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30 August, 09:20

On the basis of the Ksp values below, what is the order of the solubility from least soluble to most soluble for these compounds?

AgBr: Ksp = 5.4 x 10-13

Ag2CO3: Ksp = 8.0 x 10-12

AgCl: Ksp = 1.8 x 10-10

Ag2CO3 < AgBr < AgCl

AgBr < Ag2CO3 < AgCl

AgBr < AgCl < Ag2CO3

AgCl < Ag2CO3 < AgBr

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Answers (2)
  1. 30 August, 10:59
    0
    The order of solubility is AgBr < Ag₂CO₃ < AgCl

    Explanation:

    The solubility constant give us the molar solubilty of ionic compounds. In general for a compound AB the ksp will be given by:

    Ksp = (A) (B) where A and B are the molar solubilities = s² (for compounds with 1:1 ratio).

    It follows then that the higher the value of Ksp the greater solubilty of the compound if we are comparing compounds with the same ionic ratios:

    Comparing AgBr: Ksp = 5.4 x 10⁻¹³ with AgCl: Ksp = 1.8 x 10⁻¹⁰, AgCl will be more soluble.

    Comparing Ag2CO3: Ksp = 8.0 x 10⁻¹² with AgCl Ksp = AgCl: Ksp = 1.8 x 10⁻¹⁰ we have the complication of the ratio of ions 2:1 in Ag2CO3, so the answer is not obvious. But since we know that

    Ag2CO3 ⇄ 2 Ag⁺ + CO₃²₋

    Ksp Ag2CO3 = 2s x s = 2 s² = 8.0 x 10-12

    s = 4 x 10⁻12 ∴ s = 2 x 10⁻⁶

    And for AgCl

    AgCl ⇄ Ag⁺ + Cl⁻

    Ksp = s² = 1.8 x 10⁻¹⁰ ∴ s = √ 1.8 x 10⁻¹⁰ = 1.3 x 10⁻⁵

    Therefore, AgCl is more soluble than Ag₂CO₃

    The order of solubility is AgBr < Ag₂CO₃ < AgCl
  2. 30 August, 11:17
    0
    5.4 x 10^-13

    Explanation:

    Edmentum says it is correct
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