[Fabbyfubz]

Since I've been taking O. Chem the last few semesters, my Stoichiometry skills are getting kind of rusty. In this problem, I found the mols of KCl after it was drained (0.585 M * 0.00106 L) and then finding the Molarity after adding the water (.038/mols). I didn't think it was correct and it wasn't. Any help or an explanation on how to solve this (rather than just giving an answer) would be appreciated.

The Problem:
A 500-mL volumetric flask containing 0.585 M KCl is emptied and allowed to drain briefly, leaving 1.06 mL of liquid in the flask. The flask is then rinsed with 32.8 mL of distilled water and is then used to prepare a standard solution of KNO₃. Compute the molar concentration of Cl- in the standard KNO3 solution if the flask is rinsed only one time with distilled water.
Assume that 1.06 mL of liquid remains in the flask each time it is emptied, and that this volume of rinse water remains in the flask prior to preparation of the KNO3 solution. Although in practice rinse volumes are generally approximate, for purpose of calculation please use the exact volumes indicated.

[Dan]

In this problem, I found the mols of KCl after it was drained (0.585 M * 0.00106 L) and then finding the Molarity after adding the water (.038/mols). I didn't think it was correct and it wasn't.

Show the calculation, I can't see where you got 0.038 from.

[Fabbyfubz]

In this problem, I found the mols of KCl after it was drained (0.585 M * 0.00106 L) and then finding the Molarity after adding the water (.038/mols). I didn't think it was correct and it wasn't.

Show the calculation, I can't see where you got 0.038 from.

Must have been a typo. I meant .0189 M.  

0.585 mols/L * 0.00106 L = 0.0006201 mols    And then .0328 l of water is rinsed (I'm not sure if this is used in molar calculations or not) and so  
0.0006201 mols/.0328 L = 0.01891 M *(1mol Cl/1 mol KCl) = 0.01891 M     This answer was still incorrect

[Borek]

Try to write all steps involved in the procedure of preparing standard solution of KNO₃. Your calculations are not off, but incomplete.

[fledarmus]

Must have been a typo. I meant .0189 M.  

0.585 mols/L * 0.00106 L = 0.0006201 mols material remaining in flask after solution is poured out 

  And then .0328 l of water is rinsed (I'm not sure if this is used in molar calculations or not) and so  
0.0006201 mols/.0328 L = 0.01891 M *(1mol Cl/1 mol KCl) = 0.01891 M molarity of solution when another 32.8 mL water is added   

And then when the rinse water is poured out...?
This answer was still incorrect

[Fabbyfubz]

Does preparing it mean we fill the rest of the flask to the top with KNO₃?           
       
So it would be instead, (0.0006201 mols)/(.5 L (flask) - .0106 L(rinse)) = .001267 M

[Borek]

Does preparing it mean we fill the rest of the flask to the top with KNO₃?

More or less. But 1.06 mL is what is left, and you did a second rinsing.

SO far you have correctly calculated amount of KCl that was left after the flask was drained for the first time:

0.585 mols/L * 0.00106 L = 0.0006201 mols

That was diluted with 32.8 mL and drained again - how much KCl was left in the flask? What was the next step?

[Fabbyfubz]

Would it be:    
      
(0.0006201 mols/.0328 L) = 0.01891 M
0.01891 M * .00106 L = 2.0045*10^-6 mols
2.0045*10^-6 mols/(.5 L - .00106 L) = 4.0174*10^-5 M            
          
If this isn't correct, then I'm stumped on how to finish this problem.

[Borek]

Looks like ou are on the right track till

2.0045*10^-6 mols/(.5 L - .00106 L) = 4.0174*10^-5 M

Why do you subtract 1.06 mL? Final volume is 500 mL.

Note: following calculations where only numbers are given, is not easy. I hope I read them right, but you have not described what you are doing. I have not checked the result.

[Dan]

(0.0006201 mols/.0328 L) = 0.01891 M

In addition to the points Borek has made, there is a problem with the highlighted value - you have not taken into account the 1.06 mL to which 32.8 mL is added, the total volume is not 32.8 mL.

Step 1: We have a 0.585 M solution of KCl solution, volume is 1.06 mL. You calculate that there are 6.201 x 10^-4 mol of KCl. Fine.

Step 2: You have 6.201 x 10^-4 mol of KCl in 1.06 mL, and then you add an additional 32.8 mL of water. What is the total volume? What is the concentration (C) of the new solution?

Step 3: Now you have a solution of concentration C. Most of it is poured away, just 1.06 mL is left. Using C, calculate the moles of KCl present in 1.06 mL of the solution. Let's call that X mol.

Step 4: You dilute your X mol of KCl up to a total volume of 500 mL, what is the concentration now?

Also

0.01891 M * .00106 L = 2.0045*10^-6 mols

No, you are a factor of 10 out. 0.01891 * 0.00106 = 2.0045 x 10^-5

Be more careful with your calculator and/or keyboard.

[Fabbyfubz]

I finally got it! Thank you Dan for the written out explanation and thank you others for the help. I really appreciate it.