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Answer :
The molar solubility of Sr(OH)2 at 0°C is 0.0674 M and at 25°C is 1.2580 M, indicating that the solubility increases with temperature.
The molar solubility of strontium hydroxide, Sr(OH)2, at two different temperatures using titration data. The solubility product expression for Sr(OH)2 dissolving in water is represented by the equilibrium:
Sr(OH)2(s) \<=> Sr2+(aq) + 2OH
-(aq)
The number of moles of HCl that reacted with the dissolved Sr(OH)2 can be calculated from the volume of the HCl solution used in the titration at each temperature.
At 0oC: 0.2000 M HCl * 0.00337 L = 0.000674 moles of HCl
At 25oC: 0.2000 M HCl * 0.06290 L = 0.012580 moles of HCl
Since HCl reacts with Sr(OH)2 in a 1:1 ratio, the moles of Sr(OH)2 that dissolved are equal to the moles of HCl that reacted. Thus, the molar solubility of Sr(OH)2 at each temperature is:
At 0oC: 0.000674 moles / 0.01000 L = 0.0674M
At 25oC: 0.012580 moles / 0.01000 L = 1.2580M
This calculation shows an increase in solubility as the temperature increases from 0oC to 25oC.
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