What is the minimum concentration?
Determine the minimum concentration of the precipitating agent on the right to cause precipitation of the cation from the solution on the left.
8.0 x 10^-2 CaI2, K2SO4
1.5 x 10^-3 AgNO3, RbCl
The Ksp values for both are Ksp = 2.4*10-6 for CaSO4, and Ksp = 1.8*10-10 for AgCl.
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8.0×10−2 M CaI2;K2SO4
To determine the minimum concentration of the precipitating agent required to cause precipitation of the cation from the solution, we need to compare the ion product (Q) with the solubility product constant (Ksp).
For calcium iodide (CaI2) and potassium sulfate (K2SO4), we will be looking at the potential precipitation of calcium sulfate (CaSO4).
The balanced equation for the precipitation reaction is:
CaSO4 (s) ⇌ Ca2+ (aq) + SO4^2- (aq)
The ion product (Q) for the precipitation reaction is calculated by multiplying the concentrations of Ca2+ and SO4^2- ions:
Q = [Ca2+] * [SO4^2-]
The solubility product constant (Ksp) for CaSO4 is given as 2.4 x 10^-6.
For precipitation to occur, Q must be equal to or greater than Ksp. Therefore, we can set up the following inequality:
[Ca2+] * [SO4^2-] ≥ 2.4 x 10^-6
Now, let's look at the first scenario with CaI2 and K2SO4.
Given the concentration of CaI2 as 8.0 x 10^-2, we have:
[Ca2+] = 2 * (8.0 x 10^-2) = 1.6 x 10^-1
Since K2SO4 is the precipitating agent and doesn't contribute Ca2+ ions, [SO4^2-] would depend on the solubility of K2SO4 in water. However, K2SO4 is a soluble salt, meaning it dissociates completely in water. Therefore, we can assume [SO4^2-] to be twice the concentration of K2SO4.
Let's denote the concentration of K2SO4 as x, then:
[SO4^2-] = 2x
Substituting these values into the inequality equation, we have:
(1.6 x 10^-1) * (2x) ≥ 2.4 x 10^-6
Simplifying, we get:
3.2 x 10^-1 * x ≥ 2.4 x 10^-6
Dividing both sides by 3.2 x 10^-1, we have:
x ≥ (2.4 x 10^-6) / (3.2 x 10^-1)
x ≥ 7.5 x 10^-6
Therefore, the minimum concentration of K2SO4 required to cause precipitation of the cation from the solution is 7.5 x 10^-6.
Now, let's move on to the second scenario with AgNO3 and RbCl.
We will be looking at the potential precipitation of silver chloride (AgCl).
The balanced equation for the precipitation reaction is:
AgCl (s) ⇌ Ag+ (aq) + Cl- (aq)
The ion product (Q) for the precipitation reaction is calculated by multiplying the concentrations of Ag+ and Cl- ions:
Q = [Ag+] * [Cl-]
The solubility product constant (Ksp) for AgCl is given as 1.8 x 10^-10.
For precipitation to occur, Q must be equal to or greater than Ksp. Therefore, we can set up the following inequality:
[Ag+] * [Cl-] ≥ 1.8 x 10^-10
Now, let's look at the second scenario with AgNO3 and RbCl.
Given the concentration of AgNO3 as 1.5 x 10^-3, we have:
[Ag+] = 1 * (1.5 x 10^-3) = 1.5 x 10^-3
Similarly, RbCl is the precipitating agent and doesn't contribute Ag+ ions, so [Cl-] would depend on the solubility of RbCl in water. However, RbCl is a soluble salt, meaning it dissociates completely in water. Therefore, we can assume [Cl-] to be equal to the concentration of RbCl.
Let's denote the concentration of RbCl as y, then:
[Cl-] = y
Substituting these values into the inequality equation, we have:
(1.5 x 10^-3) * y ≥ 1.8 x 10^-10
Dividing both sides by (1.5 x 10^-3), we have:
y ≥ (1.8 x 10^-10) / (1.5 x 10^-3)
y ≥ 1.2 x 10^-7
Therefore, the minimum concentration of RbCl required to cause precipitation of the cation from the solution is 1.2 x 10^-7.
To determine the minimum concentration of the precipitating agent required to cause precipitation of the cation from the solution, we need to compare the solubility product constant (Ksp) of the precipitate with the concentrations of the ions in the precipitating agent.
Let's start with the first case: CaI2 and K2SO4.
The given Ksp value is for CaSO4, so we know that the precipitate formed is CaSO4. To determine the minimum concentration of K2SO4 needed to cause precipitation, we need to compare the concentrations of the calcium and sulfate ions in the solution with the Ksp of CaSO4.
From CaI2, we know that we have 2 moles of Ca²⁺ ions for every mole of CaI2 dissolved. Therefore, the concentration of Ca²⁺ ions in the solution is 8.0 x 10^-2 M (given concentration of CaI2).
From K2SO4, we have 1 mole of sulfate ions (SO₄²⁻) for every mole of K2SO4 dissolved. Therefore, the concentration of sulfate ions in the solution is 0.0 M (since K2SO4 is not present in the solution).
Now, comparing the concentrations of Ca²⁺ ions (8.0 x 10^-2 M) and sulfate ions (0.0 M) with the Ksp of CaSO4 (2.4 x 10^-6), we can see that the concentration of sulfate ions is much lower than the Ksp of CaSO4. Hence, no precipitation will occur in this case.
Moving on to the second case: AgNO3 and RbCl.
Again, the given Ksp value is for AgCl, so the precipitate formed is AgCl. Let's compare the concentrations of the silver and chloride ions in the solution with the Ksp of AgCl.
From AgNO3, we have 1 mole of silver ions (Ag⁺) for every mole of AgNO3 dissolved. Therefore, the concentration of Ag⁺ ions in the solution is 1.5 x 10^-3 M (given concentration of AgNO3).
From RbCl, we have 1 mole of chloride ions (Cl⁻) for every mole of RbCl dissolved. Therefore, the concentration of chloride ions in the solution is 0.0 M (since RbCl is not present in the solution).
Comparing the concentrations of Ag⁺ ions (1.5 x 10^-3 M) and chloride ions (0.0 M) with the Ksp of AgCl (1.8 x 10^-10), we see that the concentration of chloride ions is much lower than the Ksp of AgCl. Hence, precipitation will occur in this case.
Therefore, the minimum concentration of the precipitating agent (AgNO3) required to cause precipitation of the cation (Ag⁺) from the solution (CaI2) is 1.5 x 10^-3 M.