Chemistry A seemingly simple titration problem

AI Thread Summary
The discussion revolves around a titration problem where 15.4 mL of 1.0M HCl is used to neutralize NaOH. The calculated concentration of NaOH in the original 50 mL sample is determined to be 0.308M, which conflicts with the MIT OCW answer of 0.62M. The participant identifies a potential typo in their calculations, suggesting that 15.4 mL corresponds to 0.0154 moles instead of 0.00154 moles, aligning with their concentration result. They conclude that the discrepancy may be due to an error in the MIT answer or an alteration in the question without updating the answer. The participant confirms they got the correct answers for the other questions.
zenterix
Messages
774
Reaction score
84
Homework Statement
In your lab, you find an old bottle labeled “sodium hydroxide” with no concentration on it. To determine what the concentration is, 50.0 mL of the solution was diluted to 100. mL and titrated to the equivalence point with 15.4 mL of 1.0 M HCl(aq). What is the molarity of the sodium hydroxide solution in the bottle?
Relevant Equations
This seems to be a simple problem but my answer does not match the solution (on MIT OCW).
I tried to solve the problem (problem 1 on this problem set) by noting that titrating to the equivalence point means neutralizing all of the NaOH with the titrant HCl.

Since 15.4mL of HCl at 1.0M was used, then 0.00154 mol of HCl was used to neutralize the same amount of NaOH.

Therefore, in the original sample of 50mL which came from the bottle, the concentration was 0.00154mol/50mL which is 0.308M.

The available answer from MIT OCW is 0.62M.

Now, a few observations

- I know that I am not doing a good job with significant figures. I imagine my answer should be 0.31M, which means the answer from MIT OCW is double mine.

- Not sure why though. After all, whatever NaOH was neutralized in 100mL was present in the original 50mL.
 
Physics news on Phys.org
Your solution looks OK, must be mistake on the site.
 
  • Like
Likes laser1 and zenterix
If 1000ml contains 1 mole
then 100ml contains 0.1 mole
and 10ml contains 0.01 moles.
So 15.4ml contains 0.0154 moles, not 0.00154 moles.
Which is a typo on your part, of course as 0.0154 moles matches your answer of 0.308M. Which is correct answer.

Did you get the correct answers for the other questions?
 
DrJohn said:
If 1000ml contains 1 mole
then 100ml contains 0.1 mole
and 10ml contains 0.01 moles.
So 15.4ml contains 0.0154 moles, not 0.00154 moles.
Which is a typo on your part, of course as 0.0154 moles matches your answer of 0.308M. Which is correct answer.

Did you get the correct answers for the other questions?
Yes, I did.
 
So their answer of 0.62M is one of two things.
a) Wrong
or
b) They altered the question to make it slightly different, and forgot to alter the answer.
Settle for you got it right!
 
Thread 'Confusion regarding a chemical kinetics problem'
TL;DR Summary: cannot find out error in solution proposed. [![question with rate laws][1]][1] Now the rate law for the reaction (i.e reaction rate) can be written as: $$ R= k[N_2O_5] $$ my main question is, WHAT is this reaction equal to? what I mean here is, whether $$k[N_2O_5]= -d[N_2O_5]/dt$$ or is it $$k[N_2O_5]= -1/2 \frac{d}{dt} [N_2O_5] $$ ? The latter seems to be more apt, as the reaction rate must be -1/2 (disappearance rate of N2O5), which adheres to the stoichiometry of the...
I don't get how to argue it. i can prove: evolution is the ability to adapt, whether it's progression or regression from some point of view, so if evolution is not constant then animal generations couldn`t stay alive for a big amount of time because when climate is changing this generations die. but they dont. so evolution is constant. but its not an argument, right? how to fing arguments when i only prove it.. analytically, i guess it called that (this is indirectly related to biology, im...
Back
Top