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)
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.
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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