First, let's understand the dissociation of the sparingly soluble salt
A3B4. When it dissolves in water, it dissociates into its ions as follows:
A3B4(s)↔3A4+(aq)+4B3−(aq) If the solubility of
A3B4 is
s grams per liter, and the molar mass of
A3B4 is
M grams per mole, then the molar solubility (which is the concentration of the dissolved salt in moles per liter) is given by:
smol= Now, upon dissociation, one mole of
A3B4 produces 3 moles of
A4+ ions and 4 moles of
B3− ions. Therefore, if the molar solubility is
smol , the concentration of
A4+ ions will be:
[A4+]=3smol=3 And the concentration of
B3− ions will be:
[B3−]=4smol=4 The solubility product constant
K sp is given by the product of the ion concentrations, each raised to the power of their respective coefficients in the balanced equation:
K sp=[A4+]3[B3−]4 Substituting the values of the ion concentrations:
K sp=(3)3(4)4Now calculate the expression inside the parenthesis:
K sp=()3()4Which simplifies to:
Multiplying these terms together gives:
K sp=Further simplification yields:
K sp= Therefore, the correct expression for
K sp is:
K sp=6912()7The correct option is:
Option A