$Al(OH)_3 \rightarrow Al^{3+} + 3\bar{O}H$ is
$Al(OH)_3$ Solubility Product ($Ksp$) Derivation
The dissociation reaction for aluminum hydroxide ($Al(OH)_3$) in water is given as:
$Al(OH)_3 \rightarrow Al^{3+} + 3\bar{O}H$
The solubility product constant ($Ksp$) represents the equilibrium constant for the dissolution of a sparingly soluble ionic compound. For a general compound $M_xN_y$ that dissociates into $xM^{y+}$ ions and $yN^{x-}$ ions:
$M_xN_y(s) \rightleftharpoons xM^{y+}(aq) + yN^{x-}(aq)$
The expression for the solubility product is:
$K_{sp} = (M^{y+})^x (N^{x-})^y$
Where $(M^{y+})$ and $(N^{x-})$ represent the molar concentrations of the respective ions at saturation.
Applying this definition to the dissociation of $Al(OH)_3$:
- The cation is $Al^{3+}$ with a stoichiometric coefficient of 1.
- The anion is $\bar{O}H$ with a stoichiometric coefficient of 3.
Therefore, the $Ksp$ expression is:
$K_{sp} = (Al^{3+})^1 (\bar{O}H)^3$
Simplifying the expression gives:
$K_{sp} = (Al^{3+})(\bar{O}H)^3$
Important Questions from Chemical Equilibrium
The vapour pressures of water, acetone and ethanol at 293 K are 2.34 kPa, 12.36 kPa and 5.85 kPa respectively. Which of the following has the lowest boiling point?
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The equilibrium dissociation energy of a diatomic molecule is 4.75 eV and its stretching frequency corresponds to 0.5 eV. The minimum energy required to dissociate the molecule in eV is
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