ch7 acid base rev - Cal State LA

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Chapters 7 and 11

Acid-Base Titrations

Titration:Increments of reagent solution-the titrant-are added to analyte until their reaction is complete.

Titration type/endpoint determination:

•Volumetric: volume of standard is measured•Gravimetric: mass of standard is measured, e.g., Mohr method (Cl-Ag-CrO4). •Coulometric: time at constant current is measured, i.e., amount of charge.•Spectrophotometric: Beer’s Law, the absorbance

Standard Solution (Standard Titrant):Reagent of known concentration that is used to carry out the analysis.

Titration: Performed slowly by adding standard solution from a buret until the reaction is judged to be complete.

Terms

Indicators:Added to a tritration to give an observable physical change at or near the equivalence point. Usually a color or turbidity change.

Equivalence Point: The point when the amount of added titrant is chemically equivalent to the amount of analyte in the sample.

End Point: The point at which an indicator physically changes representing the estimated equivalence point (color or turbidity change).

Titration error = Vep - Veq

Terms

Back-titration:Add a known excess of one standard titrant to the analyte, and then titrate a second standard titrant to determine the excess reagent. May be required when a reaction is slow or the standard solution lacks stability.

Blank titration: Carry out the same procedure without analyte. To estimate the titration error.

Terms

Desirable Properties of Standard Solutions

1. Sufficiently stable (only need to determine concentration once).

2. React rapidly with the analyte (less time between additions).

3. React more or less completely with the analyte(satisfactory end points).

4. Undergo a selective reaction with analyte described by a simple chemical equation

Standard Solutions (Titrants)

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Methods for Establishing the Concentration of Standard Solutions

1. Direct Method: A carefully weighed quantity of primary standard is diluted to a known volume. First choice if possible.

2. Standardization: A titrant to be standardized is used to titrate a weighed quantity of a primary, secondary standard or a known volume of another standard solution.

Primary Standard: Highly purified compound that serves as a reference material.• High purity, stability (air, heat, vacuum), • Standard solution (direct method)• To standardize a standard solution (standardization)

Secondary Standard:A reference material (a compound) whose purity has been established by a chemical analysis. Often used in lieu of a primary standard because there are not many primary standards.


Summarytitrationtitrantanalyteindicatorequivalence point vs. end pointtitration errorblank titrationStandard: Primary standard, second standard, standard solution

Acid-Base Titrations

How does the pH change as titrant is added?

Titration Curve: A graph showing how the concentration of one of the reactants varies as titrant is added. (how the pH changes as titrant is added in acid and base titration)

Titration of Strong Base with Strong Acid

Titration of Strong Base with Strong Acid

1. Before the equivalence point: excess OH-

2. At the equivalence point: pH = 7.003. After the equivalence point: excess H+

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0.00, initial point10.00 ml, 50% titration19.98 ml, half-drop before equivalence point20.00 ml, 100% titration (equivalence point) 20.02 ml, half-drop after equivalence point40.00 ml, 200% titration (far-over)

EXAMPLE: Derive the titration curve for the titration of 20.00 mLof 0.1000 M HCl with 0.00, 10.00, 19.98, 20.00, 20.02 and 40.00 mLof 0.1000 M NaOH.

Titration of Strong Acid with Strong Base


At 0.00 mL of NaOH added, initial point

[H3O+] = CHCl = 0.1000 M

pH = 1.00

at 10.00 mL of NaOH added, Va * Ma > Vb * Mb

(Va Ma) - (Vb Mb)[H3O+] = ----------------------------

(Va + Vb)

((20.00mL)*(0.1000M)) - ((10.00mL)*(0.1000M))= ------------------------------------------------------------

(20.00 + 10.00)mL

= 3.33x10-2M pH =1.48


at 19.98 mL of NaOH added, Va * Ma > Vb * Mb

(Va Ma) - (Vb Mb)[H3O+] = ----------------------------

(Va + Vb)

((20.00mL)*(0.1000M)) - ((19.98mL)*(0.1000M))= ------------------------------------------------------------

(20.00 + 19.98)mL

=5.0x10-5M pH =4.30


at 20.00 mL of NaOH addedVa * Ma = Vb * Mb , equivalence pointat equivalence point of a strong acid - strong base titrationpH = 7.00


at 20.02 mL of NaOH added, Vb * Mb > Va * Ma , post equivalence point

(Vb * Mb) - (Va * Ma)[OH-] = ---------------------------(Va + Vb)

((20.02mL)*(0.1000M)) - ((20.00mL)*(0.1000M))= ------------------------------------------------------------

(20.02.00 + 20.00)mL=5.0x10-5M pOH =4.30pH=9.70


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at 40.00 mL of NaOH added, Vb * Mb > Va * Ma , post equivalence point

(Vb * Mb) - (Va * Ma)[OH-] = ---------------------------(Va + Vb)

((40.00mL)*(0.1000M)) - ((20.00mL)*(0.1000M))= ------------------------------------------------------------

(40.00 + 20.00)mL=3.33x10-2M, pH =1.48pH=12.52


Strong Acid - Strong Base Titration

0

2

4

6

8

10

12

14

0 10 20 30 40 50 60

Volume of Base Added

pH

1

0.1

0.01

0.001

0 15 25 35 45 55 65

(mL)

The Effect of Concentration• The change in pH at the equivalence point is less pronounced at decreased concentrations.• The pH at the equivalence point is still 7.

(M)

Titration of Strong Acid with Strong Base

Titration of Weak Acid with Strong Base



•The pH change at the equivalence point decreases (the end point become less distinct) as the acid becomes weaker or more dilute

•The Effect of Concentration: the pH at the equivalence point is lower for lower initial weak acid concentrations and initial pH is higher.


EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HAc with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH. Ka = 1.75 x 10-5 M

0.00, initial point15.00 ml, before equivalence point35.00 ml, 100% titration (equivalence point) 50.00 ml, after equivalence point

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at 0.00 mL of NaOH added, initial point

[H+] = Ka Ca

= (1.75 x 10-5 M) x (0.1000 M) = 1.32 x 10-3 MpH = 2.878

at 15.00 mL of NaOH added, Va Ca > Vb Cb

(Va Ca) - (Vb Cb)[HAc]ex. = ------------------------------(Va + Vb)

[(35.00 mL)x(0.1000 M)) - ((15.00 mL)x(0.1000M)]= -----------------------------------------------------------------

(35.00 + 15.00) mL= 4.000x10-2 M



[HAc]ex = 4.000x10-2 M

(Vb Cb)[Ac-] = ----------------(Va + Vb)

(15.00 mL)(0.1000 M)= ---------------------------- = 3.000x10-2 M

(35.00 + 15.00) mL



[HAc]ex = 4.000 x 10-2 M [Ac-] = 3.000 x 10-2 M

[HAc]ex[H+] = Ka = ------------------

[ Ac-]4.000 x10-2 M

= 1.75 x 10-5 M ------------------- = 2.33 x 10-5 M3.000 x10-2 M



[HAc]ex = 4.000 x 10-2 M

[Ac-] = 3.000 x 10-2 M

[H+] = 2.33 x 10-5 M pH = 4.632


at 35.00 mL of NaOH added, Va Ca = Vb Cb , equivalence point

EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HAc with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH. Ka = 1.75 x 10-5M

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Kw VaCa[OH-] = = -----------------Ka (Va + Vb)

Kb Cb


1.00 x10-14 M2 * (35.00mL)(0.1000M)[OH-] = -----------------------------------------------

1.75 x10-5 M * (35.00 + 35.00)mL

= 5.34x10-6 M

at 35.00 mL of NaOH added, Va * Ca = Vb * Cb , equivalence point



[OH-] = 5.34 x 10-6 M

pOH = 5.272 pH = 8.728


at 50.00 mL of NaOH added, Vb Cb > Va Ca , post equivalence point


at 50.00 mL of NaOH added, Vb Cb > Va Ca , post equivalence point

(Vb Cb) - (Va Ca)[OH-] = --------------------------(Va + Vb)


at 50.00 mL of NaOH added, Vb Cb > Va Ca

(Vb Cb) - (Va Ca)[OH-] = --------------------------(Va + Vb)

[(50.00mL)*(0.1000M)) - ((35.00mL)*(0.1000M)]= -------------------------------------------------------------

(35.00 + 50.00) mL= 1.765 x 10-2 M


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at 50.00 mL of NaOH added, Vb Cb > Va Ca

[OH-] = 1.765 x 10-2MpOH = 1.7533pH = 14.00 - 1.7533 = 12.25


(a) Which of the two curves represents the titration of a strong acid, and which a weak acid?(b) What is the approximate pH at the equivalence point for each of the acids?(c) What is the approximate pKa of the weak acid?

The following plot shows two titration curves, each representingthe titration of 50.0 mL of 0.100 M acid with 0.100 M NaOH:

Vb (mL)

Acid - Strong Base Titration Curve

Much the same shape, except the titration would start at high pH and decrease as acid is added

Weak Base - Strong Acid Titration

0

2

4

6

8

10

12

0 10 20 30 40 50 60

Volume of Acid Added

pH

Titration of Weak Base with Strong Acid

(mL)

EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M H2A with 0.00, 15.00, 35.00, 50.00, 70.00, and 90.00 mL of 0.1000 M NaOH. Ka1 = 7.11 x 10-3 M; Ka2 = 6.34 x 10-8 M

Diprotic Systems: Titration of Weak Acid Base with Strong Base

0.00, Initial point15.00 ml, Before the 1st equivalence point35.00 ml, At the 1st equivalence point50.00 ml, After the 1st equivalence point70.00 ml, At the 2nd equivalence point90.00 ml, After the 2nd equivalence point


at 0.00 mL of NaOH added, initial point

pH = 1.632

][1033.2

1011.71000.0][

]][[1

2

3

2

2

+−

−−+

=×=

×=−

==

Hx

xx

AHHAH

Ka

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at 15.00 mL of NaOH added, Va Ca > Vb Cb, the 1st buffer region

(Va Ca) - (Vb Cb)[H2A]ex = ----------------------------

(Va + Vb)


at 15.00 mL of NaOH added, Va · Ca > Vb · Cb, the 1st buffer region

[(35.00 mL)·(0.1000 M) - (15.00 mL)·(0.1000 M)][H2A]ex = ----------------------------------------------------------------

(35.00 + 15.00) mL= 4.000x10-2M



[H2A]ex = 4.000x10-2 M

(Vb · Cb) (15.00 mL)(0.1000 M)[HA-] = ------------ = -------------------------- = 3.000x10-2 M

(Va + Vb) (35.00 + 15.00)mL



[H2A]ex = 4.000 x 10-2 M; [HA-] = 3.000 x 10-2 M[H2A]ex 4.000 x 10-2 M

[H+] = Ka1 x -------------- = 7.11x10-3 M x ------------------[HA-] 3.000 x 10-2 M

= 9.48 x 10-3 MpH = 2.023


at 35.00 mL of NaOH added, Va · Ca = Vb · Cb, the 1st equivalence point

[H+] = Ka1 · Ka2

= (7.11 x 10-3 M)(6.34 x 10-8 M) = 2.12 x 10-5 MpH = 4.673


at 50.00 mL of NaOH added, 2(Va · Ca) > Vb · Cb, the 2nd buffer region

2 (Va · Ca) - (Vb · Cb)

[HA-]ex = -------------------------------(Va + Vb)

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at 50.00 mL of NaOH added, 2(Va · Ca) > Vb · Cb, the 2nd buffer region

[2(35.00mL)*(0.1000M)) - ((50.00mL)*(0.1000M)][HA-]ex = ---------------------------------------------------------------

(35.00 + 50.00) mL

= 2.353x10-2M


at 50.00 mL of NaOH added, 2(Va · Ca) > Vb · Cb, the 2nd buffer region[HA-]ex = 2.353x10-2M

(Vb · Cb) - (Va · Ma)[A2-] = ---------------------------

(Va + Vb)


at 50.00 mL of NaOH added, 2(Va · Ca) > Vb · Cb, the 2nd buffer region[HA-]ex = 2.353x10-2M

(50.00 mL)(0.1000 M) - (35.00 mL)(0.1000 M)[A2-] = ---------------------------------------------------------

(35.00 + 50.00) mL

= 1.764x10-2 M


at 50.00 mL of NaOH added, 2(Va · Ca) > Vb · Cb, the 2nd buffer region[HA-]ex = 2.353x10-2 M[A2-] = 1.764x10-2 M

Ka2 [HA-]ex 6.34 x 10-8 M x 2.353x10-2 M[H+] = ------------------ = ------------------------------------

[A2-] 1.764x10-2 M= 8.46x 10-8 M

pH = 7.073


at 70.00 mL of NaOH added, 2(Va · Ca) = Vb · Cb, the 2nd equivalence point

Kw · Va · Ca[OH-] = ---------------------

Ka2 (Va + Vb)


at 70.00 mL of NaOH added, 2(Va · Ca) = Vb · Cb, the 2nd equivalence point

1.00 x 10-14 M2 x (35.00 mL)x(0.1000 M)[OH-] = ---------------------------------------------------- = 7.25x10-5 M

6.34 x 10-8 M * (35.00 + 70.00) mL

pOH = 4.139 pH = 9.860

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at 90.00 mL of NaOH added, Vb · Cb > 2(Va · Ca), after 2nd equivalence point

Vb · Cb - 2(Va · Ca) [OH-] = -----------------------------------

(Va + Vb)


at 90.00 mL of NaOH added, Vb · Cb > 2(Va · Ca), after 2nd equivalence point

(90.00 mL)(0.1000 M) - 2 x (35.00 mL)(0.1000 M)[OH-] = ---------------------------------------------------------------

(35.00 + 90.00) mL

= 1.600 x 10-2 MpOH = 1.795 pH = 12.204

Diprotic Weak Acid - Strong Base Titration

0

2

4

6

8

10

12

14

0 10 20 30 40 50 60 70 80 90 100

Volume of Base Added

pH

Diprotic Systems: Titration of Weak Acid Base with Strong Base

(mL)

Diprotic Systems: Titration of Weak Base with Strong Acid

Standard Solutions for Acid/Base Titrations

Standard acid or base reagents are always strong acids or bases • HCl, HClO4, H2SO4, NaOH, and KOH• They yield sharper end points. • Weak reagents react incompletely. • Nitric acid and hot sulfuric and perchloric yield undesirable

ox/red side reactions.

Primary standards for acid standardization (see Table 11-5)• Na2CO3

• Tris (THAM) – tris(hydroxymethyl)aminomethane, (HOCH2)3CNH2

• Sodium tetraborate decahydrate, Na2B4O7.10H2O

Endpoint Determination

1. Indicators 2. Derivatives of titration curve3. Gran Plot

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Acid/Base Indicators• Weak acids or bases themselves• The conjugate forms differs in color

HIn In- + H+

acid baseColor1 Color2

pH = pKa HIn ± 1

Using Indicators to Determine the End Point

In + H2O ⇔ InH+ + OH-

base acidColor1 Color2

pOH = pKbIn ± 1

Indicators

Indicator Behavior

[In-][H+]Ka = --------------- (1)

[HIn]

HIn In- + H+

acid baseColor1 Color2

Indicators

Indicator BehaviorIn + H2O ⇔ InH+ + OH-

base acidColor1 Color2

[InH+][OH-]Kb = --------------- (2)

[In]

Indicators

Indicator Behaviorrearranging Equation (1) gives

[In-] Ka------ = ----------[HIn] [H+]

IndicatorsIndicator Behavioracid color shows when[In-] 1 [H+][In-] 1 ------ <= ---- ------------- = ---*[H+] = Ka[HIn] 10 [HIn] 10

base color shows when[In-] 10 [H+][In-] ------ >= ----- ------------- = 10*[H+] = Ka[HIn] 1 [HIn]

Indicators

Indicator Behavior

acid color shows when pH + 1 = pKa

and base color shows when pH - 1 = pKa

Color change range is pH = pKa + 1(Transition range)

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Choosing the Proper Indicator

• Color change range should be in area where titration curve is most vertical.

• Indicator error: the difference between the observed end point (color change) and the true equivalence point.

Common Indicators

See also p.215 Table 11-4

Using Derivatives of Titration Curveto Determine the End Point

Titration curve for a polyprotic acid - strong base titration.

First derivative of a weak acid- strong base titration

Second derivative of a weak acid - strong base titration

Using Derivatives of Titration Curve to Determine the End Point

Using Derivatives of Titration Curve to Determine the End Point

Using the Gran Plot to Determine the End Point

)(10

acid weak a of titrationThe

bea

A

HApHb VVKV −=⋅

−

−

γγ

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Using the Gran Plot to Determine the End Point Using the Gran Plot to Determine the End Point

)(1

10

base weak a of titrationThe

)(10

acid weak a of titrationThe

ae

BH

B

a

pHa

be

A

HAa

pHb

VVK

V

VVKV

−=⋅

−=⋅

+

−

+

−

γγ

γγ

ch7 acid base rev - Cal State LA

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