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The Ksp for the equilibrium reaction HgBr₂(s) ⇄ Hg²⁺(aq) + 2Br⁻(aq) if HgBr₂ has a molar solubility of 2.50x10⁻⁷ M is 6.25x10⁻²⁰.
The given reaction is:
HgBr₂(s) ⇄ Hg²⁺(aq) + 2Br⁻(aq) (1)
The equilibrium constant (Ksp) for the above reaction is given by:
[tex]Ksp = [Hg^{2+}][Br^{-}]^{2}[/tex] (2)
At the equilibrium, we have (eq 1):
HgBr₂(s) ⇄ Hg²⁺(aq) + 2Br⁻(aq)
s 2s
Where:
s: is the molar solubility = 2.50x10⁻⁷ M
By entering the values of molar solubility into equation (2) we can find the value of Ksp
[tex]Ksp = [Hg^{2+}][Br^{-}]^{2} = s(2s)^{2} = 4s^{3} = 4(2.50\cdot 10^{-7})^{3} = 6.25\cdot 10^{-20}[/tex]
Therefore, the Ksp is 6.25x10⁻²⁰.
You can find more about the equilibrium constant Ksp here:
I hope it helps you!