Calculate the molar solubility of mercury (I) bromide, Hg2Br2, in 1.0 M KBr. The Ksp for Hg2Br2 is 5.6 X 10−23. (Hint: How would the Br− concentration from the sparingly soluble compound itself compare to the Br− concentration that comes from the KBr?

The correct answer is 5.6 × 10⁻²³ M.

Explanation:

As a highly soluble salt, KBr dissolves easily in water, while Hg₂Br₂ is very less soluble in comparison to KBr.

Let the solubility of Hg₂Br₂ is S mol per liter.

Therefore,

KBr (s) (1.0 M) ⇒ K⁺ (aq) (1M) + Br⁻ (aq) (1M)

Hg₂Br₂ (s) (1-S) ⇔ Hg₂⁺ (aq) (S) + 2Br⁻ (aq) (2S)

Net [Br-] = (2S + 1) M

Ksp = S (2S + 1)²

Ksp = S (4S² + 1 + 4S)

Ksp = 4S³ + S + 4S²

As the solubility is extremely less, therefore, we can ignore S² and S³. Now,

Ksp = S = 5.6 × 10⁻²³ M

Hence, the solubility of Hg₂Br₂ is 5.6 × 10⁻²³ M.

i think   the answer is c hope i'm right