 # 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:

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

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:

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

$K_{ sp } \ = \ 0.01652$