The solubility of Sr(OH)2 at 298 K is 19 | Class 11 Chemistry Chapter Equilibrium, Equilibrium NCERT Solutions

Question:

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.

Answer:

Solubility of Sr(OH)2 = 19.23 g/L

Then, concentration of Sr(OH)2

= 19.23 / 121.63 M

= 0.1581 M

Sr(OH)2(aq) →  Sr2+(aq) + 2 (OH-)(aq)

∴Sr2+  = 0.1581M

[OH-] =  2 x 0.1581M = 0.3126 M

Now

Kw  =  [OH-] [H+]

10-14 / 0.3126  = [H+]

⇒ [H+] =  3.2 x 10-14

∴ pH = 13.495 = 13.50


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Welcome to the NCERT Solutions for Class 11 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 58: The solubility of Sr(OH)2 at 298 K is 19.23 g/L of solution. Calculate the concentrations of stronti....