I need help with What is the solubility of silver oxide, Ag2O, in a solution buffered at pH 10.50? The Kc for the following equilibrium is 2.0 x 10-8. Express your answer in g/L
Ag2O(s) + H2O(l) ¡ê 2Ag+ (aq) + 2OH- (aq)
To determine the solubility of silver oxide (Ag2O) in a solution buffered at pH 10.50, you can use the concept of the common ion effect and the equilibrium expression Kc.
First, let's understand the common ion effect. In a buffered solution, there is a common ion already present that affects the solubility of a compound. In this case, the buffered solution contains OH- ions, which come from the dissociation of water.
Next, let's analyze the equilibrium expression for the reaction:
Ag2O(s) + H2O(l) ⇌ 2Ag+ (aq) + 2OH- (aq)
The equilibrium expression for this reaction is written as follows:
Kc = [Ag+]^2 [OH-]^2 / [Ag2O]
Given that the Kc value is 2.0 x 10^-8, we can express it as:
2.0 x 10^-8 = ([Ag+]^2 [OH-]^2) / [Ag2O]
Since we're interested in the solubility of Ag2O, we can consider that its concentration in the solid state remains constant. Thus, we can rewrite the expression as:
2.0 x 10^-8 = [Ag+]^2 [OH-]^2 / K'
Where K' is the solubility product constant for Ag2O, and is given by:
K' = 1 / [Ag2O]
Now, let's consider the pH of the solution. Since pH = 10.50, we can determine the concentration of OH- ions using the equation:
pOH = 14 - pH
pOH = 14 - 10.50 = 3.50
To find the concentration of OH- ions, we take the antilog of -pOH:
[OH-] = 10^(-pOH)
[OH-] = 10^(-3.50)
Now, we can substitute the values into the solubility product expression to find the concentration of Ag+ ions:
2.0 x 10^-8 = [Ag+]^2 * (10^(-3.50))^2 / K'
Simplifying the equation, we obtain:
2.0 x 10^-8 = [Ag+]^2 * (10^(-7))
To solve for [Ag+], we can take the square root of both sides of the equation:
[Ag+] = sqrt((2.0 x 10^-8) / (10^-7))
Finally, you can convert the concentration of Ag+ ions into grams per liter (g/L) by multiplying it with the molar mass of Ag+ (107.87 g/mol).
So, the solubility of Ag2O in the solution buffered at pH 10.50 is:
Solubility = [Ag+] * Molar mass of Ag+
= [Ag+] * 107.87 g/mol
I suggest plugging in the values and performing the calculations to get the final answer in g/L.
Write the expression for Kc.
Kc = (Ag^+)^2(OH^-)^2
Substitute H^+ for a pH of 10.50 and solve for (Ag^+)