Calcium fluoride (CaF2) has a Ksp value of 4.0 x 10−11. Which of...
To calculate the molar solubility of CaF2 in a solution containing 0.10 M Ca(NO3)2, start with the Ksp equation and then replace the right side with actual concentration values:
4.0 x 10−11 = (Ca2+)(F⁻)2
4.0 x 10−11 = (0.1 + y)(2y)2
When the existing concentration in the solution and the Ksp value differ by more than a power of 3, which means 103 or 1000, the equation can be simplified by approximating (0.1 + y) as 0.1. Ultimately the value of y would be 1000 times less compared to 0.10 M and can be ignored; for instance, 0.1003 or 0.1009 would round to 0.10.
After the approximation, isolate the variable y by dividing both sides by 0.4 and finding the square root of both sides by taking the square root of 1 and dividing the exponent by 2:
To calculate the molar solubility of CaF2 in a solution containing 0.10 M NaF, start with the equation:
4.0 x 10−11 = (y)(0.1 + 2y)2
Simplify the equation by approximating (0.1 + 2y) to be 0.1 and solve for y by dividing both sides of the equation by 0.01:
4.0 x 10−11 = (y)(0.1)2
y = 4.0 x 10−9 M
Note the vast difference in solubility due to the common ion effect. 1.0 x 10−5 M and 4.0 x 10−9 M are the two molar solubilities, and they differ by a factor of about 2500. So CaF2 is approximately 2500 times more soluble in a solution containing 0.10 M Ca(NO3)2 than in a solution containing 0.10 M NaF.