The Ksp of mercury(II) hydroxide, Hg(OH)2, is 3.60 26. Calculate the molar solubility, s, of this compound.
The Correct Answer and Explanation is :
Calculation of Molar Solubility of Mercury(II) Hydroxide (Hg(OH)₂)
The solubility product constant (Ksp) of mercury(II) hydroxide is given as: Ksp=3.60×10−26K_{sp} = 3.60 \times 10^{-26}
Step 1: Dissociation Equation
Mercury(II) hydroxide dissociates in water as: Hg(OH)2(s)⇌Hg2+(aq)+2OH−(aq)Hg(OH)_2 (s) \rightleftharpoons Hg^{2+} (aq) + 2OH^- (aq)
Let the molar solubility of Hg(OH)₂ be ss. Then, at equilibrium:
- [Hg2+][Hg^{2+}] = ss
- [OH−][OH^-] = 2s2s (since each mole of Hg(OH)₂ releases 2 moles of OH⁻)
Step 2: Write the Ksp Expression
Ksp=[Hg2+][OH−]2K_{sp} = [Hg^{2+}][OH^-]^2
Substituting equilibrium concentrations: Ksp=(s)(2s)2K_{sp} = (s)(2s)^2 Ksp=s(4s2)=4s3K_{sp} = s (4s^2) = 4s^3
Step 3: Solve for ss
3.60×10−26=4s33.60 \times 10^{-26} = 4s^3 s3=3.60×10−264s^3 = \frac{3.60 \times 10^{-26}}{4} s3=9.00×10−27s^3 = 9.00 \times 10^{-27} s=9.00×10−273s = \sqrt[3]{9.00 \times 10^{-27}} s≈2.08×10−9 Ms \approx 2.08 \times 10^{-9} \text{ M}
Thus, the molar solubility of Hg(OH)₂ is 2.08×10−92.08 \times 10^{-9} M.
Explanation
The molar solubility of mercury(II) hydroxide is derived using its solubility product constant. Since the solid dissolves in water to release 1 mole of Hg2+Hg^{2+} and 2 moles of OH−OH^- per formula unit, we use these stoichiometric ratios in the Ksp expression. By solving the cubic equation, we determine that only a very small amount of Hg(OH)₂ dissolves in water due to the extremely low Ksp value, indicating very low solubility.
Now, let’s generate an image representing this solubility equilibrium.
Here is an illustration representing the solubility equilibrium of mercury(II) hydroxide (Hg(OH)₂) in water. It visually shows the dissociation process and the equilibrium state. Let me know if you need any modifications!
