Question 3:How can I determine the solubility of a compound using Ksp values?

  • Thread starter Thread starter Lchan1
  • Start date Start date
  • Tags Tags
    Chem
AI Thread Summary
To determine the solubility of a compound using Ksp values, users discussed calculating the remaining concentration of ions after mixing solutions, specifically AgNO3 and K2CrO4. The first question involved finding the percentage of Ag+ remaining in solution after precipitation, while the second question focused on the mass of precipitate formed from mixing saturated solutions of Ag2SO4 and PbCrO4. Participants emphasized the importance of understanding limiting reagents and the effects of dilution on Ksp calculations. Additionally, there were inquiries about calculating pK values and hydronium ion concentrations in acid-base reactions, highlighting the need for accurate equilibrium expressions. Overall, the discussion revolved around applying Ksp and equilibrium concepts to solve complex solubility problems.
Lchan1
Messages
38
Reaction score
0

Homework Statement


What percentage of the original Ag+ remains in solution when 175 mL of 0.0208 M AgNO3 is added to 250 mL 0.0380 M K2CrO4?


Homework Equations





The Attempt at a Solution



I found the initial amont of CrO4 and Ag and put that into the equilbrium expression of AgCrO4(s). I let the reation complete and used Ksp for the back reation to find the ions in solution. but i couldn't get the right answer.

Question 2:

If 100.0 mL of a clear saturated solution of Ag2SO4 is added to 266 mL of a clear saturated solution of PbCrO4 how much precipitate in mg will form?
For this problem, ignore the back reaction since it will have an insignificant effect on the amount of precipitate that forms.

The molecular weight of PbSO4 is 303.26 g/mol and the molecular weight of Ag2CrO4 is 331.73 g/mol. Give your answer to three significant figures.

Attempt to the question
So I tried finding the concentration in PbSO4 and Ag2CrO4 using Ksp expression. and used to concentration to find moles. used moles and molar weight given to find the weight. and I added the two up. And again I got an incorrect answer.
 
Physics news on Phys.org
Not sure what you mean by back reaction.

1. One of the ions is in excess. While this is not titration, calculations will be similar to those outlined at precipitation titration curve calculation - after equivalence point.

2. Have you remembered this is a twice limiting reagent question? Then you start with saturated solutions and you end with saturated solutions. Could be whatever is left in the solution can be ignored, hard to tell without looking at Kso values.

And don't forget about dilutions.
 
no, this is not an after equivalance point type question. Tell me how this question goes Lchan
 
Charlie_Brown said:
no, this is not an after equivalance point type question. Tell me how this question goes Lchan

so I did the first question right.. turns out it was just so calculation errors.
I attempted the second question again and got a different answer but it was incorrect again.

so What i did was using Ksp and Ag2SO4 -> 2Ag+ and SO4 2-
and found that there are way more Ag and SO4 ions so Pb and CrO4 ions are the limited reagent for both Ag2CrO4 and PbSO4 solid.
used the concentration of Pb and CrO4 multiply by the total volume and got the number of moles.
Got two masses and added them up. but i couldn't get the right answer.

another question:
What is the pK value for the following equilibria involving the neutralization of piperidine (C5H11N) with acid? (The Kb of piperidine is Kb = 1.3 x 10-3)
C5H11N + H3O+ <----> C5H11NH+ + H2O

I got 11.11
so H30+ presents in the equation i will need Ka.
used -logKa and got an wrong answer.
I googled the answer up and it was 11.12. I won't know the answer until i get it wrong and I am sick of trying different numbers.

also with
Calculate the H3O+ concentration at the equivalence point if the initial amount of 0.294 M NH3 was 55 mmol, and the final total volume after the titration is 99.0 mL.

So I found how much NH3 was added used the concentration and the 55mmol. Subtract the 99.0mL and get the volume of H3O+ added.
since the equation is 1 to 1. I just went 55mmol divided by the volume i got. And again the answer was incorrect.
 
i'll look at one problem first.. what do you mean by the "pk" ? There is pKa, and pKb.. can you tell me what you did, because from the "pk" I got 2.89, WHICH when you minus it from 14 (14-2.89) = 11.11, I can confirm that by solving for pKa and finding out the hydronium concencetration that the pH would be 11.1
 
Lchan1 said:
also with
Calculate the H3O+ concentration at the equivalence point if the initial amount of 0.294 M NH3 was 55 mmol, and the final total volume after the titration is 99.0 mL.

So I found how much NH3 was added used the concentration and the 55mmol. Subtract the 99.0mL and get the volume of H3O+ added.
since the equation is 1 to 1. I just went 55mmol divided by the volume i got. And again the answer was incorrect.

Is there more info to this question?
 
Charlie_Brown said:
no, this is not an after equivalance point type question.

Yes it is. Imagine you are titrating AgNO3 with K2CrO4. Imagine you started with 175 mL of 0.0208 M AgNO3, and you have added 250 mL of 0.0380 M titrant solution. Do you see the similarity now?

Sure, question doesn't mention titration nor equivalence point, but method of calculation of the Ag+ is identical to the one described at linked page.

Charlie_Brown said:
what do you mean by the "pk" ? There is pKa, and pKb.

pK is just -log(K), where K is equilbrium constant for the reaction. It doesn't have to be acid or base dissociation, any reaction will do.
 
Last edited:
What is the pK value for the following equilibria involving the neutralization of piperidine (C5H11N) with acid? (The Kb of piperidine is Kb = 1.3 x 10-3)
C5H11N + H3O+ <----> C5H11NH+ + H2O
ok I tried 2.89 and 11.11
and they were both wrong for some reason
I used Kb to find Ka. and find the concentration of H3O+
 
Show your work.

Both values are wrong.

Think about it - pyridine is a weak base, so its protonation reaction will be shifted to product side. That means equilibrium constant higher then 1. What should be the sign of pK for K > 1?
 
  • #10
ok.
so I did Ka=Kw/Kb
Ka = 7.69E-12 = C5H11NH+/[H3O] and [C5H11N]
Ka = [H3O]
K= [C5H11NH+]/H3O and C5H11N
-log k = 11.11
 
  • #11
Lchan1 said:
Ka = 7.69E-12 = C5H11NH+/[H3O] and [C5H11N]

[C5H11NH+]/([H3O+][C5H11N]) is not Ka.
 
  • #12
Ka = ([H3O+][C5H11N])/[C5H11NH+]?
 
  • #13
Lchan1 said:
Ka = ([H3O+][C5H11N])/[C5H11NH+]?

yup! you are correct. Remember: Ka=[H30+][A-]/[HA]


CB
 
  • #14
Borek said:
you start with saturated solutions and you end with saturated solutions. Could be whatever is left in the solution can be ignored, hard to tell without looking at Kso values.

As someone else pointed out, could be nothing happens at all. That will came out automatically during Kso calculations.
 
Back
Top