If the solubility constant for Ag2CO3 is equal to 8.1x10^-12 how many grams of Ag2CO3 will dissolve in a .0001 M solution of Na2CO3?
.....Ag2CO3 ==< 2Ag^+ + CO3^2-
I.....solid......0.....0.0001
C.....solid......2x.......x
E.....solid......2x...0.0001+x
Subsitute the E line into the Ksp epxression and solve for x = solubility in M. Then grams = mols x molar mass
To find out how many grams of Ag2CO3 will dissolve in a 0.0001 M solution of Na2CO3, we need to use the solubility equilibrium expression and the given solubility constant. Here's how you can calculate it:
1. Write the balanced chemical equation for the dissociation of Ag2CO3:
Ag2CO3 ⇌ 2Ag+ + CO3^2-
2. The solubility equilibrium expression for Ag2CO3 can be written as:
Ksp = [Ag+]^2 [CO3^2-]
Since Ag2CO3 dissociates into 2 Ag+ ions for every Ag2CO3 molecule, the equilibrium expression can be simplified as:
Ksp = 4[Ag+]^2 [CO3^2-]
3. Given that Ksp = 8.1 x 10^-12, substitute this value into the equation:
8.1 x 10^-12 = 4[Ag+]^2 [CO3^2-]
4. Since the concentration of Ag+ will be twice the concentration of Ag2CO3 that dissolves, let x be the concentration of Ag2CO3 in moles per liter.
[Ag+] = 2x
[CO3^2-] = 0.0001 M (given concentration of Na2CO3 solution)
Substitute these values into the equation:
8.1 x 10^-12 = 4(2x)^2 (0.0001)
5. Simplify the equation:
8.1 x 10^-12 = 0.0008x^2
6. Rearrange the equation:
x^2 = (8.1 x 10^-12) / 0.0008
7. Calculate x:
x = √((8.1 x 10^-12) / 0.0008)
8. Determine the number of grams of Ag2CO3 dissolved:
Given the volume of the solution (not provided in the question), you can multiply the concentration (x) by the volume to get the number of moles of dissolved Ag2CO3. Finally, you can convert moles to grams using the molar mass of Ag2CO3.
Please note that the calculation might vary depending on the volume of the solution provided in the question.