What is the concentration of CrO22

4 in 0.0040 M Ba(NO3)2 saturated with BaCrO4. Including the activity coefficients in the expression?
The ksp of BaCrO4 is 1.17 x10^-10

[CrO4 2-]= ? M

I want to understand the problem not just the answer please.

To find the concentration of CrO4 2- in the solution, we need to consider the dissociation of BaCrO4 and the equilibrium expression for its solubility product constant (Ksp).

The dissociation equation for BaCrO4 is:
BaCrO4(s) ⇌ Ba2+(aq) + CrO4 2-(aq)

The solubility product constant (Ksp) expression for this equilibrium reaction is:
Ksp = [Ba2+][CrO4 2-]

Given that the Ksp value for BaCrO4 is 1.17 x 10^-10, we can assign the value accordingly.

Now, let's define x as the concentration of CrO4 2- ions in the saturated solution.

Considering that the Ba(NO3)2 is already dissolved in the solution and is a strong electrolyte, it will dissociate completely into Ba2+ and NO3- ions.

Therefore, the concentration of Ba2+ ions in the solution will be equal to that of Ba(NO3)2. Given that the concentration of Ba(NO3)2 is 0.0040 M, we can assume [Ba2+] = 0.0040 M.

Now, the concentration of CrO4 2- ions can be expressed as x M, which means [CrO4 2-] = x.

Since the solution is saturated, the concentration of BaCrO4 that dissolves is equal to the concentration of CrO4 2- ions, so we can assume that [BaCrO4] = x.

Using the Ksp expression, we can write the following equation:
Ksp = [Ba2+][CrO4 2-]
1.17 x 10^-10 = (0.0040)(x)

Now, we can solve the equation for x to find the concentration of CrO4 2-:
x = (1.17 x 10^-10)/(0.0040)

Simplifying the calculation:
x = 2.93 x 10^-8 M

Therefore, the concentration of CrO4 2- in the solution is approximately 2.93 x 10^-8 M.