Hi,
I was hoping someone could double check a chemistry answer for me. I am trying to calculate the Ksp of Ag2CrO4. This question is based on a experiment to determine ksp experimentally. In the experiment the Ksp of silver chromate was calculated using a single displacement reaction between a saturated solution of silver chromate and copper metal. A 1.00 L of saturated silver chromate was used in the experiment. Before the reaction the mass of copper was 1.250 g and after it was 1.240 g. In moles this was calculated to be 0.00016 mol of Cu. In the balanced equation, the copper and silver chromate reactants have a mole ration of 1.1. The balanced reaction is Cu + Ag^2CrO^4 ----> CuCrO^4 + 2Ag. The Ksp value measured for silver chromate was given as 1.1 x 10^-12.
So given that information here is what I did.
Ksp = [Ag^+]^2 [CrO4^2-]
The molar mass of Ag^2CrO4 is 331.73 g/mol
0.00016/ 331.73 = 4.8 x 10^-7
(4.8 x 10^-7)^2(4.8 x 10^-7) = 1.10 x 10^-19
Then I calculated my percentage error.
1.10 x 10^-19 - 1.1 x 10^-12/ 1.1 x 10^-12 x 100% = 9.9 x 10^-23
Did I solve it correctly? If not could someone show me where I went wrong. Thank you in advance.
57. Determining K_sp experimentally
a. Calculate K_sp of Ag_2 CrO_4. Organize your answer so that your solution is logical. Make sure that your K_sp has correct significant figures.
a. From the experiment CU = 0.00016 mol. Also 〖Ag〗_2 〖CrO〗_4 = 0.0016 mol
〖Ag〗_2 〖CrO〗_4⇋〖2Ag〗_((aq))^++ CrO_(4 (aq))^(2-) K_(sp )=1.1*〖10〗^(-12)
[〖Ag〗_2 〖CrO〗_4 ]=n/V=0.00016mol/1.00L=1.6*〖10〗^(-4) mol/L
ICE [Ag^+ ]^2 mol/L [〖CrO〗_4^(2-) ] mol/L
Initial 0 0
Change +2x +x
Equilibrium 2(0.00016) = 0.00032 0.0016
K_sp=[Ag^+ ]^2 [〖CrO〗_4^(2-) ]
1.1*〖10〗^(-12)=(3.2*〖10〗^(-4) )^2 (1.6*〖10〗^(-4) )
=1.6*〖10〗^(-11)
b. Calculate percent error
% error= ((experimental-Theoretical))/Theoretical*100
=((1.6*〖10〗^(-11)-1.1*〖10〗^(-12)))/(1.1*〖10〗^(-12) )*100
= 13.89% = 14%
I would rethink the problem.
1.6E-4/molar mass Ag2C4O4 give I don't know what.
Cu + Ag2CrO4 ==> 2Ag + CuCrO4
If you had 1.6E-4 mols Cu you must have had 1.6E-4 mols Ag2CrO4.
CrO4^2- = 1.6E-4
Ag^+ = 2*1.6E-4
Calculate Ksp from that.
Your % error, from your calculation is 100% because of a math error (but recalculated from above it should be better than that but I don't know how much better).
To calculate the Ksp of Ag2CrO4 using the given information and experimental data, you need to follow these steps:
Step 1: Determine the moles of Ag2CrO4 used in the experiment:
Given that the mole ratio between Cu and Ag2CrO4 is 1:1, the moles of Ag2CrO4 used can be calculated as:
moles Ag2CrO4 = moles Cu = 0.00016 mol
Step 2: Calculate the concentration of Ag+ ions in the saturated Ag2CrO4 solution:
Since the moles of Ag2CrO4 and Ag+ ions are in a 1:2 ratio according to the balanced equation, you can calculate the concentration of Ag+ ions as:
[Ag+] = (2 * moles Ag2CrO4) / volume of solution
Given that the volume of solution used is 1.00 L, the concentration of Ag+ ions can be calculated as:
[Ag+] = (2 * 0.00016 mol) / 1.00 L = 0.00032 M
Step 3: Calculate the concentration of CrO4^2- ions in the saturated Ag2CrO4 solution:
Since the molar ratio between Ag2CrO4 and CrO4^2- ions is 1:1, you can calculate the concentration of CrO4^2- ions as:
[CrO4^2-] = (moles Ag2CrO4) / volume of solution
Given that the volume of solution used is 1.00 L, the concentration of CrO4^2- ions can be calculated as:
[CrO4^2-] = (0.00016 mol) / 1.00 L = 0.00016 M
Step 4: Substitute the concentrations of Ag+ and CrO4^2- ions into the Ksp expression for Ag2CrO4:
Ksp = [Ag+]^2 * [CrO4^2-]
= (0.00032 M)^2 * (0.00016 M)
= 1.024 x 10^-9 M^3
So, the calculated Ksp of Ag2CrO4 based on the given data is 1.024 x 10^-9 M^3.
To find the percentage error, you can use the following formula:
Percentage error = |measured value - calculated value| / measured value * 100%
Given that the measured value of Ksp is 1.1 x 10^-12 M^3, the percentage error can be calculated as:
Percentage error = |1.1 x 10^-12 - 1.024 x 10^-9| / 1.1 x 10^-12 * 100%
= 69.09%
Therefore, the percentage error in the calculated Ksp value is 69.09%.
It seems like you made a mistake in calculating the Ksp value and the percentage error.
To calculate the Ksp of Ag2CrO4, you need to use the given information and the equation provided. Let's break down the steps and see where you went wrong.
Step 1: Calculate the number of moles of silver:
Given: Mass of copper before reaction = 1.250 g
Mass of copper after reaction = 1.240 g
Change in mass of copper = 1.250 g - 1.240 g = 0.010 g
Now, convert this mass to moles by dividing by the molar mass of copper (63.55 g/mol):
0.010 g / 63.55 g/mol = 0.000157 mol
Step 2: Determine the moles of Ag2CrO4:
Since the mole ratio between copper and silver chromate is 1:1, the number of moles of silver chromate is the same as the number of moles of copper: 0.000157 mol.
Step 3: Calculate the concentration of Ag+:
The volume of the solution is given as 1.00 L, and the number of moles of Ag+ is 0.000157 mol. Therefore, the concentration of Ag+ is:
[Ag+] = moles / volume = 0.000157 mol / 1.00 L = 0.000157 M
Step 4: Calculate the Ksp value:
Using the equation Ksp = [Ag+]^2[CrO4^2-], we substitute the known values to get:
Ksp = (0.000157 M)^2[CrO4^2-]
Step 5: Substitute the given Ksp value for Ag2CrO4:
The given Ksp value is 1.1 x 10^-12. Therefore, we have:
1.1 x 10^-12 = (0.000157 M)^2[CrO4^2-]
Step 6: Solve for [CrO4^2-]:
[CrO4^2-] = (1.1 x 10^-12) / (0.000157 M)^2
Now, plug in the calculated value of [CrO4^2-] into the Ksp equation and solve for Ksp.
It seems that your calculation of Ksp is incorrect. Follow the correct steps above to get the accurate value of Ksp for Ag2CrO4.