do not understand this problem
The Ksp of CaSO4 is 4.93× 10–5. Calculate the solubility (in g/L) of CaSO4(s) in 0.500 M Na2SO4(aq) at 25 °C.
This is a solubility product problem with a common ion. The common ion is sulfate (SO4 in CaSO4 and SO4 in Na2SO4).l
let x = solubility in mols/L.
.........CaSO4 ==> Ca^2+ + SO4^2-
.........solid......x........x
Ksp = 4.93E-5 = (Ca^2+) (SO4^2-)
For Na2SO4 ==> 2Na^+ + SO4^2-
...............2*0.5M...0.5M
Ksp you have in the problem.
(Ca^2+) substitute x
(SO4^2-) substitute x+0.5 (NoteZ: x is from CaSO4 and 0.5 from Na2SO4.)
Solve for x = (Ca^2+) = (CaSO4) in mols/L.
You want grams/L; therefore, x in mols/L * molar mass CaSO4 = solubility CaSO4 in g/L.
thank you so much! helped me alot
To solve this problem, we need to understand the concept of the solubility product constant (Ksp). Ksp is an equilibrium constant that indicates the solubility of a compound in a solution.
Given:
Ksp of CaSO4 = 4.93 × 10^-5
Concentration of Na2SO4(aq) = 0.500 M
We are asked to calculate the solubility of CaSO4(s) in this solution.
To approach this problem, we'll set up an ICE table for the dissolution of CaSO4(s) in water:
CaSO4(s) ⇌ Ca2+(aq) + SO4^2-(aq)
In this table, "I" represents the initial concentration, "C" represents the change in concentration, and "E" represents the equilibrium concentration.
Let's assume that "x" is the solubility of CaSO4(s) in moles per liter (mol/L).
The initial concentration of Ca2+ and SO4^2- ions is zero (0) since CaSO4 is a solid before it dissolves. After dissolving, the concentration of Ca2+ and SO4^2- ions will be "x" mol/L.
Using the Ksp expression, we can write:
Ksp = [Ca2+][SO4^2-]
Substituting the equilibrium concentrations, we get:
Ksp = x * x
Ksp = x^2
Given that Ksp = 4.93 × 10^-5, we can set up the equation:
4.93 × 10^-5 = x^2
Solving this equation for "x", we find the value of the solubility "x".
x = √(4.93 × 10^-5)
Calculating this value, we find that:
x ≈ 2.22 × 10^-3 mol/L
Finally, to convert the solubility from moles per liter (mol/L) to grams per liter (g/L), we need to multiply the solubility value by the molar mass of CaSO4.
The molar mass of CaSO4 is:
40.08 g/mol (Ca) + 32.07 g/mol (S) + (4 × 16.00 g/mol) (4 × O) = 136.14 g/mol
So, the solubility of CaSO4(s) in 0.500 M Na2SO4(aq) at 25 °C is:
(2.22 × 10^-3 mol/L) * (136.14 g/mol) = 0.302 g/L
Therefore, the solubility of CaSO4(s) in 0.500 M Na2SO4(aq) at 25 °C is approximately 0.302 g/L.
To calculate the solubility of CaSO4(s) in 0.500 M Na2SO4(aq) at 25 °C, you need to consider the common ion effect. The presence of Na2SO4 will provide additional sulfate ions (SO4^2-) to the solution, affecting the solubility of CaSO4.
To solve this problem, you can follow these steps:
Step 1: Write the balanced chemical equation for the dissolution of CaSO4(s) in water:
CaSO4(s) ⇌ Ca^2+(aq) + SO4^2-(aq)
Step 2: Write the expression for the solubility product constant (Ksp) of CaSO4:
Ksp = [Ca^2+][SO4^2-]
Step 3: Determine the initial concentration of the sulfate ion (SO4^2-) in the solution. In this case, it is 0.500 M since you have a 0.500 M Na2SO4 solution.
Step 4: Substitute the Ksp value and the initial sulfate ion concentration into the solubility product constant expression. Rearrange the expression to solve for the concentration of calcium ions (Ca^2+):
Ksp = [Ca^2+][0.500 M]
[Ca^2+] = Ksp / 0.500 M
Step 5: Calculate the solubility of CaSO4 by multiplying the concentration of Ca^2+ by its molar mass. The molar mass of CaSO4 is:
Ca: 40.1 g/mol
S: 32.1 g/mol
O (4x): 16.0 g/mol
Total: 136.1 g/mol
Step 6: Convert the concentration of CaSO4 to grams per liter (g/L):
Solubility (g/L) = [Ca^2+] (mol/L) * molar mass (g/mol)
By following these steps, you can calculate the solubility of CaSO4(s) in 0.500 M Na2SO4(aq) at 25 °C.