the concentration of Ag+ in a saturated solution of Ag2Cro4 is 1.6x10^-4 M. what is the Ksp value for Ag2Cro4?
the solubility of BaCO3(s) in water is 4.0X10^-5 M. Calculate the value of Ksp for BaCO3
To determine the value of Ksp for Ag2CrO4, we need to use the given concentration of Ag+ in the saturated solution.
The formula for the solubility product constant, Ksp, for a compound is written as follows:
Ksp = [A+]^m[B-]^n
In the case of Ag2CrO4, it dissociates into Ag+ and CrO4^-2 ions. The stoichiometry of the dissociation reaction is 1:1, so we have:
Ag2CrO4(s) ⇌ 2Ag+(aq) + CrO4^-2(aq)
The Ksp expression for Ag2CrO4 can be written, considering the stoichiometry, as:
Ksp = [Ag+]^2[CrO4^-2]
We are given that the concentration of Ag+ in the saturated solution is 1.6x10^-4 M. Therefore, we can substitute this value into the Ksp expression as follows:
Ksp = (1.6x10^-4)^2[CrO4^-2]
However, we still need the concentration of the CrO4^-2 ions, which is not given in the question. Without this information, we cannot directly compute the Ksp value.
Therefore, to find the Ksp value for Ag2CrO4, we need to know either the solubility of Ag2CrO4 or the concentration of CrO4^-2 ions in the solution.
To find the Ksp (solubility product constant) value for Ag2Cro4, we need to use the given concentration of Ag+ ions in a saturated solution.
The balanced chemical equation for the dissociation of Ag2Cro4 is:
Ag2Cro4 ⇌ 2Ag+ + Cro4^2-
The expression for the Ksp of Ag2Cro4 is:
Ksp = [Ag+]^2 * [Cro4^2-]
Given that the concentration of Ag+ ions ([Ag+]) in the saturated solution is 1.6x10^-4 M, we can substitute this value into the Ksp expression:
Ksp = (1.6x10^-4)^2 * [Cro4^2-]
Now we need to determine the concentration of Cro4^2- ions. Since Ag2Cro4 is a sparingly soluble salt, it will dissociate only to a small extent. Assuming negligible dissociation, we can consider that all of the Ag2Cro4 will dissociate to give twice as many Ag+ ions as Cro4^2- ions.
Therefore, the concentration of Cro4^2- ions is half the concentration of Ag+ ions:
[Cro4^2-] = (1/2) * 1.6x10^-4 = 8x10^-5 M
Now we can substitute the value of [Cro4^2-] into the Ksp expression:
Ksp = (1.6x10^-4)^2 * (8x10^-5)
Simplifying the calculation:
Ksp = 2.048x10^-11
Therefore, the Ksp value for Ag2Cro4 is 2.048x10^-11.
For the solubility of BaCO3(s) in water, the balanced chemical equation for its dissociation is:
BaCO3 ⇌ Ba2+ + CO3^2-
The expression for the Ksp of BaCO3 is:
Ksp = [Ba2+] * [CO3^2-]
Given that the solubility of BaCO3 is 4.0x10^-5 M, we can substitute this value into the Ksp expression:
Ksp = (4.0x10^-5)^1 * (4.0x10^-5)^1 = 1.6x10^-9
Therefore, the Ksp value for BaCO3 is 1.6x10^-9.
Ag2CrO4 ==> 2Ag^+ + CrO4^-2
Ksp = (Ag^+)^2(CrO4^-2)
Set up an ICE chart, substitute into the Ksp expression and solve for Ksp.
#2 is done the same way.
Post your work if you get stuck.