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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. - Chemistry

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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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)) -> Sr_((aq))^(2+) + 2(OH^(-))_((aq))`

`:. [Sr^(2+)] = 0.1581 M`

`[OH^(-)] = 2xx 0.1581 M = 0.3126 M`

Now

`K_w = [OH^-][H^+]`

`10^(-14)/0.3126 = [H^+]`

`=> [H^+] = 3.2 xx 10^(-14)`

`:. pH = 13.495;  13.50`

  Is there an error in this question or solution?
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APPEARS IN

NCERT Class 11 Chemistry Textbook
Chapter 7 Equilibrium
Q 58 | Page 229
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