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,

K_w = [OH^-][H⁺]

Using the formula, we write

[H⁺] =

As the value of K_w 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