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.
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.
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1 Answer
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Solubility of Sr (OH)2=19.23g/L
The molecular weight of Sr (OH)2 is 87.6 + 2 (17)=121.6
Then, concentration of Sr (OH)2=19.23 /121.63M=0.1581M
Sr (OH)2 (aq)→Sr2+ (aq)+2 (OH−) (aq)
∴ [Sr2+]=0.1581M
[OH−]=2×0.1581M=0.3126M
Now, Kw= [OH−] [H+]
=> [H+] = 10−14 / 0.3126
=> [H+]=3.2×10−14
∴pH= 13.495
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