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-Dominant- [34]
3 years ago
14

f the Ksp for HgBr2 is 2.8×10−14, and the mercury ion concentration in solution is 0.085 M, what does the bromide concentration

need to be for a precipitate to occur?
Chemistry
1 answer:
goldfiish [28.3K]3 years ago
7 0

Answer:

0.057 M

Explanation:

Step 1: Given data

Solubility product constant (Ksp) for HgBr₂: 2.8 × 10⁻⁴

Concentration of mercury (II) ion: 0.085 M

Step 2: Write the reaction for the solution of HgBr₂

HgBr₂(s) ⇄ Hg²⁺(aq) + 2 Br⁻

Step 3: Calculate the bromide concentration needed for a precipitate to occur

The Ksp is:

Ksp = 2.8 × 10⁻⁴ = [Hg²⁺] × [Br⁻]²

[Br⁻] = √(2.8 × 10⁻⁴/0.085) = 0.057 M

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Answer:

The hollow ball is more buoyant than the solid ball.

Explanation:

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3 years ago
And non-flammable gases<br> Noble gases are<br> that have low chemical
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Answer:

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3 years ago
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