A saturated solution of calcium chromate (CaCrO4) freezes at -0.10 C. What is Ksp for CaCrO4? Assume complete dissociation. The density of the solution is 1.0 g/mL.
I would do this.
delta T = i*Kf*m
0.10 = 2*1.86*m
Solve m = 0.02688
m = mols/kg solvent and for such dilute solution and a density of 1.0 g/mL that is essentially 1L solution or molarity.
So Ksp = (Ca^2+)(CrO4^2-)
Ksp = (0.02688)^2
I don't have that Ksp value in any of my texts but this site is in the same park as this answer of 7.2E-4
http://users.stlcc.edu/gkrishnan/ksptable.html
Thanks so much! I never would've figured that out.
To solve this problem, we can use the freezing point depression equation:
ΔT = Kf * m
Where:
ΔT = Freezing point depression
Kf = Cryoscopic constant (constant for the solvent)
m = Molality of the solute particles in the solution
First, let's determine the molality of the solute particles in the solution.
Molar mass of CaCrO4 = (40.08 g/mol + 52 g/mol + 4 * 16 g/mol) = 116.08 g/mol
1.0 g of the solution has a volume of 1 mL, which is equivalent to 0.001 L.
Therefore, the number of moles of CaCrO4 in 1 mL of the solution is:
moles = mass / molar mass = (1.0 g / 116.08 g/mol) = 0.00861 mol
Now, we need to calculate the molality (m) of the solute particles.
Molality (m) = moles of solute / mass of solvent (in kg)
Since the volume of the solution is 1 mL and has a density of 1.0 g/mL, the mass of the solvent (water) is 0.999 g.
Therefore, the molality (m) can be calculated as:
m = 0.00861 mol / (0.999 g / 1000) = 8.623 mol/kg
Next, we need to find the freezing point depression (ΔT).
Since the solution freezes at -0.10 °C, the freezing point depression (ΔT) is:
ΔT = Tf - T
= 0.10 °C - (-0.10 °C)
= 0.20 °C
Finally, we use the freezing point depression equation to find Ksp:
Ksp = (ΔT) / (Kf * m)
Since the problem states that CaCrO4 is completely dissociated, the solute dissociates into Ca2+ and CrO4^2- ions in the solution.
The cryoscopic constant (Kf) for water is 1.86 °C/m.
Using the values we have calculated:
Ksp = (0.20 °C) / (1.86 °C/m * 8.623 mol/kg)
= 0.0139 mol^3/kg^3
Therefore, the Ksp for CaCrO4 is approximately 0.0139 mol^3/kg^3.