Calculate the solubility Ag2CrO4(s) in a 0.100 M solution of AgNO3 and compare its solubility with Ag2CrO4 solubility in pure water at room temperature. The Ksp for Ag2CrO4 is 9.0 x 10^-12 at room temperature. The Kspfor Ag2CrO4 at room temperature in pure water is 1.3 x10^-4.
Something is amiss here.
The Ksp is the Ksp. You can't have two Ksp values for Ag2CrO4.
40 mL of o.5M sulfuric acid is neutralized by 60mL of a solution of sodium hydroxide of unknown concentration. What is the concentration of sodium hydroxide? Please show the work
To calculate the solubility of Ag2CrO4(s) in a 0.100 M solution of AgNO3, we need to consider the common ion effect.
Step 1: Write the balanced equation for the dissociation of Ag2CrO4(s) in water:
Ag2CrO4(s) ⇌ 2Ag+(aq) + CrO4^2-(aq)
Step 2: Write the expression for the solubility product constant (Ksp) for Ag2CrO4:
Ksp = [Ag+]^2 [CrO4^2-]
Step 3: Identify the concentration of Ag+ ion in the 0.100 M solution of AgNO3. Since AgNO3 dissociates completely in water:
[Ag+] = 0.100 M
Step 4: Substitute the known values into the Ksp expression and solve for the solubility of Ag2CrO4 in the AgNO3 solution:
9.0 x 10^-12 = (0.100 M)^2 [CrO4^2-]
Solve for [CrO4^2-]:
[CrO4^2-] = (9.0 x 10^-12) / (0.100 M)^2
Step 5: Calculate the solubility of Ag2CrO4 in the AgNO3 solution:
[CrO4^2-] = (9.0 x 10^-12) / (0.100)^2
[CrO4^2-] = 9.0 x 10^-10 M
The solubility of Ag2CrO4 in the 0.100 M solution of AgNO3 is 9.0 x 10^-10 M.
To compare the solubility of Ag2CrO4 in pure water, we can use the Ksp value of 1.3 x 10^-4.
Step 6: Write the Ksp expression for Ag2CrO4 in pure water:
Ksp = [Ag+]^2 [CrO4^2-]
Step 7: Substitute the known values into the Ksp expression and solve for [CrO4^2-]:
1.3 x 10^-4 = [Ag+]^2 [CrO4^2-]
Since it is pure water, [Ag+] = [CrO4^2-] (stoichiometrically)
1.3 x 10^-4 = [CrO4^2-]^2
Step 8: Solve for [CrO4^2-]:
[CrO4^2-] = √(1.3 x 10^-4)
[CrO4^2-] = 3.6 x 10^-3 M
The solubility of Ag2CrO4 in pure water at room temperature is 3.6 x 10^-3 M.
Comparing the two solubilities:
The solubility of Ag2CrO4 in the 0.100 M AgNO3 solution is 9.0 x 10^-10 M,
while the solubility of Ag2CrO4 in pure water is 3.6 x 10^-3 M.
Therefore, the presence of Ag+ ions from AgNO3 decreases the solubility of Ag2CrO4 compared to its solubility in pure water.
To calculate the solubility of Ag2CrO4(s) in a 0.100 M solution of AgNO3, we can use the common ion effect. The common ion effect states that the solubility of a slightly soluble salt is decreased by the presence of a common ion from a strong electrolyte.
Here's how you can calculate the solubility:
1. Write the balanced equation for the dissolution of Ag2CrO4(s):
Ag2CrO4(s) ⇌ 2Ag+(aq) + CrO4^2-(aq)
2. Determine the initial concentration of Ag+ ion in the solution. Since the AgNO3 is 0.100 M, the concentration of Ag+ ion is also 0.100 M.
3. Use an ICE (Initial Change Equilibrium) table to set up the equilibrium expression for Ag2CrO4:
Ksp = [Ag+]^2 [CrO4^2-]
4. Substitute the known values into the equilibrium expression:
9.0 x 10^-12 = (0.100)^2 [CrO4^2-]
5. Solve for [CrO4^2-]:
[CrO4^2-] = (9.0 x 10^-12) / (0.100)^2 ≈ 9.0 x 10^-10 M
So, the solubility of Ag2CrO4(s) in a 0.100 M solution of AgNO3 is approximately 9.0 x 10^-10 M.
To compare this solubility with the solubility of Ag2CrO4 in pure water at room temperature, we can compare their respective Ksp values.
For Ag2CrO4 in pure water at room temperature, the Ksp is given as 1.3 x 10^-4. The Ksp represents the solubility product constant, which is the equilibrium constant for the dissolution of a slightly soluble salt. Therefore, a higher Ksp value indicates a higher solubility.
Comparing the Ksp values:
Ksp (Ag2CrO4 in 0.100 M AgNO3) = 9.0 x 10^-12
Ksp (Ag2CrO4 in pure water) = 1.3 x 10^-4
Since 1.3 x 10^-4 is significantly larger than 9.0 x 10^-12, it means that the solubility of Ag2CrO4 in pure water at room temperature is higher than in a 0.100 M solution of AgNO3.