- #1
zenterix
- 774
- 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.
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.