*the molar solubility of BaCo3 in 0.50M Bacl2 solution IS?ksp(1.6*10^-6)
I CALCULATED AND GOT 3.2*10^-9
WANT TO MAKE SURE IF IT CORRECT
I don't think so.
1.6E-6/0.5 = ?
To find the molar solubility of BaCO3 in a 0.50 M BaCl2 solution, you can use the concept of the solubility product constant (Ksp).
The balanced chemical equation for the dissociation of BaCO3 is:
BaCO3(s) ⇌ Ba2+(aq) + CO3^2-(aq)
Using the stoichiometry of the equation, we can see that the ratio between BaCO3 and Ba2+ ions is 1:1. Therefore, the molar solubility of BaCO3 is equal to the concentration of Ba2+ ions formed in the solution.
Let's denote the molar solubility of BaCO3 as "x" (in mol/L). Then, the concentration of Ba2+ ions will also be "x" (in mol/L) because of the 1:1 stoichiometry.
The concentration of Ba2+ ions coming from the dissociation of BaCl2 will be 0.50 M. Since the BaCl2 is a strong electrolyte, it completely dissociates to give Ba2+ and Cl- ions.
Now, we can write the expression for the solubility product constant (Ksp):
Ksp = [Ba2+][CO3^2-]
Substituting the concentrations with "x" (since it is the molar solubility of BaCO3 = concentration of Ba2+), we get:
Ksp = x * x
Given that Ksp = 1.6 * 10^-6, we can solve for "x":
1.6 * 10^-6 = x^2
Taking the square root of both sides to solve for "x":
x = sqrt(1.6 * 10^-6)
Calculating this value, we get:
x ≈ 3.99 * 10^-4
Therefore, the molar solubility of BaCO3 in a 0.50 M BaCl2 solution is approximately 3.99 * 10^-4 M.
If you obtained a different value (3.2 * 10^-9), please double-check your calculations to ensure accuracy.
To determine the molar solubility of BaCO3 in a 0.50M BaCl2 solution, we can use the solubility product constant (Ksp) for BaCO3, which is given as 1.6 * 10^(-6).
The balanced equation for the dissociation of BaCO3 is:
BaCO3(s) <=> Ba2+(aq) + CO3^(2-)(aq)
Let's assume the molar solubility of BaCO3 is 'x'. Then the concentration of Ba2+ ions in the solution will also be 'x', while the concentration of CO3^(2-) ions will be 'x' as well.
Based on the stoichiometry of the balanced equation, we know that the concentration of Ba2+ ions in solution is twice that of BaCO3, and the concentration of CO3^(2-) ions is also equal to that of BaCO3.
Using this information, we can set up the expression for the Ksp and solve for 'x':
Ksp = [Ba2+][CO3^(2-)] = (x)(x) = x^2
Substituting the given Ksp value:
1.6 * 10^(-6) = x^2
Taking the square root of both sides, we find:
x = sqrt(1.6 * 10^(-6))
Evaluating this expression gives us approximately 1.26 * 10^(-3) M.
Therefore, the molar solubility of BaCO3 in a 0.50M BaCl2 solution is approximately 1.26 * 10^(-3) M.