 0.233 M in bromide ion?

12.0 mL

Explanation:

AlBr₃ is a strong electrolyte that ionizes according to the following equation.

AlBr₃ (aq) → Al³⁺ (aq) + 3 Br⁻ (aq)

If the concentration of AlBr₃ is 0.580 M, the concentration of Br⁻ is 3 * 0.580 M = 1.74 M.

This is the initial concentration C₁. We want to prepare a diluted solution with C₂ = 0.233 M and V₂ = 89.3 mL. To find out V₁, we will use the dilution rule.

C₁ * V₁ = C₂ * V₂

1.74 M * V₁ = 0.233 M * 89.3 mL

V₁ = 12.0 mL

The initial volume is 35, 9 ml

Explanation:

We use the formula:

V initial x C Initial = V Final x C Final, being V = volume and C = concentration

V initial x 0, 580 M = 89, 3 ml x 0, 233 M

V initial = (89, 3 ml x 0, 233 M) / 0, 580 M = 35, 9 ml