what is the solubility of silver carbonate in water in 25 degrees celcius if Ksp=8.4X10-12? i don't know what to do
Write the equation.
Ag2CO3(s) ==> 2Ag^+ + CO3^=
Ksp = (Ag^+2)(CO3^=)
Let x = solubility of Ag2CO3. At equilibrium,(Ag^+) = 2x; (CO3^=) = x. Plug those into the Ksp expression and solve for x. The answer comes out in mols/L. If you want grams, then M x molar mass Ag2CO3 = grams Ag2CO3.
Post your work if you get stuck.
To determine the solubility of silver carbonate (Ag2CO3) in water at 25 degrees Celsius, you can use the solubility product constant (Ksp) and the equation for the dissociation of silver carbonate in water.
First, write the equation for the dissociation of silver carbonate:
Ag2CO3(s) → 2Ag^+ + CO3^2-
Next, use the expression for Ksp:
Ksp = [Ag^+]^2 * [CO3^2-]
Since the concentration of Ag^+ and CO3^2- ions is twice the solubility (x) due to the stoichiometry of the equation, rewrite the expression using x:
Ksp = (2x)^2 * (x)
Ksp = 4x^3
Now solve for x:
x = ³√(Ksp / 4)
Substitute the given value of Ksp (8.4 × 10^-12) into the equation:
x = ³√(8.4 × 10^-12 / 4)
Calculate the cube root of the result to find the solubility in mols/L. If you want to convert it to grams, you can multiply the molar mass of Ag2CO3 by the solubility in mols/L.
Don't hesitate to ask if you need further assistance or clarification on any step of the process!
To determine the solubility of silver carbonate in water at 25 degrees Celsius, we can use the given value of Ksp and the equation provided:
Ag2CO3(s) ⇌ 2Ag+ + CO3^2-
The equation shows that for every 1 mole of Ag2CO3 that dissolves, it produces 2 moles of Ag+ ions and 1 mole of CO3^2- ions.
The Ksp expression for silver carbonate is as follows:
Ksp = [Ag+]^2 * [CO3^2-]
In this case, the solubility of silver carbonate is represented by "x", so we can substitute the concentrations into the Ksp expression:
Ksp = (2x)^2 * (x)
8.4 x 10^-12 = 4x^2 * x
8.4 x 10^-12 = 4x^3
Now, rearrange the equation and solve for x:
x^3 = (8.4 x 10^-12) / 4
x = ∛ [(8.4 x 10^-12) / 4]
Using a calculator, we can find that x ≈ 2.17 x 10^-4 mol/L.
If you want to convert the solubility from moles per liter to grams, you can multiply the molar mass of Ag2CO3 (which is 275.75 g/mol) by the solubility in moles per liter:
Solubility (in grams) = (2.17 x 10^-4 mol/L) * (275.75 g/mol)
So, the solubility of silver carbonate in water at 25 degrees Celsius is approximately 0.06 grams per liter.
Ag2CO3= 2Ag^1+CO3^-2
Ksp=[Ag^+1]^2[CO3^-2]
. =(2x)^2(x)
8.4*10^-12=4x^3
8.4*10^-12/4= 4x^3/4
Cube root of 2.1*10^-12 =x
0.0001280
1.28*10^-4M=x