[nerdchick]

Hello, I recently finished a chem lab trying to figure how much NH3 was in a cleaning solution. First we had to determine the M of HCl that we were using and figured .135 M. For the next part we put 20.00 mL of the cleaning solution in an erlenmeyer and added 80.00 ml of deionized water and mixed it. We then took 10.00 mL of that and added 9.70 mL of HCl to it until we had a color change. I initially did the calculation as M2=(.135 x 9.70)/ 10.00mL for .131 M of NH3 but my lab partner later said that we have to take into account the dilution for the cleaning product into our calculation. How would I go about that?
 Thank you in advance

[Borek]

0.131 M is concentration of ammonia in which solution - original, or the one that was prepared by diluting the original solution (by mixing 20/80)?

[nerdchick]

The NH3 cleaning solution was an unknown concentration/molarity. At the start of the experiment we took 20 mL of it then added 80 mL of water to dilute it for a total of 100mL. We then took out 10mL of the NH3/water diluted solution and added 9.70 mL of HCl to it (the HCl being .135 M). Someone said I should take the molarity I got from the original calculation (.131 M NH3) and multiply it by 5 because we diluted it by 5, however I am not sure if that is correct.
Sorry if I am confusing, this is my first chem lab in over 10 years and have no clue what is going on.
Thank you again

[Borek]

Basically you repeated what you already posted without even trying to address the question I asked. 0.131 M is a correct concentration - but of what?

[nerdchick]

Sorry, it was the first of three calculations I did for the molarity of the NH3. The first time I used the formula -
(1.35 M HCl x 9.75 mL of HCl used in reaction)/10.00 mL of the diluted ammonia cleaning solution.
That is how I came to .131 M of NH3
I repeated it two more times using 9.30mL HCl and 9.05mL HCl
Using the average of those I came to .126 M NH3. (which I guess I should have gave that value in first post instead of the .131 M of NH3, but I figured I would just do those calculations the same way.

[Borek]

You are still ignoring my question. OK, you calculated 0.126 M. This is an average concentration of ammonia in which solution?

[nerdchick]

I'm sorry I'm not purposely ignoring it, I think I just don't understand what you are asking. What I'm thinking is that the .131 M of NH3 is from the solution of the cleaning product containing ammonia that we diluted with water.
Thank you again.

[lokifenrir96]

First, the stoichiometric equation for the reaction between NH3(aq) and HCL(aq) is:

NH4OH + HCL -> NH4Cl + H2O

It shouldn't actually be written as NH4OH, but when you dissolve NH3 in water, you get NH4+ and OH- ions.

Anyway, since your concentration of HCL was 0.135 M (this means 0.135 mol dm-3, which means 0.135 mol in every 1000cm3, or 1 litre), and your volume of HCL used was 9.70mL, then the amount of HCL was 0.135 x 9.70/1000 = 0.0013095 mol.

So your amount of NH3 in that 10 mL was also 0.0013095 mol, since NH3:HCL is 1.1.

You took the 10mL of diluted NH3 from 100cm3 of stock solution, so the stock solution contains 0.0013095x10 = 0.013095 mol of NH3.

Now, this is the exact same amount of NH3 present in your 20 mL of UNdiluted NH3, since you just added water but the amount of NH3 remains the same.

So that means that 20mL of your original cleaning solution contained 0.013095 mol of NH3.

And hence your cleaning solution's concentration is 0.013095 x 1000 / 20 = 0.65475M.

I think this is correct (can someone check, thanks), and I noticed that your calculations did not divide the volumes by 1000, which you must do since they are in mL and M means mol per 1000mL.

So, to answer that last question, yes you have to take into account dilution if you want to find out the concentration of your original cleaning solution. But if you just want to find the concentration of the diluted NH3, then it would be (.135 x 9.70/1000)/(10/1000). Which is which Borek wanted to know