Let’s solve the problem step-by-step.
1. Determine the concentrations of the ions:
- We are given that 0.0260 moles of the salt [tex]\( AB_3 \)[/tex] are soluble in 1.00 L of water.
- When [tex]\( AB_3 \)[/tex] dissolves in water, it dissociates according to the equation:
[tex]\[
AB_3(s) \rightleftharpoons A^{3+}(aq) + 3B^-(aq)
\][/tex]
- For each mole of [tex]\( AB_3 \)[/tex] that dissolves, it releases 1 mole of [tex]\( A^{3+} \)[/tex] ions and 3 moles of [tex]\( B^- \)[/tex] ions.
2. Calculate the concentration of [tex]\( A^{3+} \)[/tex]:
- The concentration of [tex]\( A^{3+} \)[/tex] ions will be equal to the number of moles of [tex]\( AB_3 \)[/tex] dissolved, since 1 mole of [tex]\( AB_3 \)[/tex] produces 1 mole of [tex]\( A^{3+} \)[/tex].
[tex]\[
[A^{3+}] = 0.0260 \, \text{M}
\][/tex]
3. Calculate the concentration of [tex]\( B^- \)[/tex]:
- The concentration of [tex]\( B^- \)[/tex] ions will be three times the number of moles of [tex]\( AB_3 \)[/tex] dissolved, since 1 mole of [tex]\( AB_3 \)[/tex] produces 3 moles of [tex]\( B^- \)[/tex].
[tex]\[
[B^-] = 3 \times 0.0260 \, \text{M} = 0.0780 \, \text{M}
\][/tex]
4. Write the expression for the solubility product constant ([tex]\( K_{sp} \)[/tex]):
- The [tex]\( K_{sp} \)[/tex] expression for the salt [tex]\( AB_3 \)[/tex] is given by:
[tex]\[
K_{sp} = [A^{3+}] \times [B^-]^3
\][/tex]
5. Substitute the concentrations into the [tex]\( K_{sp} \)[/tex] expression:
- We have [tex]\( [A^{3+}] = 0.0260 \, \text{M} \)[/tex] and [tex]\( [B^-] = 0.0780 \, \text{M} \)[/tex].
[tex]\[
K_{sp} = (0.0260) \times (0.0780)^3
\][/tex]
6. Calculate [tex]\( K_{sp} \)[/tex]:
- Perform the calculations:
[tex]\[
K_{sp} = 0.0260 \times (0.0780)^3 = 0.0260 \times 0.000474552 = 1.2338352 \times 10^{-5}
\][/tex]
So, the solubility product constant [tex]\( K_{sp} \)[/tex] of the salt [tex]\( AB_3 \)[/tex] at [tex]\( 25^{\circ}C \)[/tex] is [tex]\( 1.2338352 \times 10^{-5} \)[/tex].
