0.10 M potassium chromate is slowly added to a solution containing 0.50 M AgNO3 and 0.50 M Ba(NO3)2. What is the Ag+ concentration when BaCrO4 just starts to precipitate? The Ksp for Ag2CrO4 and BaCrO4 are 1.1 x 10-12 and respectively.

To determine the Ag+ concentration when BaCrO4 just starts to precipitate, we need to compare the solubility product constants (Ksp) for Ag2CrO4 and BaCrO4.

Step 1: Write the balanced equation for the precipitation reaction.
BaCrO4 (s) ↔ Ba2+ (aq) + CrO4^2- (aq)

Step 2: Use the Ksp expression to set up an equilibrium expression for the precipitation reaction.
Ksp = [Ba2+][CrO4^2-]

Step 3: Determine the initial concentrations of Ba2+ and CrO4^2- ions.
The question states that the initial concentrations of AgNO3 and Ba(NO3)2 are both 0.50 M. Since BaCrO4 is just starting to precipitate, initially all of the Ba2+ ions and CrO4^2- ions will be in the form of their aqueous salts.

[Ba2+] = 0.50 M
[CrO4^2-] = 0.10 M (0.10 M K2CrO4)

Step 4: Substitute the initial concentrations into the equilibrium expression.
Ksp = (0.50 M)(0.10 M) = (0.50)(0.10)

Step 5: Calculate the value of Ksp for BaCrO4.
Ksp = 0.05

Step 6: Compare Ksp values for Ag2CrO4 and BaCrO4.
Given:
Ksp(Ag2CrO4) = 1.1 x 10^-12
Ksp(BaCrO4) = 0.05

Since Ksp(Ag2CrO4) < Ksp(BaCrO4), the solubility of Ag2CrO4 is less than BaCrO4. Therefore, Ag2CrO4 will start to precipitate before BaCrO4.

Step 7: Calculate the Ag+ concentration when BaCrO4 just starts to precipitate.
Since Ag2CrO4 is the first solid to begin forming, we can assume that all Ag+ ions will be used to form Ag2CrO4.

[Ag+] = initial concentration - concentration used for Ag2CrO4
[Ag+] = 0.50 M - concentration used for Ag2CrO4

Step 8: Calculate the concentration of Ag+ used for Ag2CrO4.
Using the stoichiometry of the balanced equation, we know that for every 2 Ag+ ions, we form 1 Ag2CrO4 molecule. Therefore, the concentration used for Ag2CrO4 can be calculated as:

Concentration used for Ag2CrO4 = (0.10 M CrO4^2-) / 2

Step 9: Substitute the concentration used for Ag2CrO4 into the equation for [Ag+].
[Ag+] = 0.50 M - (0.10 M / 2)

Step 10: Calculate the Ag+ concentration.
[Ag+] = 0.50 M - 0.05 M
[Ag+] = 0.45 M

Therefore, the Ag+ concentration when BaCrO4 just starts to precipitate is 0.45 M.

To find the Ag+ concentration when BaCrO4 just starts to precipitate, we need to determine the point at which the concentration of Ag+ ions exceeds the solubility product constant (Ksp) for Ag2CrO4.

Let's set up the balanced chemical equation for the precipitation reaction:

BaCrO4(s) ⇌ Ba2+(aq) + CrO42-(aq)

From the reaction equation, we can see that one mole of BaCrO4 dissociates to give one mole of Ba2+ ions and one mole of CrO42- ions.

The solubility product expression for BaCrO4 can be written as:
Ksp(BaCrO4) = [Ba2+][CrO42-]

Given that the potassium chromate concentration is 0.10 M, we can assume that the concentration of CrO42- ions will also be 0.10 M when the reaction reaches equilibrium.

However, we don't know the initial concentration of Ba2+ ions or Ag+ ions. Let's assign them as "x" for simplicity.

Now let's consider the AgNO3 in the solution:

AgNO3(aq) ⇌ Ag+(aq) + NO3-(aq)

Since 0.50 M AgNO3 is present initially, the concentration of Ag+ ions before the reaction occurs is also 0.50 M.

To determine the Ag+ concentration at equilibrium, we need to examine the formation of Ag2CrO4:

2Ag+(aq) + CrO42-(aq) ⇌ Ag2CrO4(s)

The balanced equation shows that two moles of Ag+ ions react with one mole of CrO42- ions to form one mole of Ag2CrO4 solid.

Now, let's write the equilibrium expression for Ag2CrO4 solubility product:

Ksp(Ag2CrO4) = [Ag+]^2[CrO42-]

Since we want to determine the Ag+ concentration when BaCrO4 just starts to precipitate, we can set up an equation that relates the Ag+ ions in the two reactions:

2x = [Ag+]^2

Now, let's consider the reaction of BaCrO4:

BaCrO4(s) ⇌ Ba2+(aq) + CrO42-(aq)

From the balanced equation, we know that for every BaCrO4 molecule that dissolves, one Ba2+ ion and one CrO42- ion are formed. So, the concentration of Ba2+ ions is equal to the concentration of the dissolved BaCrO4.

However, we don't know the concentration of BaCrO4 at this point. Let's assign it as "y" for simplicity.

The solubility product expression for BaCrO4 becomes:
Ksp(BaCrO4) = y[CrO42-]

Since the concentration of CrO42- ions is 0.10 M and the concentration of Ba2+ ions is equal to y, the solubility product expression becomes:
Ksp(BaCrO4) = y(0.10)

Given that the Ksp value for BaCrO4 is 3.0 x 10^-5, we can set up the equation:

3.0 x 10^-5 = y(0.10)

Solving for y, we get:
y = (3.0 x 10^-5) / (0.10)

Now, let's substitute the value of y into the equation we obtained earlier for the Ag+ ions:

2x = [Ag+]^2

Since we know the value of x and found the value of y, we can solve for the Ag+ concentration at equilibrium:

2(0.50) = [Ag+]^2

1 = [Ag+]^2

Taking the square root of both sides, we find that:
[Ag+] = 1

Therefore, the Ag+ concentration when BaCrO4 just starts to precipitate is 1 M.

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