A student prepares a solution of hydrochloric acid that is approximately 0.1 M and wanted to determine its precise concentration. A 25.00 mL portion of the HCl solution is transferred to a flask, and after a few drops of indicator are added, the HCl solution is titrated with 0.0619 M NaOH solution. The titration requires exactly 38.90 mL of the standard NaOH solution to reach end point. What is the molarity of the HCl solution?
C(HCl) =
2 answers
same ol' same ol'. Look at your aspirin, acetic acid, etc titrations. I shall be happy to address specific questions you have about these but the method of working them is the same.
M1V1=M2V2
M*25= 0.0619*38.9 = MHCl= 2.408/25
MHCl= 0.0963 M
M*25= 0.0619*38.9 = MHCl= 2.408/25
MHCl= 0.0963 M