What is the approximate ratio of the solubility of CaSO4 (Ksp = 2.4 x 10−5) in pure water to the solubility of CaSO4 in 0.10 M Na2SO4 at the same temperature?
Dr. Bob is the greatest person of all time if he answers this
CaSO4 ===> Ca^2+ + SO4^2-
For pure water, Ksp = (Ca^2+)[(SO4)^2-]
So (Ca^2+) = x and [(SO4)^2-] = x.
Solve for x and that is the solubility of CaSO4 in pure water.
For the 0.1M Na2SO4 solution.
Ksp is the same expression.
(Ca^2+) = x
[(SOr)^2-] = 0.1 + x.
[Note: that is 0.1M from the Na2SO4 and xc from the CaSO4).
Solve for x and that is the solubility in the 0.1M Na2SO4.
This is the common ion effect. It ALWAYS decreases the solubility of a slightly soluble salt.
Your other question yesterday was tougher than this one. This one is a general chemistry question.
You have the two solubilities. You will need to calculate the ratio from that.
To find the approximate ratio of the solubility of CaSO4 in pure water to the solubility of CaSO4 in 0.10 M Na2SO4, we can use the concept of the common ion effect.
The common ion effect states that the solubility of a slightly soluble salt in a solution containing a common ion is reduced compared to its solubility in pure water.
In this case, CaSO4 is the slightly soluble salt, and Na2SO4 provides the common ion, SO4^2-.
To calculate the solubility ratio, we need to find the solubility of CaSO4 in pure water and in 0.10 M Na2SO4 and then divide the former by the latter.
First, let's consider the solubility of CaSO4 in pure water. The solubility product constant (Ksp) of CaSO4 is given as 2.4 x 10^-5. The equilibrium expression for the dissolution of CaSO4 is:
CaSO4(s) ⇌ Ca2+(aq) + SO4^2-(aq)
The solubility, represented by the concentration of Ca2+ and SO4^2-, can be assumed to be "x". Therefore, the equilibrium expression can be written as:
Ksp = [Ca2+][SO4^2-] = x * x = x^2
Since the molar ratio of Ca2+ to SO4^2- in CaSO4 is 1:1, the concentrations of both ions equal "x".
Now, solve the equilibrium expression for x:
x^2 = 2.4 x 10^-5
Taking the square root of both sides:
x ≈ √(2.4 x 10^-5) ≈ 0.0049 M
So, the solubility of CaSO4 in pure water is approximately 0.0049 M.
Next, let's consider the solubility of CaSO4 in 0.10 M Na2SO4. Since Na2SO4 provides the common ion SO4^2-, the solubility will be reduced compared to that in pure water.
The presence of 0.10 M Na2SO4 means the concentration of SO4^2- is 0.10 M. To calculate the solubility of CaSO4 in this solution, we again assume it to be "x" and write the equilibrium expression as:
Ksp = [Ca2+][SO4^2-] = x * 0.10
Since the molar ratio of Ca2+ to SO4^2- in CaSO4 is 1:1, the concentration of Ca2+ is also "x".
Now, solve the equilibrium expression for x:
(0.10 * x) * x = 2.4 x 10^-5
0.10x^2 = 2.4 x 10^-5
x^2 = (2.4 x 10^-5) / 0.10
x ≈ √(2.4 x 10^-5 / 0.10) ≈ 0.0022 M
So, the solubility of CaSO4 in 0.10 M Na2SO4 is approximately 0.0022 M.
Finally, calculate the ratio of the solubility in pure water to the solubility in 0.10 M Na2SO4:
Solubility ratio = (Solubility in pure water) / (Solubility in 0.10 M Na2SO4)
= 0.0049 M / 0.0022 M
≈ 2.23
Therefore, the approximate ratio of the solubility of CaSO4 in pure water to the solubility of CaSO4 in 0.10 M Na2SO4 is approximately 2.23.