If the solubility of Ag2CO3 is 1.3×10^-4 mol/L what is the Ksp?
I'm trying to do it but my answer I keep on getting is wrong and doesn't match the text book which is 8.8×10^-12
Instead of me guessing what you wrote why not show your work and let me find the error (if there is one)?
To determine the Ksp (solubility product constant) of Ag2CO3, we need to use the given solubility value. Remember that the solubility product constant represents the equilibrium constant for the dissociation of a sparingly soluble compound into its ions in a solution.
The balanced chemical equation for the dissociation of Ag2CO3 is as follows:
Ag2CO3(s) ⇌ 2Ag+(aq) + CO3^2-(aq)
To find the Ksp, we need to use the solubility of Ag2CO3. In this case, the solubility is 1.3×10^-4 mol/L. Since Ag2CO3 dissociates into two Ag+ ions, the concentration of Ag+ ions produced can be determined by multiplying the solubility by 2:
[Ag+] = 2 × (1.3×10^-4) mol/L
The concentration of CO3^2- ions is determined by the stoichiometry of the balanced equation, which is equal to the concentration of Ag+ ions:
[CO3^2-] = [Ag+] = 2 × (1.3×10^-4) mol/L
Now, we can substitute these values into the Ksp expression for Ag2CO3:
Ksp = [Ag+]^2 × [CO3^2-]
= (2 × (1.3×10^-4))^2 × (2 × (1.3×10^-4))
= 8 × (1.3×10^-4)^3
≈ 8.128 × 10^-12
The calculated value for Ksp is approximately 8.128 × 10^-12, which is close to the textbook value of 8.8 × 10^-12. Keep in mind that there might be slight variations due to different rounding methods or experimental determinations in different sources.