Question

Question 7.58 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.

Class XI Equilibrium Page 229

Answer 1

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Answer 2

given, solubility of Sr(OH)2 is 19.23g/L
molecular weight of Sr(OH)2 is 121.6 g/mol
          we know,
 molarity = solubility/molecular weight
               = 19.23 g/L/121.6g/mol
               =0.1581 M
Sr(OH)2   ⇔ Sr²⁺   +             2OH⁻
0.1581M    0.1581M         2×0.1581M = 0.3162M
[Sr²⁺ ] = 0.1581M and [OH⁻] = 0.3162M

we know, ionic product(Kw) = [H⁺][OH⁻]  = 10⁻¹⁴
                        [H⁺] = 10⁻¹⁴/[OH⁻] = 10⁻¹⁴/0.3162  = 3.16 × 10⁻¹⁴
now Arrhenius formula,
     pH = -log[H⁺] = -log(3.16 × 10⁻¹⁴) = 14 -0.4997 = 13.5003