I work in a lab, normally we evaporate our acid and dilute with water, but I have 300 samples and this would destroy the fume hood so I need to raise the pH using ammonia.
Samples are 40ml of 9M HCl
I need to add Xml of XM ammonium hydroxide (aka aqueous ammonia)
The desired final pH needs to be between 0.8 and 2.
First you need the reaction of HCl + NH3--> NH4+ + Cl-
we have n = (.04L)(9mol/L) = 0.36 mol HCl
The molarity of aqueous ammonia is about 14.5M but I think it cannot be exactly determined.
Once we go through this equation with the known moles of ammonia, we can subtract that amount from the reactants and add it to the products
THen we use the equation HCl + H20 --> H30+ + Cl-
using a Ka = Kw/Kb = the concentration of products over reactants we get
5.56x10^-10 = [H30+][Cl-]/[HCl]
where [H3O+] is x, we solve for x, then take the negative log to get the pH.
The Attempt at a Solution
I completed the problem based on the protocol written above and got 2 microliters of 0.001M NH3 to obtain a pH of ~1. However, when I tried to neutralize a test sample in the lab, it took about 21ml of concentrated ammonia to raise the pH of 40ml of HCl to about 0.4. I am not sure if I am doing something wrong with the calculation or with the practice. I did think that perhaps the immense amount of vapors produced from adding concentrated NH3 (aq) to 9M HCl is reducing the efficiency of the buffer. Perhaps if I try to do this with the diluted NH3 it will work. I will try this tomorrow but in the meantime, can anyone verify this calculation or steer me in the right direction? Thanks.
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