SELECT * FROM question_mgmt as q WHERE id=3282 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=2 AND subId=9 AND chapterId=48 and ex_no='1' AND status=1 ORDER BY CAST(question_no AS UNSIGNED) CBSE Free NCERT Solution of 11th chemistry Equilibrium the solubility of sr oh 2 at 298 k is 19 23 g l of

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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