NCERT Class XI Chemistry Equilibrium Solutions

Question : 58

Total: 73

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.

Solution:

Molar mass of Sr(OH)2 = 87.6 + 34 = 121.6 gmol–1
Solubility of Sr(OH)2 in moles L–1 =

19.23gL−1

121.6gmol−1

= 0.1581 M
Assuming complete dissociation, Sr(OH)2 → Sr2++2OH–
∴ [Sr2+] = 0.1581 M, [OH–] = 2 × 0.1581 = 0.3162 M
pOH = –log (0.3162) = 0.5,
∴ pH = 14 – 0.5 = 13.5

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