Calculate the solubility (in g/L) of silver carbonate in water at 25°C if the Ksp for Ag2CO3 is 8.4 x 10^-12
Ag2CO3(s) ==> 2Ag^+ + CO3^-2
Ksp = (Ag^+)^2(CO3^-2)
Set up an ICE chart and solve which gives the solubility in moles/L, then convert that to grams. Post your work if you get stuck.
3.34716475*10^-04
To calculate the solubility (in g/L) of silver carbonate (Ag2CO3) in water at 25°C, we need to use the solubility product constant (Ksp) value.
The balanced chemical equation for the dissociation of silver carbonate is:
Ag2CO3(s) ⇌ 2Ag+(aq) + CO3^2-(aq)
The Ksp expression for this reaction is:
Ksp = [Ag+]^2 * [CO3^2-]
Given that the Ksp for Ag2CO3 is 8.4 x 10^-12 and assuming that the dissociation is complete (which is usually a reasonable assumption for very low Ksp values), we can set up the equation for the solubility as:
8.4 x 10^-12 = (2x)^2 * x
Where 'x' represents the molar solubility of Ag2CO3 in moles per liter.
Simplifying the equation, we get:
8.4 x 10^-12 = 4x^3
Rearranging the equation, we have:
x^3 = (8.4 x 10^-12) / 4
x^3 = 2.1 x 10^-12
Taking the cube root of both sides, we get:
x ≈ 1.24 x 10^-4 M
Since 1 mole of Ag2CO3 weighs 275.75 grams, the solubility in g/L can be calculated as:
Solubility (g/L) = (1.24 x 10^-4 M) * (275.75 g/mol)
Solubility (g/L) ≈ 3.41 x 10^-2 g/L
Therefore, the solubility of silver carbonate in water at 25°C is approximately 3.41 x 10^-2 g/L.
To calculate the solubility of silver carbonate (Ag2CO3) in water at 25°C using the given Ksp value, we can follow the steps below:
Step 1: Write the balanced equation for the dissolution of silver carbonate.
The balanced equation for the dissolution of Ag2CO3 in water is:
Ag2CO3(s) ⇌ 2Ag+(aq) + CO3^2-(aq)
Step 2: Define the solubility.
Let's assume the solubility of Ag2CO3 in water is 's' grams per liter (g/L).
Step 3: Write the solubility product expression.
The solubility product expression (Ksp) for Ag2CO3 is:
Ksp = [Ag+]^2 * [CO3^2-]
Step 4: Substitute values into the solubility product expression.
We substitute the concentration of Ag+ and CO3^2- ions, which are equal to the solubility 's', into the Ksp equation:
Ksp = (2s)^2 * (s) = 8.4 x 10^-12
Step 5: Solve for 's'.
Simplifying the equation:
4s^3 = 8.4 x 10^-12
Taking the cube root:
s^3 = (8.4 x 10^-12) / 4
Calculating the cube root:
s^3 ≈ 7.35 x 10^-4
Taking the cube root once again to solve for 's':
s ≈ ∛(7.35 x 10^-4)
s ≈ 0.094 g/L
Therefore, the solubility of silver carbonate (Ag2CO3) in water at 25°C is approximately 0.094 g/L.