Sure, let's solve this problem step-by-step:
1. Understanding the problem:
We have the solubility product constant, [tex]\(K_{\text{sp}}\)[/tex], for [tex]\( \text{AlPO}_4 \)[/tex], which is [tex]\(9.84 \times 10^{-21}\)[/tex]. We need to determine the molar solubility of [tex]\( \text{AlPO}_4 \)[/tex].
2. Express the dissociation reaction:
The compound [tex]\( \text{AlPO}_4 \)[/tex] dissociates in water as follows:
[tex]\[
\text{AlPO}_4 (s) \leftrightarrow \text{Al}^{3+} (aq) + \text{PO}_4^{3-} (aq)
\][/tex]
From this reaction, we can see that for every 1 mole of [tex]\(\text{AlPO}_4\)[/tex] that dissolves, we get 1 mole of [tex]\(\text{Al}^{3+}\)[/tex] and 1 mole of [tex]\(\text{PO}_4^{3-}\)[/tex].
3. Set up the expression for [tex]\(K_{\text{sp}}\)[/tex]:
The solubility product [tex]\(K_{\text{sp}}\)[/tex] is given by:
[tex]\[
K_{\text{sp}} = [\text{Al}^{3+}] \times [\text{PO}_4^{3-}]
\][/tex]
4. Define molar solubility:
Let [tex]\( s \)[/tex] be the molar solubility of [tex]\(\text{AlPO}_4\)[/tex]. This means:
- [tex]\([\text{Al}^{3+}] = s\)[/tex]
- [tex]\([\text{PO}_4^{3-}] = s\)[/tex]
Using the dissociation equation, the concentration terms can be replaced with [tex]\( s \)[/tex]:
[tex]\[
K_{\text{sp}} = s \times s = s^2
\][/tex]
5. Solve for the molar solubility [tex]\( s \)[/tex]:
[tex]\[
9.84 \times 10^{-21} = s^2
\][/tex]
To find [tex]\( s \)[/tex], we take the square root of both sides:
[tex]\[
s = \sqrt{9.84 \times 10^{-21}}
\][/tex]
6. Calculate [tex]\( s \)[/tex]:
Upon calculating the square root:
[tex]\[
s \approx 9.92 \times 10^{-11} \, \text{M}
\][/tex]
So, the molar solubility of [tex]\( \text{AlPO}_4 \)[/tex] at [tex]\( 25^{\circ}C \)[/tex] is approximately [tex]\( 9.92 \times 10^{-11} \, \text{M} \)[/tex].
Therefore, the correct answer is:
[tex]\[
\boxed{9.92 \times 10^{-11} \, \text{M}}
\][/tex]
