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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 lons and the pH of the solution. |
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Answer» Solution :Molecular MASS of `Sr(OH)_2`= Sr+2[OH] =87.6+34= 121.6 g `mol^(-1)` Concentration of `Sr(OH)_2 (M)` `="WEIGHT (g)"/"Volume (L) X Molecular mass"` `=(19.23 gL^(-1))/(121.6 "g mol"^(-1))=0.1581 "mol L"^(-1)` or M `Sr(OH)_2` which is totally soluble is COMPLETELY ionized. `Sr(OH)_2=Sr_((aq))^(2+) + 2OH_((aq))^(-)` `THEREFORE [Sr^(2+)]=[Sr(OH)_2] `=0.1581 M `[ObarH]=2[Sr(OH)_2] = 2xx0.1581`=0.3162 M pOH=-log `[OH^-]` =-log (0.3162) =-(-0.5) =+0.5 pH=14.0-pOH=(14.0-0.5) =13.5 |
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