What's a 'prac'?
You clearly describe your objective. Good.
Let me see if I understand what you did. First, you measured the mass of a flask + rubber stopper. Then you took a store bought ammonia product and put some in the flask. You weighed the flask + stopper + ammonia. From these two mass measurements you were able to determine the mass of the ammonia solution, which you did correctly.
Then you put a little indicator (phenolthylein probably) into the ammonia solution and titrated it with a 0.1M hydrochloric acid solution. You used 6.75 mL of acid before the indicator showed titration was complete (it turned slightly pink).
You calculated that you used 6.75x10^(-4) mol HCl.
Here's where I get lost in what you're doing. You wrote this:
C (diluted) = n/v
= 0.000675/0.02
=0.03375M
n (NH4OH) = m/Mr
= 7.642/(14.01+4+16.01)
= 0.218343mol
What you need to do at this point is realize that by the titration reaction equation (which you copied down incorrectly) you know that 6.75x10^(-4) mol of ammonium ion was present in the commercial ammonia solution you were analyzing.
The problem with your equation is that you weren't paying attention when you copied it down. There is no element symbolized by A! The reaction is actually written
NH4OH (aq) + HCl(aq) -> NH4Cl (aq) + H2O (l)
Now, you know that in the 7.642g of ammonia solution, there are 6.75x10^(-4) mol NH4OH.
The next thing you do is calculate what the mass of a mol of NH4OH is. Then calculate the mass of the 6.75x10^(-4) mol NH4OH. Also calculate the mass of NH3, as NH3 (ammonia) in water becomes NH4OH (ammonium hydroxide), and the manufacturer might have used either one for their calculations.
Well, no, the aim you wrote down refers to an ammonium compound, so go with the NH4OH.
Once you've calculated the mass of NH4OH in the solution, you can subtract that mass of the ammonium hydroxide from the total mass of the solution. Assuming the solution is nothing but ammonium hydroxide and water, you now know how much of the solution is NH4OH and how much is water by mass. Using the density of water you can calculate how much water was used to make the portion of solution you titrated.
Divide the mass of ammonium hydroxide by this volume of water. If I remember correctly, when the concentration of a solution is written in g/L, it's grams of solute in liters of solvent, not grams of solute in liters of solution.
This will get you the concentration of ammonium hydroxide in the solution in g/L. Make sure your units are correct; don't calculate g/mL and submit the answer like that or anything.
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