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The solubility product of Bacro4 is 2.4×10-10 m2 .The maximum concentration of bano3 possible with out precipitation in a 6×10-4 m k2cro4 solution is?
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The solubility product of Bacro4 is 2.4×10-10 m2 .The maximum concentr...
 Ksp = [Ba2+][CrO4^2-] = 2.4 x 10^-10M^2 
In 5 x 10^-4M K2CrO4 the [CrO4^2-] = 5 x 10^-4M 
So [Ba2+] x 5 x 10^-4 = 2.40 x 10^-10 
Hence [Ba2+] = 4.80 x 10^-7M
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The solubility product of Bacro4 is 2.4×10-10 m2 .The maximum concentr...
Solubility Product of BaCrO4:
The solubility product (Ksp) of a compound is the equilibrium constant for the dissociation of that compound into its ions in a saturated solution. In the case of BaCrO4 (barium chromate), the solubility product is given as 2.4×10-10 m^2.

Calculating the Maximum Concentration of Ba(NO3)2:
To determine the maximum concentration of Ba(NO3)2 (barium nitrate) that can be added to a solution without precipitation, we need to consider the common ion effect. The common ion effect states that the solubility of a salt is reduced when a common ion is present in the solution.

In this case, the common ion is the chromate ion (CrO4 2-). The solubility of BaCrO4 will be affected by the presence of K2CrO4 (potassium chromate) in the solution.

Writing the Dissociation Equations:
1. BaCrO4 ⇌ Ba2+ + CrO4 2-
2. K2CrO4 ⇌ 2K+ + CrO4 2-

Setting up the Solubility Product Expression:
The solubility product expression for BaCrO4 can be written as:
Ksp = [Ba2+][CrO4 2-]

Using the Common Ion Effect:
When K2CrO4 is added to the solution, it will provide additional CrO4 2- ions. These additional ions will shift the equilibrium of the dissociation reaction for BaCrO4 to the left, decreasing the solubility of BaCrO4.

To determine the maximum concentration of Ba(NO3)2 that can be added, we need to find the concentration of CrO4 2- ions in the solution when precipitation occurs.

Calculating the Concentration of CrO4 2- ions:
Let's assume the concentration of Ba(NO3)2 is x M. Since Ba(NO3)2 dissociates into Ba2+ and NO3- ions, the concentration of Ba2+ ions will also be x M.

From the dissociation equation of BaCrO4, we know that the concentration of CrO4 2- ions will also be x M.

Setting up the Solubility Product Expression:
Using the concentrations of Ba2+ and CrO4 2- ions, we can write the solubility product expression as:
Ksp = (x)(x) = x^2

Solving for the Concentration:
Now, we can solve for x by equating the solubility product expression to the given Ksp value:
x^2 = 2.4×10-10

Taking the square root of both sides, we get:
x = √(2.4×10-10)

Calculating this value, we find:
x ≈ 1.55×10-5 M

Therefore, the maximum concentration of Ba(NO3)2 that can be added without precipitation in a 6×10-4 M K2CrO4 solution is approximately 1.55×10-5 M.
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The solubility product of Bacro4 is 2.4×10-10 m2 .The maximum concentration of bano3 possible with out precipitation in a 6×10-4 m k2cro4 solution is?
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