Friday may 6 th a day agenda
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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:

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Friday, May 6 th : “A” Day Agenda

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Friday may 6 th a day agenda

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”


Section 15 2 quiz acidity basicity and ph

Section 15.2 Quiz “Acidity, Basicity, and pH”

  • You may use your guided notes, your book, and a partner to complete the quiz.


Neutralization

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*


Neutralization1

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.


Titrations

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.


Titrations1

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.


Titration

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.


Titration1

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.


Titration2

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.


Equivalence point

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.


Indicators

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.


Selecting an indicator

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.


How to perform a titration

How to Perform a Titration


How to perform a titration1

How to Perform a Titration


Titration calculations

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


Titration calculations1

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


Sample problem d pg 555 calculating concentration from titration data

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)


Sample problem d continued

Sample Problem D, continued…

  • NaOH is a strong base so:

    NaOH Na + + OH-

    0.5500 M0.5500 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


Additional practice

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


Additional practice continued

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


Homework

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!


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