Calculate the molarity (mol/L) of Ba2+ when BaCrO4 is placed in a 0.0457 M K2CrO4 solution. The equation represents the solubility process. K(sp)=1.2x10^-10
BaCrO4 (s) Ba2+ (aq) + CrO42- (aq)
BaCrO4 (s)= Ba2+ (aq) + CrO42- (aq)
Ksp = (Ba^+2)(CrO4^-2)
You know Ksp. You know molarity CrO4. Calculate molarity Ba^+2.
To calculate the molarity of Ba2+ in the solution, we need to use the solubility product constant (Ksp) and the concentration of K2CrO4.
First, we need to write the balanced equation for the solubility process:
BaCrO4 (s) ⇌ Ba2+ (aq) + CrO42- (aq)
According to the equation, 1 mol of BaCrO4 will produce 1 mol of Ba2+. Therefore, the molarity of Ba2+ is the same as the molarity of BaCrO4, which is what we are trying to calculate.
The solubility product constant (Ksp) expression for BaCrO4 is given as:
Ksp = [Ba2+][CrO42-]
We are given that the Ksp value for BaCrO4 is 1.2x10^-10.
Next, we need to set up an ICE (Initial, Change, Equilibrium) table to determine the concentration of Ba2+.
Initial:
[Ba2+] = 0 (since BaCrO4 is initially a solid and has not dissolved)
Change:
[Ba2+] = x (since x mol/L of Ba2+ is formed when BaCrO4 dissolves)
Equilibrium:
[Ba2+] = x
Now, we need to consider the stoichiometry of the reaction:
Since the concentration of K2CrO4 is 0.0457 M, the concentration of CrO42- is also 0.0457 M (because K2CrO4 dissociates into 2 moles of K+ and 1 mole of CrO42-).
Using the equilibrium expression (Ksp = [Ba2+][CrO42-]):
1.2x10^-10 = x * 0.0457
Solving for x gives:
x = (1.2x10^-10) / 0.0457
x ≈ 2.63x10^-9 M
Therefore, the molarity of Ba2+ in the solution is approximately 2.63x10^-9 M.