Please! This is urgent!
I don't know how to do this.
Determine the solubility at 25°C of Cd(IO3)2 in water and in 0.20 M CdCl2. Ksp of Cd(IO 3) 2 is 2.5 × 10 –8 at 25°C.
Please help!
Oh, dear desperate human! Fear not, for Clown Bot is here to put a smile on your face and help you out!
To determine the solubility of Cd(IO3)2 at 25°C in water and in 0.20 M CdCl2, we need to use the concept of Ksp. Ksp stands for "solubility product constant," which tells us the equilibrium concentration of ions in a saturated solution.
Given that the Ksp of Cd(IO3)2 is 2.5 × 10–8:
1. Solubility in water:
The solubility of Cd(IO3)2 in pure water can be represented as x (mol/L). We can use the Ksp expression to set up an equation and solve for x:
Ksp = [Cd2+][IO3-]^2
2.5 × 10–8 = (x)(x^2) [As there are two IO3- ions and one Cd2+ ion in the formula]
Solving this equation will give us the solubility of Cd(IO3)2 in water at 25°C.
2. Solubility in 0.20 M CdCl2:
In a solution of 0.20 M CdCl2, we need to consider the common ion effect. The Cd2+ ion from CdCl2 will already be present, so we'll have to adjust the equation:
Ksp = [Cd2+] * (IO3-)^2
Since we know the concentration of Cd2+ is 0.20 M, we can substitute this value and the Ksp into the equation to solve for (IO3-).
Don't worry, I won't leave you hanging! Give me just a moment to crunch the numbers and compute the solubilities for Cd(IO3)2 in water and 0.20 M CdCl2.
*dances around with a calculator*
To determine the solubility of Cd(IO3)2 in water and in 0.20 M CdCl2, we can use the concept of the solubility product constant (Ksp).
Step 1: Write the balanced chemical equation for the dissolution of Cd(IO3)2 in water.
Cd(IO3)2 (s) ↔ Cd2+ (aq) + 2IO3- (aq)
Step 2: Write the expression for Ksp.
Ksp = [Cd2+][IO3-]^2
Step 3: Determine the solubility in water.
Since the solubility is x, the concentration of Cd2+ and IO3- ions will also be x.
- [Cd2+] = x
- [IO3-] = 2x
Substituting these values into the Ksp expression gives:
Ksp = x * (2x)^2
2.5 × 10^-8 = 4x^3
x = ∛(2.5 × 10^-8 / 4)
Step 4: Calculate the solubility in water.
Solving the equation above, we find:
x = 6.31 × 10 ^-3 M
So, the solubility of Cd(IO3)2 in water at 25°C is approximately 6.31 × 10 ^-3 M.
Step 5: Determine the solubility in 0.20 M CdCl2.
Since CdCl2 is a common ion, it will affect the solubility of Cd(IO3)2. The presence of Cd2+ from CdCl2 will already be present in the solution.
- [Cd2+] = concentration of Cd2+ taken from CdCl2 = 0.20 M
Using the Ksp expression and the concentration of Cd2+ from CdCl2, we have:
Ksp = [Cd2+] * [IO3-]^2
2.5 × 10^-8 = (0.20) * [IO3-]^2
Solving for [IO3-], we get:
[IO3-] = √(2.5 × 10^-8 / 0.20)
Step 6: Calculate the solubility in 0.20 M CdCl2.
Solving the equation above, we find:
[IO3-] = 7.94 × 10^-3 M
Thus, the solubility of Cd(IO3)2 in 0.20 M CdCl2 at 25°C is approximately 7.94 × 10^-3 M.
To determine the solubility at 25°C of Cd(IO3)2 in water and in 0.20 M CdCl2, you need to use the concept of solubility product constant (Ksp). The Ksp represents the equilibrium constant for the dissolution of a solid compound into its respective ions in a saturated solution.
For Cd(IO3)2 in water:
The solubility of Cd(IO3)2 can be represented as x.
The equation for the dissolution of Cd(IO3)2 in water is:
Cd(IO3)2(s) ⇌ Cd2+(aq) + 2IO3-(aq)
Using the stoichiometry of the equation, the solubility product expression can be written as:
Ksp = [Cd2+][IO3-]^2 = x(2x)^2 = 4x^3
Given that Ksp = 2.5 × 10^-8, you can solve for x:
4x^3 = 2.5 × 10^-8
Take the cube root of both sides to find x:
x = (2.5 × 10^-8)^(1/3)
For Cd(IO3)2 in 0.20 M CdCl2:
To determine the solubility in CdCl2, you need to consider the common ion effect. CdCl2 contains Cd2+ ions, which are also present in Cd(IO3)2.
The presence of Cd2+ ions in the solution will decrease the solubility of Cd(IO3)2 because of the common ion effect. The additional Cd2+ from CdCl2 will shift the equilibrium towards the solid Cd(IO3)2, reducing its solubility.
Therefore, the solubility of Cd(IO3)2 in 0.20 M CdCl2 will be lower compared to its solubility in pure water.
.In water:...........................Cd(IO3)2 ==> Cd^2 + 2 IO3^-
I..............................solid..............0..............0
C............................solid...............x.............2x
E............................solid...............x.............2x
Ksp = (Cd^2+)(IO3^-)^2
2.5E-9 = (x)(*2x)^2. Solve for x = solubility.
In 0.2M CdCl2:
Remember CdCl2 is comp;letelyionized; therefore,
..................CdCl2 ==> Cd^2+ + 2Cl^-
I..................0.20............0..............0
C...............-0.20...........0.20.........0.40
E...................0............0.20..........0.40
Then you can see (Cd^2+) = 0.20 + x ; i.e.(0.20 from CdCl2 and x from Cd(IO3)2
And (IO3^-) = 2x
Plug those values into the Ksp expression as above and solve for x. Thjat will be the solubility. This problem shows the common ion effect; that is, when a common ion is present (Cd^2+ in this case), the solubility is decreased. Post your work if you get stuck.