Find the molar solubility of BaCrO4 (Ksp= 2.1 × 10−10) in 3.5 × 10−3 M Na2CrO4
To find the molar solubility of BaCrO4 in 3.5 × 10−3 M Na2CrO4, we will use the Ksp expression and the common ion effect.
The balanced chemical equation for the dissociation of BaCrO4 is:
BaCrO4(s) ⇌ Ba2+(aq) + CrO4^2-(aq)
The Ksp expression for this equilibrium is:
Ksp = [Ba2+][CrO4^2-]
We are given that Ksp = 2.1 × 10−10. Let's assume that the molar solubility of BaCrO4 is "x" mol/L.
From the balanced equation, we can see that the concentration of Ba2+ ion is also "x" mol/L, and the concentration of CrO4^2- ion is also "x" mol/L.
Since Na2CrO4 is a soluble salt, it will dissociate completely in water to give 2 moles of CrO4^2- ions for each mole of Na2CrO4. Therefore, the concentration of CrO4^2- ions from Na2CrO4 is 2 times the concentration of Na2CrO4, which is 2 × 3.5 × 10−3 M = 7 × 10−3 M.
Now, substitute the values into the Ksp expression:
2.1 × 10−10 = (x)(x + 7 × 10−3)
Since x is a small value compared to 7 × 10−3, we can approximate (x + 7 × 10−3) as 7 × 10−3.
2.1 × 10−10 = 7 × 10−3 * x
Divide both sides of the equation by 7 × 10−3:
x = (2.1 × 10−10) / (7 × 10−3) ≈ 3 × 10−8
Therefore, the molar solubility of BaCrO4 in 3.5 × 10−3 M Na2CrO4 is approximately 3 × 10−8 mol/L.
To find the molar solubility of BaCrO4 in a solution of Na2CrO4, we need to start by writing the balanced chemical equation for the dissolution of BaCrO4 in water.
The balanced equation is:
BaCrO4 (s) ⇌ Ba2+ (aq) + CrO42- (aq)
According to the balanced equation, each mole of BaCrO4 dissociates into one mole of Ba2+ ions and one mole of CrO42- ions.
Let's assume that the molar solubility of BaCrO4 in the solution is "s." So, the equilibrium concentrations of Ba2+ and CrO42- ions can be represented as "s."
Since Na2CrO4 is a soluble salt, it dissociates completely in water. Therefore, the concentration of CrO42- ions in the solution can be calculated using the initial molarity (3.5 × 10^-3 M) of Na2CrO4.
Now, we can write the expression for the solubility product constant (Ksp) of BaCrO4 as:
Ksp = [Ba2+][CrO42-]
Substituting the equilibrium concentrations for Ba2+ and CrO42-, we get:
Ksp = s × s
Given that the value of Ksp for BaCrO4 is 2.1 × 10^-10, we can now set up the equation:
2.1 × 10^-10 = s × s
To solve for "s," take the square root of both sides of the equation:
s = √(2.1 × 10^-10)
Calculating the square root using a calculator, we find that the molar solubility of BaCrO4 in the 3.5 × 10^-3 M Na2CrO4 solution is approximately 4.59 × 10^-6 M.
........BaCrO4 ==> Ba^2+ + CrO4^2-
I.......solid.......0.......0
C.......solid.......x.......x
E.......solid.......x.......x
Substitute the E line into Ksp expression and solve for x = solubility