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Given the Ksp of Mecury (I) Bromide is 5.6 x 10-23 Calculate the concentration of Mecury ions (the salt exist as a dimer)

User Sabin
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Final answer:

The concentration of Hg2²+ ions in a saturated solution of Mercury(I) bromide is 1.12 x 10­8 M, and the concentration of Br­ ions is 2.24 x 10­8 M, calculated using the cube root of the Ksp value.

Step-by-step explanation:

To estimate the concentration of Hg22+ and Br­ ions using the provided Ksp of Mercury(I) bromide (Hg2Br2), we need to understand that the salt exists as a dimer. Therefore, when it dissolves, the dissociation equation is represented as:

Hg2Br2 (s) → Hg22+ (aq) + 2 Br­ (aq)

Let's designate the molar solubility of Hg2Br2 as 's'. This means [Hg22+] = s and [Br­] = 2s. Inserting these into the expression for the solubility product constant, Ksp, we get:

Ksp = [Hg22+][Br­]2 = s(2s)2 = 4s3

Substituting the given Ksp, we have:

5.6 x 10­23 = 4s3

Dividing both sides by 4:

1.4 x 10­23 = s3

Now, to find the value of s (the molar solubility), we take the cube root:

s = ∛(1.4 x 10­23)

≈ 1.12 x 10­8 M

Finally, we get the concentrations:

  • [Hg22+] = 1.12 x 10­8 M
  • [Br­] = 2(1.12 x 10­8)
  • = 2.24 x 10­8 M

User Searock
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