The solubility product of Al(OH)3 is 2.7 x 10-11. Calculate its solubility in g–L and also find out pH of this solution. (Atomic mass of A1 = 27 u).

To find the solubility of Al(OH)3Al(OH)3 in grams per liter (g/L), we first need to determine the concentration of Al3+Al3+ ions in the saturated solution. Then, we can use stoichiometry to find the concentration of Al(OH)3Al(OH)3 and finally convert it to grams per liter.

Given:

• Solubility product (KspKsp) of Al(OH)3Al(OH)3 = 2.7×10−112.7×10−11
• Atomic mass of Al (AlAl) = 27 u

1. Calculate the concentration of Al3+Al3+ ions:

The dissolution of Al(OH)3Al(OH)3 proceeds as follows:

Al(OH)3⇌Al3++3OH−Al(OH)3⇌Al3++3OH−

Let xx be the solubility of Al(OH)3Al(OH)3 in moles per liter (M). Then, the concentration of Al3+Al3+ ions will also be xx M.

Using the solubility product expression:

Ksp=[Al3+]×[OH−]3Ksp=[Al3+]×[OH−]3

Given that [OH−]=3x[OH−]=3x, we can substitute these values into the expression:

2.7×10−11=x×(3x)32.7×10−11=x×(3x)3

Solving for xx:

2.7×10−11=27x42.7×10−11=27x4

x4=2.7×10−1127x4=272.7×10−11

x4=1×10−12x4=1×10−12

x=1×10−124x=41×10−12

x=0.01 Mx=0.01M

2. Calculate the molar mass of Al(OH)3Al(OH)3:

Molar mass of Al(OH)3=Molar mass of Al+3×Molar mass of O+3×Molar mass of HMolar mass of Al(OH)3=Molar mass of Al+3×Molar mass of O+3×Molar mass of H

=27+3×16+3×1=27+3×16+3×1

=27+48+3=27+48+3

=78 g/mol=78g/mol

3. Calculate the solubility in grams per liter (g/L):

The solubility of Al(OH)3Al(OH)3 in grams per liter (g/L) is the molar mass of Al(OH)3Al(OH)3 multiplied by its molar solubility:

Solubility=Molar solubility×Molar massSolubility=Molar solubility×Molar mass

Solubility=0.01 M×78 g/molSolubility=0.01M×78g/mol

Solubility=0.78 g/LSolubility=0.78g/L

4. Calculate the pH of the solution:

Since Al(OH)3Al(OH)3 is a sparingly soluble salt, we can assume that it dissociates completely in solution. Therefore, the concentration of Al3+Al3+ ions is equal to the solubility, which is 0.01 M0.01M.

To find the pH of the solution, we need to calculate the concentration of H+H+ ions. Since Al3+Al3+ ions are neutral, for each mole of Al3+Al3+ ions, three moles of H+H+ ions are produced.

[H+]=3×0.01 M=0.03 M[H+]=3×0.01M=0.03M

Now, we can use the formula for pH:

pH=−log⁡[H+]pH=−log[H+]

pH=−log⁡(0.03)pH=−log(0.03)

pH≈−log⁡(3×10−2)pH≈−log(3×10−2)

pH≈−(−1.5229)pH≈−(−1.5229)

pH≈1.5229pH≈1.5229

Therefore, the solubility of Al(OH)3Al(OH)3 in g/L is 0.78 g/L0.78g/L, and the pH of the solution is approximately 1.521.52.