# Homework Help: Calculating Ksp of Ag2CrO4

1. Apr 9, 2014

### Ritzycat

1. The problem statement, all variables and given/known data
I have to calculate the KSP of Ag2CRo4. I did this lab several days ago and I'm completely stumped. If anyone could help me out here, at the least guide me in the right direction so I can get this done as I am getting very frustrated :P

Here is the data I have derived thus far.

2.5x10-5 = concentration of AgNO3
1.5x10-5 = concentration of K2CrO4

Using a concentration / absorbance chart, we found the conc. of the CrO4-2 ion, 1.24x10-3.

My teacher says that the concentration of the Ag+ ion can be found stiochiometrically by using the CrO4-2 ion, then we can calculate KSP from that. I have no idea of how to get to the concentration of the Ag+ ion though.

The instructions say this:

1.
"Determine the value of Q for your original mixture of 5.0 mL of 0.0050 M AgNO3 mixed with 5.0 mL 0.0030 M K2CrO4. Keeping in mind that a precipitate forms from this mixture, is the actual Ksp for Ag2CrO4 greater or less than this value?"

the diluted concentrations of those molecules are listed above.

2.
Perform necessary calculations in order to determine the Ksp for Ag2CrO4.
This would include:
a. calculation of chromate ion concentration using the equation of your line from day 1.
b. Calculation of the silver ion concentration.
c. Calculation of Ksp for Ag2CrO4."

The chromate ion concentration is 1.24x10-3.

2. Relevant equations
AgNO3 + K2CRO4 <-> Ag2CrO4 + KNO3
Ag2CrO4 <-> 2Ag+ + CrO4-2
Ksp = [Ag+][CrO4-2]

All are aqueous, except Ag2CrO4 is solid.

I don't really have an "attempted solution" for this exact calculation because I really don't know what to do next. I've been able to calculate all of the necessary ingredients up until now. Thanks in advance for anyone who is willing to give me some guidance / show some calculations.

2. Apr 10, 2014

### Staff: Mentor

Hint: how much CrO42- disappeared from the solution? Where is it now? What does it tell you about Ag+?

3. Apr 10, 2014

### Ritzycat

am I supposed to make an ICE thing ?

4. Apr 10, 2014

### Staff: Mentor

In a way. ICE is just a nice method of following stoichiometry.

5. Apr 10, 2014

### Ritzycat

If I were to set it up though I'd have two unknown variables since I don't know either Ksp or KNO concentration...

6. Apr 10, 2014

### Staff: Mentor

7. Apr 10, 2014

### Ritzycat

Yes but I am not sure how. The only info I have so far about the chromate ion is its concentration as a result of mixing the 5 mL of compounds together.

8. Apr 10, 2014

### Staff: Mentor

No, you know how much was there initially, and you know - from the spectroscopy - how much was left.

9. Apr 10, 2014

### Ritzycat

How do I know how much there was initially, do I use stoichiometry using the moles of my reactants

10. Apr 10, 2014

### Ritzycat

A correction to the OP since I seem unable to edit it:

2.5x10-5 = mols of AgNO3
1.5x10-5 = mols of K2CrO4

NOT the concentrations, at least so I think.

(5 mL) X (1 L/1000 mL) x (.0050 mol / 1 L) = 2.5x10-5 mols.

The concentrations of silver nitrate and potassium chromate are .0050 and .0030M respectively

11. Apr 10, 2014

### Ritzycat

Hang on, I think you are right. On the graph that we calcualted on the spectrometer the original concentration of the chromate ion was 2.2x10-4

Also, the Solubility-Product constant of the Chromate ion is .44, I think.
I guess I could use that to find the concentration of the Ag ion with this, right?

.44 = [1.24x10-3][Ag+]2

Then the AG concentration would be 18.3. That seems unusually high.

Some other useful datas.

trial | volume of .003 M K2CrO4 | Total Volume | [CrO4] | Absorbance (430 nm)
1. | 5 mL | 2.2x10^-4 | 100 mL | .06
2. | 10 mL | 2.7x10^-4 | 100 mL | .10
3. | 15 mL | 4.7x10^-4 | 100 mL | .17
4. | 18 mL | 5.3x10^-4 | 100 mL | .19

Last edited: Apr 10, 2014
12. Apr 10, 2014

### Staff: Mentor

How much chromate have you put into sample? You prepared the solution, didn't you? And to do so, you measured some volume of chromate? Can't you use this information to calculate how much chromate was initially present?

What is the final volume of the solution? What is the measured concentration? How many moles of chromate does it mean?

Please try to answer these questions, I am trying to guide you to an answer, but you are taking every possible tangential road which is why you don't move ahead.

13. Apr 10, 2014

### Ritzycat

We did not directly measure/prepare the chromate itself. We first created a Concentration vs. Absorbance graph, since chromates are yellowish we prepared 4 different concentrations of K2CrO4, then put it in the spectrometer to measure the absorbance vs. the concentrations at 430 nm.

Then we put 5.0 mL .0050M AgNO3 and 5.0 mL .0030M K2CrO4 into two test tubes each. We shook the tubes, put them in the centrifuge for 2 minutes, dumped it out in the sink, added 3mL water, and repeated that process. After three rounds we then decanted it into the cuvette to be measured in the spectrometer. We took the value for the absorbance (.44) and plugged it into the formula we made for our graph.

.44 = 350.3x + .003878

So the concentration of it would be 1.2x10^-3

Sorry Im not able to directly answer your questions. I am completely stumped though as to what I need to do to get the concentration of the Ag+ ions using all of my information.

14. Apr 10, 2014

### Staff: Mentor

How much chromate is that?

15. Apr 10, 2014

### Ritzycat

(5 mL) x (1 L/1000 mL) x (.0030 mol / 1 L) = 1.5x10-5 mol K2CrO4

K2CrO4 <-> 2K+ + CrO42-

1.5x10-5 mol K2CrO4 = 1.5x10-5 mol CrO42-

1.5x10-5 mol CrO42- X (115.996g/1 mol) = 1.7x10-3 g CrO4

Does this all look right to you?
I think I need the initial concentration though, how should I go about getting that from this?

16. Apr 11, 2014

### Staff: Mentor

Sigh. In you opening post you omitted whole procedure, and what you wrote suggested something completely different than what you really did. Next time post EVERYTHING at once to not waste everyone's time (including yours).

From your description - which is still not necessarily clear - you measured concentration of CrO42- in the saturated solution over precipitate. Do I understand the procedure correctly?

17. Apr 11, 2014

### Ritzycat

Sorry. I haven't been sure what exactly is important to solving the problem so I have not been sure what info to give.

Yes we measured the concentration of chromate ion after the solution was very dilute.

18. Apr 11, 2014

### Staff: Mentor

Solution was very dilute, or it was a SATURATED solution in contact with the precipitate? What it was that you were throwing away after centrifuging - solid, or just the liquid above?

19. Apr 11, 2014

### Ritzycat

We only threw away the liquid above

20. Apr 11, 2014

### Staff: Mentor

So the solution was always in contact with the solid, and we can safely assume it was saturated. If so, it is pretty easy. Write reaction equation of dissolution. How many moles of Ag+ produced per each mole of CrO42-?

21. Apr 13, 2014

### Ritzycat

2 moles Ag+ for every 1 mole of CrO4-2

Ag2CrO4-2 <-> 2Ag+ + CrO4-2

22. Apr 13, 2014

### Staff: Mentor

So what is concentration of Ag+?

23. Apr 13, 2014

### Ritzycat

What can I do to find that out without having directly measured it?

24. Apr 14, 2014

### Staff: Mentor

Look at the reaction equation, stoichiometry of the dissolution tells you how much silver got in the solution together with chromate ion.

How many moles of Ag+ are dissolved with each mole of CrO42-? Let's say the volume at the moment of the measurement was 10 mL. How many moles of chromate were dissolved> How many moles of silver were dissolved? What was the Ag+ concentration?

25. Apr 14, 2014

### Ritzycat

Would the concentration of the Ag+ simply be double the CrO4 -2? In this case, if the CrO4 concentration was 1.2x10^-3 then the Ag+ concentration would be 2.4x10^-3.

Also, part of the write-up for this lab includes writing calculations for finding the Q of silver chromate. I'd assume this would be the following.

Q = [1.2x10^-3][2.4x10^-3] = 2.9x10^-6
Since in the main reaction equation a precipitate formed, K > Q (moved right).

Last edited: Apr 14, 2014