"The osmotic pressure of a saturated solution of strontium sulfate at 25C is 21 torr. What is the solubility product of this salt at 25C (Hint: calculate molarity from the colligative properties eqn of osmotic pressure and use it in the equilibrium expression"
I converted torr to atm and solved P=MRT for molarity (M) = 0.00113M
but I'm not sure what to do next...if you can tell me how to do this problem that would be much appreciated!!!
Thank you
To find the solubility product of strontium sulfate (SrSO4) at 25°C, you can use the molarity you calculated (M = 0.00113) and the equilibrium expression for the solubility product.
The equilibrium expression for the solubility product of SrSO4 is:
Ksp = [Sr2+][SO42-]
Since strontium sulfate dissociates into one strontium ion (Sr2+) and one sulfate ion (SO42-), the solubility product can be written as:
Ksp = [Sr2+][SO42-]
To find the solubility product, you need to determine the concentration of the strontium ion ([Sr2+]) and sulfate ion ([SO42-]) in the saturated solution.
From the colligative properties equation of osmotic pressure, you calculated the molarity to be 0.00113 M.
Since a saturated solution of strontium sulfate will dissociate into one strontium ion and one sulfate ion, the concentration of strontium ion and sulfate ion in the saturated solution will also be 0.00113 M.
Therefore, you can substitute this molarity value into the equilibrium expression:
Ksp = [Sr2+][SO42-] = (0.00113)(0.00113)
Now you can solve for the solubility product (Ksp):
Ksp = 0.00113^2 = 1.28 x 10^-6
Therefore, the solubility product of strontium sulfate at 25°C is 1.28 x 10^-6.
To find the solubility product of strontium sulfate (SrSO4) at 25°C, we need to use the colligative properties equation for osmotic pressure and then apply it to the equilibrium expression for the solubility product.
Step 1: Calculate the molarity (M) of the solution using the given osmotic pressure.
Given: OsMotic Pressure (π) = 21 torr = 21/760 atm = 0.02763 atm
R = 0.0821 L·atm/(K·mol) (ideal gas constant)
T = 25°C + 273.15 K = 298.15 K (temperature in Kelvin)
P = MRT
M = P/(RT)
M = 0.02763 atm / (0.0821 L·atm/(K·mol) * 298.15 K)
M = 0.001255 M (rounded to 4 decimal places)
Step 2: Write the balanced chemical equation for the dissolving of strontium sulfate (SrSO4) in water.
SrSO4(s) ↔ Sr2+(aq) + SO4^2-(aq)
Step 3: Write the solubility product expression for SrSO4.
Ksp = [Sr2+][SO4^2-]
Step 4: Substitute the calculated molarity (M) of the solution into the equilibrium expression and solve for the solubility product (Ksp).
Ksp = (0.001255 M)(0.001255 M)
Ksp = 1.57703 × 10^-6 (rounded to 4 decimal places)
Therefore, the solubility product of strontium sulfate (SrSO4) at 25°C is approximately 1.57703 × 10^-6.
I think you omitted the van't Hoff factor for SrSO4. There are two particles so
P = iMRT which makes the molarity just 1/2 of your number = ??
Since that is the solubility of SrSO4, then
SrSO4(s) ==> Sr^+ + SO4^-2
Ksp = (Sr^+)(SO4^-2)
Sr^+2 = ??
SO4^-2 = ??
The internet gave me a Ksp 3.44 x 10^-7 which makes this fairly close.