Hi, I have a few chemistry problems that I'm having trouble solving. I put my attempted solutions after the problems in *'s. There was one problem I had no idea how to solve.
What concentration of silver chromate (Ksp = 9.0 x 10-12) will dissolve to make a saturated solution in water?
*For this one the formula is Ag2CrO4, which is [Ag]^2[CrO4]. I plugged x in for CrO4, so x^3 is equal to Ksp. I then got that x = 2.0e-4. However, that x is for CrO4, not for silver chromate. Is the answer for silver chromate just the Ksp value?*
What concentration of silver chromate (Ksp = 9.0 x 10-12) will dissolve in 500. mL of 0.01 M aqueous silver nitrate?
A saturated solution of CaSO4(aq) is made in a beaker until there is excess calcium sulfate resting at the bottom. Then solid potassium sulfate is added. Which of the following is true? (The Ksp for potassium sulfate is larger than the Ksp for calcium sulfate)
a. More calcium sulfate will start to precipitate out of solution
b. More calcium sulfate will start to dissolve in solution
*I know that when a smaller Ksp is added to a bigger Ksp it precipitates out of solution. Is that the same in the reverse?*
Ag2CrO4 ==> 2Ag + CrO4^-2
Ksp = (Ag^+)^2(CrO4^-2)
What you do is let S = solubility of Ag2CrO4 in molarity. Then S is concn CrO4 and 2S = concn
(2S)^2(S) = Ksp
4S^3 = Ksp
Solve for S. 2S will be Ag^+, S will be Ag2CrO4 and S will be CrO4^-2.
#2. Here you have two equilibria.
Ag2CrO4 ==> 2Ag^+ + CrO4^-2
Ksp = (Ag^+)^2(CrO4^-2)
And AgNO3 ==> Ag^+ + NO3^-
AgNO3 is completely soluble BUT it contributes Ag^+. So if
S = solubility Ag2CrO4, then S = CrO4^-2, 2S+0.01 = Ag^+. Plug these into Ksp for Ag2CrO4. (This is called a common ion problem and the common ion in this problem is the Ag^+. It's another example of Le Chatelier's Principle; adding AgNO3 forces the solubility to the left in the reaction and makes it more insoluble. In most cases, 2S + 0.01 = 0.01; i.e., 2S is small and contributes only slightly to the total Ag^+ and it makes it easier to solve the equation. So you end up with
9.0 x 10^-12 = (0.01)^2(S) and solve for S which is CrO4 and Ag2CrO4. You will see that the solubility is much less in this problem than in the first problem solved.
#3. I really hope no one is trying to tell you that K2SO4 has a solubility product. It doesn't. Not even close! This is a case of common ion, also. Here is how it works.
CaSO4(s) ==> Ca^+2 + SO4^-2
Ksp = (Ca^+2)(SO4^-2).
K2SO4 ==> 2K^+ + SO4 (completely soluble).
The SO4^-2 is the common ion. By Le Chatelier's principle adding sulfate (by way of K2SO4) forces the CaSO4 equilibrium to the left which means that CaSO4 is LESS soluble with K2SO4 added than when it isn't added. Therefore, some of the CaSO4 in solution will ppt. Talk of Ksp for K2SO4 has nothing to do with it.
thank you so much!! I got a 100 on the test I was studying for!
To find the concentration of silver chromate that will dissolve to make a saturated solution, you need to find the concentration of the ions in solution.
For the equation Ag2CrO4 ⇌ 2Ag+ + CrO4^2-, the solubility product expression is Ksp = [Ag+]^2[CrO4^2-].
Since the stoichiometric coefficient of Ag+ is 2, the concentration of Ag+ in the saturated solution will be 2x, where x represents the concentration of CrO4^2- (since Ksp = x^3).
Let's solve this step-by-step:
1. Set up the solubility product expression: Ksp = (2x)^2(x) = 4x^3.
2. Substitute the given value for Ksp into the equation: 9.0 x 10^-12 = 4x^3.
3. Solve for x by taking the cube root of both sides: x = (9.0 x 10^-12)^(1/3).
4. Calculate the value of x using a calculator: x ≈ 4.92 x 10^-4.
Therefore, the concentration of CrO4^2- in the saturated solution is approximately 4.92 x 10^-4 M.
Regarding your attempted solution for the first problem, you correctly obtained x for CrO4^2-. However, the concentration of silver chromate (Ag2CrO4) is 2x since there are two moles of Ag+ for every mole of CrO4^2-. Thus, the concentration of silver chromate is 2(4.92 x 10^-4) = 9.84 x 10^-4 M.
Moving on to the second problem:
To determine the concentration of silver chromate that will dissolve in 500 mL of 0.01 M aqueous silver nitrate, you can use the common ion effect to find the shift in equilibrium.
The balanced chemical equation for the dissolution of silver chromate is:
Ag2CrO4(s) ⇌ 2Ag+(aq) + CrO4^2-(aq)
When silver nitrate (AgNO3) is added, it dissociates to give Ag+ ions. These Ag+ ions will contribute to the concentration of the common ion in the reaction, which affects the equilibrium shift.
1. Since the concentration of Ag+ in the silver nitrate solution is 0.01 M, the common ion effect will decrease the solubility of silver chromate.
2. Assume that x mol/L of Ag2CrO4 dissolves. This will result in the formation of 2x mol/L of Ag+ and x mol/L of CrO4^2-.
3. Write the solubility product expression: Ksp = (2x)^2(x) = 4x^3.
4. Substitute the given value for Ksp and solve for x: 9.0 x 10^-12 = 4x^3.
Take the cube root of both sides: x ≈ (9.0 x 10^-12)^(1/3) ≈ 4.92 x 10^-4.
5. The concentration of Ag+ is 2x = 2(4.92 x 10^-4) = 9.84 x 10^-4 M.
6. Since the initial concentration of Ag+ ions from silver nitrate was 0.01 M, the additional Ag+ ions from the dissolution of silver chromate will be 9.84 x 10^-4 M.
7. Subtract this value from the initial concentration of Ag+ to find the effective concentration: 0.01 M - 9.84 x 10^-4 M = 0.00902 M.
Therefore, the concentration of silver chromate that will actually dissolve in 500 mL of 0.01 M aqueous silver nitrate is approximately 0.00902 M.
Regarding the final question about the solubility of calcium sulfate and potassium sulfate:
The statement "The Ksp for potassium sulfate is larger than the Ksp for calcium sulfate" indicates that the solubility product constant (Ksp) of potassium sulfate (K2SO4) is greater than the Ksp of calcium sulfate (CaSO4).
In general, when two chemical species with different solubility product constants are mixed, the substance with the smaller Ksp will precipitate out of solution. In this case, the smaller Ksp belongs to calcium sulfate (CaSO4).
Therefore, when solid potassium sulfate (K2SO4) is added to a saturated solution of calcium sulfate (CaSO4), more calcium sulfate (CaSO4) will start to precipitate out of solution. Hence, option a. "More calcium sulfate will start to precipitate out of solution" is the correct answer.
For the first problem, the concentration of silver chromate that will dissolve to make a saturated solution in water can be determined using the concept of Ksp (solubility product constant). The Ksp expression for silver chromate can be written as [Ag2CrO4] = [Ag]^2[CrO4].
To find the concentration of silver chromate, you need to first determine the concentration of either silver ions ([Ag]) or chromate ions ([CrO4]) in the saturated solution. From your attempted solution, you correctly considered the equilibrium expression for silver chromate and set up the equation x^3 = Ksp, where x represents the concentration of chromate ions.
However, to find the concentration of silver chromate, you need to find the concentration of silver ions [Ag], which can be obtained by back-substituting the calculated concentration for chromate ions [CrO4] into the equilibrium expression as [Ag]^2[x]. Since [Ag] = 2x (from the stoichiometric coefficients), you should solve the equation (2x)^2[x] = Ksp to find the value of x.
Regarding the second problem, the concentration of silver chromate that will dissolve in 500 mL of 0.01 M aqueous silver nitrate can be determined by examining the common ion effect. In this case, both silver ions (from silver nitrate) and chromate ions (from silver chromate) are present, and the solubility will be affected by the common ion (silver ion) in solution.
To determine the concentration of silver chromate, you need to calculate the concentration of silver ions in the solution and then use that value to evaluate the remaining concentration of chromate ions. Since the silver nitrate concentration is given as 0.01 M, the concentration of the silver ions [Ag] will be 0.01 M as well. You can then substitute this value into the solubility equation as [Ag]^2[x] = Ksp to solve for x, the concentration of chromate ions.
For the third problem, the addition of solid potassium sulfate can potentially impact the solubility equilibrium between calcium sulfate and potassium sulfate. The statement "The Ksp for potassium sulfate is larger than the Ksp for calcium sulfate" suggests that the solubility product constant, Ksp, for potassium sulfate is greater than the Ksp for calcium sulfate.
When a compound with a larger Ksp is added to a solution containing a compound with a smaller Ksp, the common ion (in this case, the sulfate ion) will have a greater tendency to stay in solution and form more of the compound with the larger Ksp (potassium sulfate). As a result, there will be a shift in the equilibrium towards the formation of more potassium sulfate. This means that more calcium sulfate will start to precipitate out of the solution, option a is correct.