calculate the molar solubility of BaCO3 in a 0.10M solution of NaCo3 (aq). (ksp(BaCO3)=8.1x10^-9)
To calculate the molar solubility of BaCO3 in a 0.10M solution of NaCO3, we can use the concept of the common ion effect. The common ion effect states that the solubility of a salt decreases in the presence of a common ion. In this case, we have NaCO3 as a common ion present.
Let's assume that x is the molar solubility of BaCO3 in mol/L. The dissociation of BaCO3 can be represented by the equation:
BaCO3(s) ⇌ Ba2+(aq) + CO32-(aq)
The solubility product expression for BaCO3 is: Ksp = [Ba2+][CO32-]. According to the given Ksp value, we have:
8.1x10^-9 = [Ba2+][CO32-]
Since NaCO3 is a strong electrolyte, it dissociates completely into Na+ and CO32-. However, since the concentration of NaCO3 is much higher than x (the molar solubility of BaCO3), we can assume that the concentration of CO32- contributed by BaCO3 is negligible compared to the concentration of CO32- contributed by NaCO3.
Therefore, the total concentration of CO32- is approximately 0.1M (the concentration of NaCO3). Therefore, we can write:
[CO32-] ≈ 0.1M
Now, substituting the values into the solubility product expression:
8.1x10^-9 = [Ba2+](0.1)
Solving for [Ba2+], we get:
[Ba2+] = 8.1x10^-9 / 0.1 = 8.1x10^-8 M
Therefore, the molar solubility of BaCO3 in a 0.10M solution of NaCO3 is approximately 8.1x10^-8 M.
To calculate the molar solubility of BaCO3 in a 0.10M NaCO3 solution, we need to set up an equilibrium expression based on the solubility product constant (Ksp) for BaCO3.
The balanced equation for the dissolution of BaCO3 is:
BaCO3(s) ⇌ Ba2+(aq) + CO32-(aq)
The Ksp expression for BaCO3 is:
Ksp = [Ba2+][CO32-]
Since NaCO3 is a soluble salt, it dissociates completely in water:
NaCO3(s) → Na+(aq) + CO32-(aq)
This means that the concentration of the carbonate ion (CO32-) is equal to the initial concentration of NaCO3, which is 0.10 M.
To find the molar solubility of BaCO3, we assume x moles of BaCO3 dissolve per liter, so the concentrations of Ba2+ and CO32- are both equal to x.
Using the Ksp expression, we can substitute the concentrations:
Ksp = [Ba2+][CO32-]
Ksp = (x)(x)
Ksp = x^2
Now, substitute the value of the Ksp constant:
8.1 x 10^-9 = x^2
To solve for x, take the square root of both sides:
√(8.1 x 10^-9) = x
The molar solubility of BaCO3 in a 0.10M solution of NaCO3 is approximately 9 x 10^-5 M.