Question

A\(_3\)B\(_2\) is a sparingly soluble salt of molar mass M (g mol\(^{-1}\)) and solubility \(x\text{ g L}^{-1}\). The solubility product satisfies \(K_{sp} = a \left(\frac{x}{M}\right)^5\). The value of \(a\) is \(\dots\dots\dots\). (Integer answer)

Hint:

For a general salt \(A_xB_y\), the solubility product is \(K_{sp} = x^x y^y S^{(x+y)}\). For \(A_3B_2\), \(K_{sp} = 3^3 2^2 S^5 = 108 S^5\).

Answer 1

Correct Answer: 108

To find the integer value of \(a\) in the solubility product expression \(K_{sp} = a \left(\frac{x}{M}\right)^5\) for the salt A\(_3\)B\(_2\), we proceed as follows:
1. Identify the dissociation reaction for A\(_3\)B\(_2\):
A\(_3\)B\(_2(s)\) ⇌ 3A\(^+\) + 2B\(^-\).
2. If \(x\) grams per liter is the solubility, then the molar concentration of A\(_3\)B\(_2\) is \(\frac{x}{M}\) mol L\(^{-1}\).
3. From the stoichiometry of the dissociation, the concentrations are:
[A\(^+\)] = 3\(\frac{x}{M}\), [B\(^-\)] = 2\(\frac{x}{M}\).
4. Substitute these values into the expression for the solubility product:
\(K_{sp} = [A\(^+\)]^3[B\(^-\)]^2 = (3\frac{x}{M})^3(2\frac{x}{M})^2\).
5. Simplify:
\((3\frac{x}{M})^3 = 27\frac{x^3}{M^3}\),
\((2\frac{x}{M})^2 = 4\frac{x^2}{M^2}\).
6. Thus, \(K_{sp} = 27\frac{x^3}{M^3} \cdot 4\frac{x^2}{M^2} = 108\frac{x^5}{M^5}\).
7. By comparing with \(K_{sp} = a\left(\frac{x}{M}\right)^5\), it is clear that \(a = 108\).
8. Validate \(a\) against the given range 108,108, confirming \(a = 108\) is correct.