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.

Asked by Pragya Singh | 1 year ago |  110

##### Solution :-

Solubility of Sr(OH)2 = 19.23 g/L

Then, concentration of Sr(OH)2

$$=\dfrac{19.23}{121.63}M\\ \\$$

$$=0.1581M$$

Sr(OH)2(aq)→  Sr2+(aq) + 2(OH)(aq)

[ Sr2+]= 0.1581M

[OH– ]=  2 × 0.1581M

=0.3126

Now,

Kw=[OH][H+]

⇒ $$\dfrac{10^{-14}}{0.3126}=[H^{+}]$$

⇒ $$[H^{+}]=3.2\times 10^{-14}$$

pH = 13.50

Answered by Abhisek | 1 year ago

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