All Questions
7
questions
1
vote
1
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NaOH + atmospheric CO2: Unexpected results troubleshooting
I've conducted an experiment where a solution of $\pu{4.2 M}$ $\ce{NaOH}$ was left exposed to atmospheric $\ce{CO2}$ for a period of time and titrated using potentiometric titration and potassium ...
0
votes
2
answers
84
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Writing a chemical equation with a reactant of unspecified chain length [closed]
Would appreciate it if I got some general tips and hints and not the solution cause I'm trying to learn from this question :)
The question is as follows:
FA1 is a solution containing $\pu{5.00 g dm-3}...
0
votes
2
answers
151
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Is this stoichiometric question answerable? (unknown variable)
This question kind of borders on the fields of both semantics and math, as well as chemistry, but I still believe this is the right place to ask.
On a test I took, a question was posed as such:
Via ...
0
votes
1
answer
1k
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Why is finding [H+] when given [OH-]= 3.57e-5 M wrong when I use -log[OH-]?
Question/Problem
I know that [H+][OH-]=$10^{-14}$
And the problem asks to find out the concentration of H+ when the concentration of OH- is $3.57*10^{-5}$M
By using the equation above i find that ...
5
votes
1
answer
1k
views
Back titration of excess HCl used for dissolving limestone
$15.0~\mathrm{mL}$ of $1.4~\mathrm{M}\ \ce{HCl}$ was mixed with $1.00~\mathrm{g}$ of limestone (impure $\ce{CaCO3}$) until all the solid had dissolved. The solution was then transferred to a conical ...
3
votes
1
answer
11k
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How to find the concentration of an of ammonia solution by titration with nitric acid?
I am stuck with this question I am doing:
To determine the ammonia ($\ce{NH3}$) concentration in a cleaning product, a chemist diluted 20 mL of the product to 500 mL. Of the 500 mL solution, 25 mL ...
8
votes
1
answer
805
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Why do these two calculations give me different answers for the same acid-base titration?
In class, I performed an weak acid-strong base titration using commercial white vinegar and sodium hydroxide with the aim of finding the concentration of of ethanoic acid in the vinegar, and I used ...