Asked by holly
your lab possesses a stock solution of NaOH whose concentration is approx. 0.1M. you prepare a secondary solution by diluting 15.00mL of this stock solution to a total volume of 250.0 mL. you titrate a 50.00mL aliquot of the secondary solution with a hydrochloric acid solution that is 0.0252M. the initial burette reading is 12.38mL and the final burrette reading (at the equivalence point) is 25.01mL. Express the derived concentration of the stock solution.
Answers
Answered by
DrBob222
mL used = 25.01-12.38 = 12.63 mL of 0.0252M HCl.
mols HCl = 0.0252 x 0.01238 = about 0.00031 (but you need to be more careful with these numbers)
You titrated 50 mL of the diluted NaOH; therefore, the M of the titrated solution is
0.00031/0.050L = about 0.0062M
That solution came from the stock solution that was diluted 15 to 250 mL; therefore, 0.00624 x 250/15 = about 0.104M for the stock solution.
mols HCl = 0.0252 x 0.01238 = about 0.00031 (but you need to be more careful with these numbers)
You titrated 50 mL of the diluted NaOH; therefore, the M of the titrated solution is
0.00031/0.050L = about 0.0062M
That solution came from the stock solution that was diluted 15 to 250 mL; therefore, 0.00624 x 250/15 = about 0.104M for the stock solution.
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