Looking for Ksp -------
Compound: SrCO3
[Cation] : 2.4 x 10^-5
[Anion] : 2.4 x 10^-5
Ksp ??
Assume all concentrations are equilibrium constant in M, (mol/L)
To calculate the solubility product constant (Ksp) for SrCO3, you need to know the concentrations of the cation (Sr2+) and the anion (CO32-).
Given:
[Cation] = 2.4 x 10^-5 M
[Anion] = 2.4 x 10^-5 M
The balanced equation for the dissociation of SrCO3 is:
SrCO3(s) ⇌ Sr2+(aq) + CO32-(aq)
Using the coefficients in the balanced equation, we can assume that the concentration of Sr2+ and CO32- ions formed are equal to each other. Therefore, we can use the given concentrations for both the cation and anion.
Ksp can be calculated by multiplying the concentrations of the cation and anion raised to the power of their respective stoichiometric coefficients.
Ksp = [Sr2+] * [CO32-]
Now, substituting the given concentrations in the equation:
Ksp = (2.4 x 10^-5 M) * (2.4 x 10^-5 M)
Ksp = 5.76 x 10^-10 M^2
Therefore, the solubility product constant (Ksp) for SrCO3 is 5.76 x 10^-10 M^2.
To find the solubility product constant (Ksp) for SrCO3, we need to determine the concentrations of the cation Sr2+ and the anion CO32-.
Given:
[Cation] = 2.4 x 10^-5 M
[Anion] = 2.4 x 10^-5 M
The balanced chemical equation for the dissociation of SrCO3 is:
SrCO3(s) ⇌ Sr2+(aq) + CO32-(aq)
Since the stoichiometry of SrCO3 is 1:1, the concentration of Sr2+ and CO32- will be the same.
Now, we can set up the expression for Ksp, using the concentrations at equilibrium:
Ksp = [Sr2+][CO32-]
Substituting the given concentrations, we have:
Ksp = (2.4 x 10^-5 M) * (2.4 x 10^-5 M)
Ksp = 5.76 x 10^-10
Therefore, the solubility product constant (Ksp) for SrCO3 is 5.76 x 10^-10.