Question

the solubility of the ionic compound m₂x₃, having a molar mass of 273 g/mol, is 3.60 × 10⁻⁷ g/l. calculate the ksp of the compound.
ksp = 3.5×10^-43
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Explanation:

Step1: Convert solubility to molarity

The solubility is given as \(3.60\times 10^{-7}\space g/L\) and the molar mass is \(273\space g/mol\). Molarity (\(s\)) is calculated by dividing solubility by molar mass:
\(s=\frac{3.60\times 10^{-7}\space g/L}{273\space g/mol}\approx1.32\times 10^{-9}\space mol/L\)

Step2: Write the dissociation equation

The compound \(M_2X_3\) dissociates as:
\(M_2X_3(s)
ightleftharpoons 2M^{3+}(aq)+3X^{2-}(aq)\)

Step3: Determine ion concentrations

From the dissociation, \([M^{3+}]=2s\) and \([X^{2-}]=3s\). Substituting \(s = 1.32\times 10^{-9}\space mol/L\):
\([M^{3+}]=2\times1.32\times 10^{-9}=2.64\times 10^{-9}\space mol/L\)
\([X^{2-}]=3\times1.32\times 10^{-9}=3.96\times 10^{-9}\space mol/L\)

Step4: Calculate \(K_{sp}\)

The solubility product constant \(K_{sp}\) for \(M_2X_3\) is given by:
\(K_{sp}=[M^{3+}]^2[X^{2-}]^3\)
Substitute the ion concentrations:
\(K_{sp}=(2.64\times 10^{-9})^2\times(3.96\times 10^{-9})^3\)
First, calculate \((2.64\times 10^{-9})^2 = 6.9696\times 10^{-18}\)
Then, calculate \((3.96\times 10^{-9})^3\approx6.156\times 10^{-26}\)
Multiply these two results:
\(K_{sp}\approx6.9696\times 10^{-18}\times6.156\times 10^{-26}\approx4.29\times 10^{-43}\) (Note: The initial incorrect \(K_{sp}\) in the image might be a placeholder; the correct calculation follows the steps above.)

Answer:

\(\approx4.3\times 10^{-43}\) (or more precisely \(4.29\times 10^{-43}\))