# Chemistry

25 ml of cloudy ammonia solution is added to a 250 ml Erlenmeyer flask. 50 ml of .1 M HCl is then immediately added to the flask, which reacted with the ammonia in the solution. Next, the solution was titrated with .025 M NaOH to neutralize the excess HCl that was not neutralized by the ammonia. 21.50 ml of NaOH was added to neutralize the excess HCl. Calculate the concentration of ammonia in the cloudy cleaning solution.

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1. NH3 + HCl ==>NH4Cl
Excess HCl + NaOH ==> NaCl + H2O

Mols HCl added initially = M x L = 0.1M x 0.050L = ?
mols NaOH to neutralize excess HCl = M x L = 0.025M x 0.02150 = ?
mols HCl used by ammonia = mols initially - mols excess = ? = mols NH3 initially.
M NH3 = mols NH3/L NH3 = mols from above/0.025L = ?

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2. I understand everything except how to calculate the mols NH3? Do the mols of HCL used by ammonia equal the mols of NH3 initially? Also, what is the mols excess that you used to calculate the mols of HCl used by ammonia?

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3. yes.
The problem in words works out this way. You have so many mols NH3 at the beginning. You had an excess of HCl which reacts with all of the NH3 and leaves extra HCl in the solution. The first thing you want to do is to find the how much excess HCl is there. You do that by titrating the HCl with a known volume and concn of NaOH. The NaOH is used to determined how much excess HCl you had. The mols NaOH used = mols HCl the NH3 DID NOT USE. So the difference in mols HCl you started with minus the mols HCl the NH3 did not use = mols NH3 used.

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4. Oh, okay. I understand now, thank you so much for your help.

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