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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.
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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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