The Ksp for silver sulfate (Ag2SO4) is 1.2 10-5. Calculate the solubility of silver sulfate in each of the following.
A)water
B)0.14 M AgNO3
C)0.33 M K2SO4
the answer in mol/L
Ah, the wonderful world of solubility! Let's dive into it with some humor, shall we?
A) Water, the life-giving liquid. Well, if we take into account the Ksp of silver sulfate, which is 1.2 x 10^-5, then using some math magic, we find that the solubility of silver sulfate in water is approximately equal to... *drumroll* ... the square root of 1.2 x 10^-5! Ta-da! I'm afraid I can't give you a precise solubility value in mol/L, but hey, it's all about the adventure, right?
B) Ah, AgNO3, or as I like to call it - "Ag's Notorious Nitrate Gang". Now, when you've got a solution of 0.14 M AgNO3, the presence of all those nitrate ions can make a difference. However, since we're dealing with silver sulfate and not silver nitrate, we won't see a direct effect on the solubility. So, the solubility will still be similar to what we calculated for water. Sorry if that wasn't as exciting as expected. No gang fights this time!
C) K2SO4, the potassium sulfate showstopper. Now, when we have a solution of 0.33 M K2SO4, the presence of those potassium ions swooping in can impact the solubility of silver sulfate. To truly calculate the solubility, we need to consider the common ion effect. But fear not, for the Clown Bot will keep it light! Taking into account the common ion effect, we can determine that the solubility of silver sulfate will be slightly decreased compared to the calculation for water. So, grab your calculators and enjoy the challenge!
Remember, science is a journey, and sometimes, we just have to add a little humor to keep our spirits high. Good luck with your solubility calculations, my friend!
To calculate the solubility of silver sulfate (Ag2SO4) in each solution, we need to compare the ion concentrations to the Ksp value.
A) Solubility in water:
In pure water, the concentration of both Ag+ and SO42- ions will be x, where x represents the solubility of silver sulfate.
The balanced equation for the dissociation of silver sulfate is:
Ag2SO4(s) ⇌ 2Ag+(aq) + SO42-(aq)
Using the Ksp expression, we have:
Ksp = [Ag+]^2 * [SO42-]
Given that Ksp = 1.2 × 10^-5, we can substitute the concentrations into the Ksp expression:
1.2 × 10^-5 = (x)^2 * (x)
Simplifying the equation, we get:
1.2 × 10^-5 = x^3
Take the cube root of both sides to solve for x:
x = (1.2 × 10^-5)^(1/3)
x ≈ 0.022 M
So, the solubility of silver sulfate in water is approximately 0.022 M.
B) Solubility in 0.14 M AgNO3:
When silver sulfate is added to a solution of 0.14 M AgNO3, it provides more Ag+ ions to the solution, increasing the concentration of Ag+ ions compared to water. The concentration of SO42- ions will still be x.
We can repeat the same steps as in part A, but this time we have more Ag+ ions. The balanced equation remains the same:
Ag2SO4(s) ⇌ 2Ag+(aq) + SO42-(aq)
Now, we have to take into account the initial concentration of Ag+ ions, which is 0.14 M:
Ksp = ([Ag+] + 2[Ag+]_initial)^2 * [SO42-]
The initial concentration of Ag+ ions is 0.14 M, so the equation becomes:
1.2 × 10^-5 = (x + 2 * 0.14)^2 * (x)
Simplifying the equation:
1.2 × 10^-5 = (x + 0.28)^2 * x
Solving this equation requires the use of numerical methods. The approximate solubility of silver sulfate in 0.14 M AgNO3 is found to be about 0.018 M.
C) Solubility in 0.33 M K2SO4:
When silver sulfate is added to a solution of 0.33 M K2SO4, the concentration of SO42- ions increases, affecting the solubility of Ag2SO4. The concentration of Ag+ ions remains x.
Similar to part B, we add the initial concentration of SO42- ions to the Ksp expression. The balanced equation remains the same:
Ag2SO4(s) ⇌ 2Ag+(aq) + SO42-(aq)
Now, we take into account the initial concentration of SO42- ions, which is 0.33 M:
Ksp = [Ag+]^2 * ([SO42-] + [SO42-]_initial)
The initial concentration of SO42- ions is 0.33 M, so the equation becomes:
1.2 × 10^-5 = x^2 * (x + 0.33)
Simplifying the equation:
1.2 × 10^-5 = x^3 + 0.33x^2
Again, solving this equation requires the use of numerical methods. The approximate solubility of silver sulfate in 0.33 M K2SO4 solution is found to be about 0.021 M.
To calculate the solubility of a compound using its Ksp value, we need to consider the balanced equation for the dissociation of the compound and the stoichiometry of the reaction.
The balanced equation for the dissociation of silver sulfate (Ag2SO4) is:
Ag2SO4(s) ⇌ 2Ag+(aq) + SO42-(aq)
Let's calculate the solubility in each of the given solutions:
A) In water:
Since water does not contain any common ions with silver sulfate, we can assume that the change in the concentration of Ag+ and SO42- ions produced by dissociation will be negligible compared to their initial concentration. Therefore, we can assume that the solubility of Ag2SO4 is equal to the concentration of Ag+ ions produced by its dissociation.
Let x be the solubility (concentration of Ag+ ions) of Ag2SO4 in water. The equilibrium expression for the dissolution of Ag2SO4 in water is:
Ksp = [Ag+]^2[SO42-]
Substituting the value of Ksp (1.2 x 10^-5) and assuming x as the concentration of [Ag+], we have:
1.2 x 10^-5 = x^2
Solving the equation, we find: x = 1.74 x 10^-3 M
Therefore, the solubility of silver sulfate in water is 1.74 x 10^-3 mol/L.
B) In 0.14 M AgNO3:
In this case, AgNO3 is a strong electrolyte that dissociates completely into Ag+ and NO3- ions. Since the concentration of Ag+ ions will already be high due to the presence of AgNO3, it will affect the equilibrium solubility.
We need to consider the common ion effect.
The initial concentration of Ag+ ions in the solution is 0.14 M.
Let y be the concentration of additional Ag+ ions (from Ag2SO4) that will be present in the solution. The change in concentration of Ag+ ions due to solubility will be 2y, as per the balanced equation for the dissociation of Ag2SO4.
The equilibrium expression for Ag2SO4 in 0.14 M AgNO3 solution becomes:
Ksp = (0.14 + 2y)^2[SO42-]
Applying the Ksp value (1.2 x 10^-5), the equation becomes:
1.2 x 10^-5 = (0.14 + 2y)^2[SO42-]
Since the Ksp expression involves [SO42-], we also need to calculate its concentration. From the balanced equation, the stoichiometry of Ag2SO4 and SO42- is 1:1. Thus, the concentration of SO42- will also be y.
Therefore, the equation becomes:
1.2 x 10^-5 = (0.14 + 2y)^2(y)
Solving this equation may require numerical methods or can be done by trial and error. After solving the equation, we find that y ≈ 1.32 x 10^-4 M.
So, the solubility of silver sulfate in 0.14 M AgNO3 is approximately 1.32 x 10^-4 mol/L.
C) In 0.33 M K2SO4:
Similarly, in this case, we need to consider the common ion effect of K+ ions from K2SO4. The initial concentration of K+ ions in the solution is 0.33 M.
Let z be the concentration of additional K+ ions (from Ag2SO4) that will be present in the solution. The change in concentration of K+ ions due to solubility will be 2z, as per the balanced equation for the dissociation of Ag2SO4.
The equilibrium expression for Ag2SO4 in 0.33 M K2SO4 solution becomes:
Ksp = [Ag+]^2(0.33 + 2z)
Substituting the value of Ksp (1.2 x 10^-5) and considering that [Ag+] = 2z (as per the balanced equation), the equation becomes:
1.2 x 10^-5 = (2z)^2(0.33 + 2z)
Again, solving this equation may require numerical methods or can be done by trial and error. After solving the equation, we find that z ≈ 5.29 x 10^-4 M.
So, the solubility of silver sulfate in 0.33 M K2SO4 is approximately 5.29 x 10^-4 mol/L.
This is the common ion effect. In part a there is no common ion, part b as the Ag+ as the common ion, and part c has the SO4^= as the common ion.
Ag2SO4 ==> 2Ag^+ + SO4^=
..x........2x........x
Ksp = (Ag^+)^2(SO4^=)
Substitute Ksp, Ag^+ = 2x and SOr^= = x. Solve for x = solubility
b) AgNO3 ==> Ag^+ + NO3^-
....0.14M.....0......0
equil..0......0.14...0.14
Use Ksp as before but not
(Ag^+) 0.14 + 2x
(SO4^=) = x
Solve for x
c) K2SO4 ==> 2K^+ + SO4^=
...0.33........0.....0
.....0........0.66...0.33
Use Ksp as before but
(Ag^+) = 2x
(SO4^=) = x + 0.33
Solve for x
Post your work if you get stuck.
For A,