The solubility of Sr(OH)2 at 298 K is 19.23 g/L of solution. Calculate the concentrations of strontium and hydroxyl ions and the pH of the solution.
Given:
Solubility of Sr(OH)2 = 19.23 g/L
Molar mass of Sr(OH)2 = 87.6 + 34 = 121.6 g/mol
To calculate concentration of Sr(OH)2, we apply the formula,
Concentration= 
⇒[Sr(OH)2] = 
⇒[Sr(OH)2] = 0.1581 M
Ionisation of Sr(OH)2
Sr(OH)2 ↔ Sr2+ + 2(OH-)
As it is completely ionised,
∴ [Sr2+] = 0.1581 M
[OH-] = 2× 0.1581 = 0.3126M
Thus, the concentration of strontium and hydroxyl ions are 0.1581 and 0.3126
Now,
Kw = [OH-][H+]
Using the formula, we write
[H+] = 
As the value of Kw is taken as 10-14
[OH-]= 0.3126 M (calculated)
∴ [H+] = 
⇒[H+] = 3.2 × 10-14 M
To calculate pH of the solution, we apply the formula,
pH= -log(H+)
⇒pH= -log (3.2 × 10-14)
⇒pH = -log 3.2 – (-14) log 10
⇒pH= 13.49
Thus, the pH of the solution is 13.49