1 / 22

Friday, May 6 th : “A” Day Agenda

Friday, May 6 th : “A” Day Agenda. Homework questions/problems/collect Quiz over section 15.2: “Acidity, Basicity, and pH” Section 15.3: “Neutralizations and Titrations” Neutralization reaction, equivalence point, titration, titrant, standard solution, transition range, end point Homework:

lexiss
Télécharger la présentation

Friday, May 6 th : “A” Day Agenda

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Friday, May 6th: “A” DayAgenda • Homework questions/problems/collect • Quiz over section 15.2: “Acidity, Basicity, and pH” • Section 15.3: “Neutralizations and Titrations” • Neutralization reaction, equivalence point, titration, titrant, standard solution, transition range, end point • Homework: • Sec. 15.3 review, pg. 556: #1-10 • Concept Review: “Neutralizations and Titrations” • Lab Write-Up: “Titration of an Acid and a Base”

  2. Section 15.2 Quiz “Acidity, Basicity, and pH” • You may use your guided notes, your book, and a partner to complete the quiz.

  3. Neutralization • Neutralization reaction: the reaction of hydronium ions and hydroxide ions to form water molecules and a salt. • When solutions of a strong acid and a strong base, having exactly equal amounts of H3O+(aq) and OH−(aq) ions, are mixed, almost all of the hydronium and hydroxide ions react to form water. H3O+(aq) + OH−(aq)  2 H2O(l) *correct*

  4. Neutralization • Supposethat hydrochloric acid, HCl, and sodium hydroxide, NaOH are mixed. • The result will be a solution of only water and the spectator ions sodium and chlorine. • This is just a solution of sodium chloride. HCl + NaOH  NaCl + H2O • This representation can be misleading because the only reactants are H3O+(aq) and OH−(aq) ions and the only product is H2O.

  5. Titrations • If an acidic solution is added gradually to a basic solution, at some point the neutralization reaction ends because the hydroxide ions are used up. • Likewise, if a basic solution is added to an acid, eventually all of the hydronium ions will be used up. • The point at which a neutralization reaction is complete is known as the equivalence point. • Equivalence point: the point at which the two solutions used in a titration are present in chemically equivalent amounts.

  6. Titrations • When a solution of a strong base is added to a solution of a strong acid, the equivalence point occurs when the amount of added hydroxide ions equals the amount of hydronium ions originally present. • Titration: the gradual addition of one solution to another to reach an equivalence point. • The purpose of a titration is to determine the concentration of an acid or a base.

  7. Titration • In addition to the two solutions, the equipment needed to carry out a titration usually includes two burets, a titration flask, and a suitable indicator. • One buret is for the acid solution, the other is for the basic (alkaline) solution. • Titrant: a solution of known concentration that is used to titrate a solution of unknown concentration.

  8. Titration • To find the concentration of the solution being titrated, you must already know the concentration of the titrant. • Standard solution: a solution of known concentration. • The concentration of a standard solution has usually been determined by reacting the solution with a precisely weighed mass of a solid acid or base.

  9. Titration • A distinctively shaped graph, called a titration curve, results when pH is plotted against titrant volume. • Because the curve is steep at the equivalence point, it is easy to locate the exact volume that corresponds to a pH of 7.00. • A titration is exact only if the equivalence point can be accurately detected.

  10. Equivalence Point • This graph of pH versus the volume of 1.000 M NaOH added to an HCl solution indicates that the equivalence point occurred after 38.6 mL of titrant was added.

  11. Indicators • Transition range: the pH range through which an indicator changes color. • End point: the point in a titration at which a marked color change takes place. • If an appropriate indicator is chosen, the end point and the equivalence point will be the same.

  12. Selecting an Indicator • In titrations of a strong acid by a strong base, the equivalence point occurs at pH 7. • When a weak acid is titrated by a strong base, the equivalence point is at a pH greater than 7. • The titration of a weak base and a strong acid, the equivalence point is at a pH less than 7.

  13. How to Perform a Titration

  14. How to Perform a Titration

  15. Titration Calculations • At the equivalence point in a titration of a strong acid by a strong base, the amount of hydroxide ion added equals the initial amount of hydronium ion. • C: concentration (in moles per liter) • V: volume (in liters) of the solution

  16. Titration Calculations • An easier way to think of this: (C Acid)(V Acid) = (C Base) (V Base) • C: concentration (in moles per liter) • V: volume (in liters) of the solution

  17. Sample Problem D, pg. 555Calculating Concentration from Titration Data A student titrates 40.00 mL of an HCl solution of unknown concentration with a 0.5500 M NaOH solution. The volume of base solution needed to reach the equivalence point is 24.64 mL. What is the concentration of the HCl solution in moles/liter? (CAcid) (VAcid) = (CBase) (VBase)

  18. Sample Problem D, continued… • NaOH is a strong base so: NaOH Na + + OH- 0.5500 M 0.5500 M • C(acid) = ? • V(acid) = 40.00 mL = .04L • C(base) = 0.5500 M • V(base) = 24.64 mL = .02464 L (CAcid) (VAcid) = (CBase) (VBase) • C (Acid) (.04 L) =(0.5500 M) (.02464 L) C(acid) = 0.3388 mol/L

  19. Additional Practice If 72.1 mL of 0.543 M H2SO4 completely titrates 39.0 mL of KOH solution, what is the molarity of the KOH solution? • H2SO4 is a strong acid so: H2SO4 + 2 H2O SO42- + 2 H3O + 0.543 M 1.086 M • Because of the 1:2 ratio, 1 mole of H2SO4 makes 2 moles of H3O +. • [H3O+] = 2 [H2SO4] = 2 (0.543 M) = 1.086 M

  20. Additional Practice, continued… (CAcid) (VAcid) = (CBase) (VBase) C(acid) = 1.086 M V(acid) = 72.1 mL = .0721 L C(base) = ? V(base) = 39.0 mL = .0390 L (1.086 M) (.0721 L) = (C Base) (.0390L) C(base) = 2.01 M

  21. Homework • Section 15.3 review, pg. 556: #1-10 • Concept Review: “Neutralizations and Titrations” • Lab Write-Up: “Titration of an Acid and a Base” Looking Ahead: U of I 3D-Printing Presentation on Tuesday! Titration lab on Thursday No Flip-Flops!

More Related