91影视

First, we need to write the solubility reaction of SrSO4 and the expression for its solubility product Ksp. The solubility reaction is as follows: SrSO鈧� (s) 鈬� Sr虏鈦� (aq) + SO鈧劼测伝 (aq) The solubility product Ksp is equal to the product of the concentrations of the dissolved ions: Ksp = [Sr虏鈦篯 [SO鈧劼测伝]

We are given the osmotic pressure (蟺) and need to find the concentration of the ions in the dissolved SrSO鈧�. Using the osmotic pressure formula, we can solve for the molar concentration C: 蟺 = nCRT Since SrSO鈧� dissociates into 2 ions, the number of particles (n) is 2. The osmotic pressure is given as 21 torr. We need to convert this to the same unit as the ideal gas constant (R). Since 1 atm = 760 torr, we have: 蟺 = 21 torr 脳 \(\frac{1\, \text{atm}}{760\, \text{torr}}\) = 0.02763 atm The temperature (T) is given in Celsius and will need to be converted to Kelvin for the purpose of this calculation: T = 25掳C + 273.15 = 298.15 K The ideal gas constant R is given as 0.0821 L atm / K mol. Now we can plug in the values and solve for the concentration (C): 0.02763 atm = 2 脳 C 脳 0.0821 L atm / K mol 脳 298.15 K C = 0.02763 atm / (2 脳 0.0821 L atm / K mol 脳 298.15 K) = 5.618 脳 10鈦烩伌 M

Now that we have the concentration of the ions, we can find the solubility product Ksp. Since the stoichiometry of the reaction is 1:1, the concentrations of Sr虏鈦� and SO鈧劼测伝 are equal, and both are equal to the solubility of SrSO鈧� in the dissolved state. Thus: Ksp = [Sr虏鈦篯 [SO鈧劼测伝] = (5.618 脳 10鈦烩伌)虏 = 3.15 脳 10鈦烩伔 So, the solubility product (Ksp) of SrSO鈧� at 25掳C is 3.15 脳 10鈦烩伔.