{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Lab 2 - Ryan O'Hearn Chemical Principles II Lab 1011.206.43...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Ryan O’Hearn Chemical Principles II Lab - 1011.206.43 Acid-Base Titrations 22:43:53 Purpose Learn how to do acid-base titrations in order to find the concentration of acid in an unknown liquid. In this process learn how to properly do an acid-base titration with an indicator. Procedure See “Chemistry 206 Lab: Chemical Principles II Laboratory,” Lab #2: “Acid-Base Titrations.” With the exception of Part IV, because due to a lack of enough unknown solution, we were only able to complete four trials instead of five. Data Tables, Graphs Part III: Standardization of NaOH Titrant Trial Mass KHP Moles KHP mL NaOH Final mL NaOH Initial Ml NaOH Used M NaOH A 0.640 .0090 14.25 0.0 14.25 0.632 B 0.602 .0084 27.6 14.25 13.35 0.629 C 0.613 .0086 41.4 27.6 13.80 0.623 D 0.624 .0088 16.8 3.0 13.80 0.638 E 0.611 .0086 30.35 16.8 13.55 0.635 Average M NaOH = 0.6314 Part IV: Determination of Unknown Acid Molarity Unknown = 28 Trial mL NaOH Final mL NaOH Initial mL NaOH Used M Unknown Acid A 41.5 30.35 11.15 0.704 B 1.15 0.0 11.15 0.704 C 22.1 11.15 10.95 0.691 D 33.4 22.1 11.3 0.713 Average M Unknown = 0.703 Sample Calculations Average Molarity = ∑ Molarities ÷ Total # of Trials Average M NaOH = (0.632+0.629+0.623+0.638+0.635) ÷ 5 = 0.6314 1
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Moles KHP = Grams KHP ÷ Molar Mass KHP Moles KHP = 0.640g KHP ÷ 71.078 = .009004 moles Molarity of NaOH Solution = M 1 x V 1 = M 2 x V 2 M 1 x V 1 = M 2 x V 2 => =
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern