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Enter the solubility-product expression for Al(OH)3 (s)

 

This question aims to develop the understanding of solubility product $ k_{ sp } $ which is related to the solubility reactions and proportions.

To solve this question, we may use a four-step process.

Step (1) – Molar mass estimation of the subject compound employing its chemical formula.

Step (2) – Mass (in grams) estimation of the subject compound that is dissolved per unit liter of the solution.

Step (3) – Number of moles estimation of the subject compound that is dissolved per unit liter of the solution.

Step (4) – Finally the solubility product estimation of the subject solution.

Let’s consider the following solubility equation:

\[ A_{(s)} \longleftrightarrow a \ A_{(a)} \ + \ b \ B_{(a)} \]

Where the ions A and B are the ionic decompositions of C. Factors a and b are the proportions involved in the reaction. The solubility product can be estimated through the use of following equation:

\[ K_{ sp } \ = \ [ A ]^a \ \times \ [ B ]^b \]

Expert Answer

Step (1) – Molar mass estimation of Aluminum Hydroxide $ Al ( OH )_3 $:

\[ \text{Molar mass of } Al ( OH )_3 \ = \ 27 \ + \ 3 \bigg ( 1 \ + \ 16 \bigg ) \]

\[ \Rightarrow \text{Molar mass of } Al ( OH )_3 \ = \ 27 \ + \ 3 \bigg ( 17 \bigg ) \]

\[ \Rightarrow \text{Molar mass of } Al ( OH )_3 \ = \ 27 \ + \ 51 \]

\[ \Rightarrow \text{Molar mass of } Al ( OH )_3 \ = \ 78 \ g/mole \]

Step (2) – Mass (in grams) estimation of Aluminum Hydroxide $ Al ( OH )_3 $ dissolved per unit liter or 1000 milliliter solution:

Since its not given, lets assume it as $ x $.

Step (3) – Number of moles estimation of Aluminum Hydroxide $ Al ( OH )_3 $ dissolved per unit liter or 1000 milliliter solution:

\[ \text{ Moles dissolved in 1 L solution } = \ \dfrac{ \text{ Mass dissolved in 1 L solution } }{ \text{ Molar Mass } } \]

\[ \Rightarrow \text{ Moles dissolved in 1 L solution } = \ \dfrac{ x }{ 78 } \ moles \]

Step (4) – Solubility product estimation.

The solubility equation of given reaction can be written as follows:

\[ Al ( OH )_3 (s) \longleftrightarrow \ Al^{ +3 } ( aq ) \ + \ 3 \ OH^{ -1 } ( aq ) \]

This means that:

\[ [ Al ( OH )_3 ] \ = \ [ Al^{ +3 } ] \ = \ 3 [ OH^{ -1 } ] \ = \ \dfrac{ x }{ 78 } \ mole \]

\[ \Rightarrow  [ OH^{ -1 } ] \ = \ \dfrac{ x }{ 26 } \ mole \]

So:

\[ K_{ sp } \ = \ [ Al^{ +3 } ]^1 \ \times \ [ OH^{ -1 } ]^3 \]

\[ \Rightarrow K_{ sp } \ = \ \dfrac{ x }{ 78 } \ \times \ \bigg ( \dfrac{ x }{ 26 } \bigg )^3 \]

Numerical Result

\[ K_{ sp } \ = \ \dfrac{ x }{ 78 } \ \times \ \bigg ( \dfrac{ x }{ 26 } \bigg )^3 \]

Where x is the grams dissolved per unit liter of solution.

Example

For the same scenario given above, calculate the $ K_{ sp } $ if 100 g is dissolved in a 1000 mL solution.

Calculating the number moles of copper chloride $ Cu Cl $ dissolved in 1 L = 1000 mL solution:

\[ x \ = \ \dfrac{ \text{ Mass in 1000 mL solution } }{ \text{ Molar Mass } } \]

\[ \Rightarrow x \ = \ \dfrac{ 100 }{ 78 \ g/mole } \]

\[ \Rightarrow x \ = \ 1.28 \ mole/L \]

Recall the final expression:

\[ K_{ sp } \ = \ \dfrac{ x }{ 78 } \ \times \ \bigg ( \dfrac{ x }{ 26 } \bigg )^3 \]

Substituting values:

\[ K_{ sp } \ = \ \dfrac{ 1.28 }{ 78 } \ \times \ \bigg ( \dfrac{ 1.28 }{ 26 } \bigg )^3 \]

\[ K_{ sp } \ = \ 0.01652 \]

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