Ammonia Testing
 

Hi, I'm new to here, so it would be great if you would be able to help me out here.

I did a prac to test for ammonina. But i can't find out the percentage by mass.

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.

I can't really follow what you are doing. Ammonia content is generally done by a back titration (due to volatility of ammonia) with bromocresol green indicator (offsets shift in pH at equivalence point). There is considerable uncertainty due to the volatility of HCl. Here is an example procedure with units in g NH3 / g solution (wt %):

Ex calculation:
To a 100 mL bottle was added 2.007 g of the ammonia solution, 3.156 g of concentrated HCl (36.5 - 38 wt %), 25 mL of water and 4 drops of bromocresol green indicator solution. The yellow solution was titrated until blue (16.80 mL) with 0.5998 M sodium hydroxide solution.

(0.5998 mol NaOH / 1 L) (0.0185 L) = 0.011096 mol NaOH used = 0.011096 mol HCl ---> (1 mol HCl / 1 mol NaOH)

added HCl = 3.156 g solution (0.37 g HCl / 1 g solution)(1 mol HCl / 36.5 g HCl) = 0.03199 mol

0.03199 mol HCl added - 0.011096 mol neutralized = 0.02089 mol consumed (1 mol NH3 / 1 mol HCl)*(17 g NH3 / 1 mol) / 2.007 g ammonia solution =

17.7 wt % NH3

Solutions
 

Just a little clarification...

Solution concentration is measured in the units per volume of solution NOT water (i.e. moles/litre solution or g/litre solution or g/100cm3 solution)

This is (I imagine) purely down to the convenience factor of solution preparation. Which usually goes as follows:

1. Measure out xg of a solute.
2. Dissolve in a quantity of water in a volumetric flask (100cm2, 250cm3, 1dm3 etc) .
3. Make up to the mark with distilled water.

The end result is units per litre of solution.

Re: Solutions
 

You are partially mistaken charco. Molarity is indeed moles of solute per liter of solution, and mass percent is mass of solute per mass of solution, but molality is moles of solute per kg of solvent. Whether or not solution or solvent is in the denominator varies. It's not always solution.

My general chemistry textbook does not describe mass per volume concentrations. It has moles per liter, moles per kg, mass per mass, volume per volume, and moles per moles.

I'll try to look this up.

This website suggests that it's grams of solute per liter of solution, like you suggest, charco:

http://dbhs.wvusd.k12.ca.us/webdocs/...GramSolub.html

This website seems to agree:

Percentage weight in volume (w/v) expresses the number of grams of solute in 100 ml of solution.

http://www.cop.ufl.edu/safezone/prok...solubility.htm

In this case, it looks like charco is right - it's grams of solute over volume of solution.

If it is indeed grams of solute per liter of solution, you're going to have to assume one liter of solution is made from one liter of water, which should be correct for the most part. Therefore the method I gave you would basically work.

If you have worked out the number of moles of ammonia in a sample then you can calculate its mass (moles x RMM)

If you know the total mass of the solution then mass of water = mass of solution - mass of ammonia

If you know the mass of water you can calculate the volume of water from a density conversion table for the given temperature (it will be approximately 1g = 1cm3

You then have the required info to do the w/v calculation...