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Karnataka Board PUCPUC Science Class 11

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

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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.

Numerical
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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

\[\ce{Sr(OH)_{2(aq)} -> Sr^{2+}_{(aq)} + 2(OH^-)_{(aq)}}\]

`therefore ["Sr"^(2+)] = 0.1581  "M"`

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

Now

`"K"_"w" = ["OH"^-]["H"^+]`

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

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

`therefore "pH" = 13.495;  13.50`

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