What mass of Ag2CO3 would be found in 1.4 L of a saturated solution if the Ksp of Ag2CO3 is 8.2 x 10-12?
My Work:
SInce mass = number of mols x Molar mass, I first look for the molar mass of Ag2CO3, which is 275.75 g/mol. Multiplying .0000000000082 by 275.5 gives a value of 2.0 x 10^-9 g, which becomes the mass.
Is this correct?
No. You haven't used Ksp at all except in a superficial manner. First you must determine the solubility of Ag2CO3. Use Ksp for that.
Ag2CO3 ==> 2Ag^+ + CO3^=
Ksp = (Ag^+)^2(CO3^=) = 8.2 x 10^-12
Solubility Ag2CO3 in mols/L = y
Then (Ag^+) = 2y
(CO3^=) = y
Now substitute those variables into Ksp expression.
(2y)^2(y) = 8.2 x 10^-12
solve for y which is (Ag2CO3)in mols/L.
Convert to grams by grams = mols x molar mass. That is the mass in 1 L.
Then that mass x 1.4 L/1.0 L = the mass of Ag2CO3 in 1.4 L. Post your work if you need additional assistance.
Can u solve for y
Well, I hate to burst your bubble, but your calculation is not quite correct. Let's see where things went a little haywire.
To find the mass of Ag2CO3 in the saturated solution, we need to find the number of moles of Ag2CO3 first. We can do this by rearranging the Ksp expression:
Ksp = [Ag+]^2 x [CO3^2-]
Since Ag2CO3 dissociates into 2 Ag+ ions and 1 CO3^2- ion, we have:
Ksp = 4x^3
Now, we can solve for the concentration of Ag+ ions (x). Taking the cube root of the Ksp value (8.2 x 10^-12) gives us the concentration of Ag+ ions in the saturated solution.
Concentration of Ag+ ions = ∛(8.2 x 10^-12)
Once we have the concentration of Ag+ ions, we can multiply it by 2 to get the concentration of Ag2CO3 in the solution. Finally, we can multiply the concentration by the volume (1.4 L) to find the mass.
Now that we've taken a slightly more serious approach, would you like me to help you with the actual calculation?
No, your calculation is incorrect. To find the mass of Ag2CO3 in the saturated solution, you need to use the equilibrium expression for the solubility product (Ksp) and the molar solubility of Ag2CO3.
The balanced equation for the dissociation of Ag2CO3 is:
Ag2CO3(s) ⇌ 2Ag+(aq) + CO3^2-(aq)
The Ksp expression for Ag2CO3 is:
Ksp = [Ag+]^2[CO3^2-]
Since the coefficient of Ag2CO3 is 1, the molar solubility of Ag2CO3 is equal to [Ag+] in the solution. Let's assume the molar solubility of Ag2CO3 is x.
Now, we can set up the Ksp expression using the given value of Ksp and the molar solubility:
8.2 x 10^-12 = (2x)^2 * x
Simplifying the equation:
8.2 x 10^-12 = 4x^3
Now solve for x:
x^3 = (8.2 x 10^-12) / 4
x^3 = 2.05 x 10^-12
x ≈ 1.28 x 10^-4 M
The molar solubility of Ag2CO3 in the saturated solution is approximately 1.28 x 10^-4 M.
To find the mass of Ag2CO3 in the solution, multiply the molar solubility by the molar mass of Ag2CO3:
Mass = molar solubility x molar mass
Mass = (1.28 x 10^-4 M) x (275.75 g/mol)
Mass ≈ 3.53 x 10^-2 g
Therefore, the mass of Ag2CO3 found in 1.4 L of the saturated solution is approximately 3.53 x 10^-2 g.
Your approach is almost correct, but there is a small mistake in your calculation.
To find the mass of Ag2CO3 in the given saturated solution, we need to consider the concentration of Ag2CO3 in the solution and multiply it by the volume of the solution.
The solubility product constant (Ksp) is given as 8.2 x 10^-12, which represents the product of the concentrations of the individual ions in the saturated solution. In the case of Ag2CO3, it dissociates into two Ag+ ions and one CO3 2- ion.
Let's assume that 'x' represents the concentration of Ag2CO3 in mol/L. Therefore, the concentration of Ag+ ions and CO3 2- ions would also be 'x' mol/L.
Since each molecule of Ag2CO3 dissociates into two Ag+ ions, the concentration of Ag+ ions would be 2x mol/L. Similarly, the concentration of CO3 2- ions would be 'x' mol/L.
Now, we can write the expression for the solubility product constant (Ksp) as follows:
Ksp = [Ag+]^2 * [CO3 2-]
Substituting the expressions for [Ag+] and [CO3 2-]:
8.2 x 10^-12 = (2x)^2 * x (Since [Ag+] = 2x and [CO3 2-] = x)
Expanding and simplifying:
8.2 x 10^-12 = 4x^3
Rearranging the equation:
x^3 = (8.2 x 10^-12) / 4
x^3 = 2.05 x 10^-12
Now, we can solve for 'x' by taking the cube root of both sides of the equation:
x = (∛(2.05 x 10^-12))
Calculating 'x' gives: x ≈ 1.26 x 10^-4 mol/L
To find the mass of Ag2CO3 in the saturated solution, we can multiply the concentration 'x' by the volume of the solution (1.4 L):
Mass = concentration x volume
Mass = (1.26 x 10^-4 mol/L) * (1.4 L)
Mass ≈ 1.76 x 10^-4 mol
Finally, to convert the mass from moles to grams, we multiply it by the molar mass of Ag2CO3 (275.75 g/mol):
Mass ≈ (1.76 x 10^-4 mol) * (275.75 g/mol)
Mass ≈ 0.0484 g
Therefore, the mass of Ag2CO3 in 1.4 L of the saturated solution would be approximately 0.0484 grams.