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23 August, 04:38

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?

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Answers (2)
  1. 23 August, 05:22
    0
    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
  2. 23 August, 05:23
    0
    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
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