03b_Buffer Ws Answers and Titration Notes

8/12/2019 03b_Buffer Ws Answers and Titration Notes

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Entry Task: Feb 12thTuesday

Define Buffer capacity

You have 5 minutes


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Agenda

Discuss Buffer ws 1

In-class notes little more buffer info and

practice on Titrations

HW: Buffers ws #2


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Explain why a mixture of HCl and KCl does not

function as a buffer, whereas a mixture of

HC2H3O2and NaC2H3O2does?

HCl and KCl are conjugate-pairs, problem is that

potassium is an alkali metal and will stay

dissociated and add more + to the system

making it more acidic.


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2. What factors determine a) the pH, and b) the

buffer capacity of a buffer solution?

A) The pH of a buffer is determined by Ka forthe conjugate acid present and the ratioofthe conjugate base concentration to the

conjugate acid concentration.B) The buffering capacity of buffer is

determined by the concentrations of theconjugate acid and conjugate base present.Higher the concentration the higher thecapacity.


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3. In a solution, when the concentrations of a weak

acid and its conjugate base are equal,

A) the system is not at equilibrium.

B) the buffering capacity is significantly decreased.

C) the -log of the [H+] and the -log of the Ka are equal.

D) all of the above are true.


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4. Of the following solutions, which has the greatest

buffering capacity?

A) 0.821 M HF and 0.217 M NaF

B) 0.821 M HF and 0.909 M NaF

C) 0.100 M HF and 0.217 M NaF

D) 0.121 M HF and 0.667 M NaF

E) They are all buffer solutions and would all have the

same capacity.


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5. The addition of hydrofluoric acid and __________

to water produces a buffer solution.

A) HCl

B) NaNO3

C) NaCl

D) NaOH

E) NaBr


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6. Which of the following could be added to a solution

of sodium acetate to produce a buffer?

A) acetic acid only

B) acetic acid or hydrochloric acid

C) hydrochloric acid only

D) potassium acetate only

E) sodium chloride or potassium acetate


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7. A solution is prepared by dissolving 0.23 mol ofhydrazoic acid and 0.27 mol of sodium azide in watersufficient to yield 1.00 L of solution. The addition of 0.05mol of NaOH to this buffer solution causes the pH toincrease slightly. The pH does not increase drasticallybecause the NaOH reacts with the __________ present inthe buffer solution. The Ka of hydrazoic acid is 1.9 10

-5.

A) H2OB) H3O+

C) azide

D) hydrazoic acidE) This is a buffer solution: the pH does not change uponaddition of acid or base.


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8. What is the pH of a buffer solution that is 0.211 M

in lactic acid and 0.111 M in sodium lactate? The Ka of

lactic acid is 1.4 10-4.

Ka = [x][0.111]

[0.211]

1.4 x 10-4= [x][0.111]

[0.211]

Rearrange to

get X by itselfx= (1.4 x 10-4)(0.211)

0.111

x = [H+]= 2.66 x 10-4

pH = -log(2.66 x 10-4) = 3.57


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Buffer Problems

Hurdles-

1st -Which species in problem is acid or base?

2ndKb value for OH-or Ka for H+

3rdSetting up Ka or Kb expression correctly (1st)

4thIs there any changes in concentrations from given:

*GramsMolesMolarity

*Mixing two different volumes (M1V1= M2V2)

*Addition of a new species (ICE table)


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Different Volumes

Treat as dilution problems


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1a. Calculate the pH of a buffer that is 0.100 M in NaHCO3

and 0.125M in Na2CO3. b. Calculate the pH of a solution

formed by mixing 55mls of 0.20M NaHCO3with 65mls of 0.15

M Na2CO3.

Ka = [H+][CO3-2

][HCO3

-]5.6 x 10

-11

= [x][0.125][0.100]

Who is the conjugate in this reaction?

x= (5.6 x 10-11)(0.100)

0.125 x = [H+]= 4.48 x 10

-11

pH = -log(4.48 x 10-11) = 10.35

HCO3-so we use H2CO3Ka value= 5.6x10

-11

Provide the Ka expression Provide the Ka express with #

Rearrange to get X by itself


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formed by mixing 55mls of 0.20M NaHCO3with 65mls of 0.15

M Na2CO3.Ka = [H+][CO3

-2]

[HCO3-]

5.6 x 10-11= [x][0.125]

[0.100]

OR use the H-H equationpH = pK

a+ log

[base]

[acid]

pH = 10.25 + log [0.125][0.100] pH = 10.25 + 0.0969

10.35 = 10.25 + 0.0969


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formed by mixing 55mls of 0.20M NaHCO3with 65mls of

0.15 M Na2CO3. NOTICE we have different volumes!!

What will be the TOTAL VOLUME? 120 mls

We have diluted our mixture so we set up a dilution

problem- MAKE SURE volume units are the same!!M1V1= M2V2

(0.20M)(0.055L) = (x) (0.120L)=

(0.15M)(0.065L) = (x)(0.120L)=

9.17x10-2M of NaHCO3(new M)

8.125x10-2 Mof Na2CO3 (new M)

x= (5.6 x 10-11)(9.17x10-2)

8.125x10-2 x = [H+]= 6.32 x 10-11

pH = -log(6.32 x 10-11) = 10.20


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formed by mixing 55mls of 0.20M NaHCO3with 65mls of

0.15 M Na2CO3.

5.6 x 10-11= [x][8.125x10-2][9.17x10-2]

NOTICE we have different volumes!!


H-H equation

pH = 10.25 + log[8.125 x10-2]

[9.17x10-2

10.20= 10.25 + -0.0525


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2a. Calculate the pH of a buffer that is 0.12 M in lactic acid

and 0.11M in sodium lactate. b. Calculate the pH of a

solution formed by mixing 85mls of 0.13M lactic acid with

95mls of 0.15M sodium lactate. Ka= 1.4 10-4

1.4 x 10-4= [x][0.11]

[0.12]

pH = pKa+ log [base]

[acid]

pH = 3.85 + log [0.11][0.12]

3.8 = 3.85 + -0.0378

l l h f b ff h l d d


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2a Calculate the pH of a buffer that is 0.12 M in lactic acid and

0.11M in sodium lactate. B. Calculate the pH of a solution

formed by mixing 85mls of 0.13M lactic acid with 95mls of

0.15M sodium lactate.


We have diluted our mixture so we set up a dilution

problem- MAKE SURE volume units are the same!!(0.13M)(0.085L) = (x)(0.180L)= 6.14x10-2M oflactic acid (new M)(0.15M)(0.095L) = (x)(0.180L)=7.92x10-2 M of sodium lactate (new M)

pH = 3.85 + log[7.92 x10-2]

[6.14x10-2]

3.96= 3.85 + 0.011


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Addition of strong acid-base


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Addition of strong acid

NaHCO3 + HCl H+ + CO3-2

NaHCO3 H+ CO3

-2

Before 0.100 mol 0.010

mol

0.125 mol

Change +0.010

mol

-0.010 -0.010

molAfter Reaction 0.110 mol 0.000

mol

0.115 mol

Using the same information given in 1a calculate the pH after

the addition of 0.010M of HCl.

Added neutralizes any CO3-2ions


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Calculating pH Changes in Buffers

Ka expression with NEW molarities to calculate new pH

Ka = [H+][CO3-2]

[HCO3-]

5.6 x 10-11= [x][0.115]

[0.110]

x= (5.6 x 10-11)(0.110)

0.115

x = [H+]= 5.83 x 10-11

pH = -log(3.97 x 10-11) = 10.3



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Calculating pH Changes in BuffersUsing the same information given in 1a calculate the pH after the

addition of 0.050M of NaOH

NaHCO3 OH CO2

2

Before 0.100 mol 0.050

mol

0.125

molChange -0.050

mol

-0.050 +0.050

mol

After Reaction 0.050 mol 0.000

mol

0.175

mol

Added

neutralizes

any H+ ions


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Calculating pH Changes in Buffers

Ka expression with NEW molarities to calculate new pH

Ka = [H+][CO3-2]

[HCO3-]

5.6 x 10-11= [x][0.175]

[0.050]

x= (5.6 x 10-11)(0.050)

0.175

x = [H+]= 1.6 x 10-11

pH = -log(1.6 x 10-11) = 10.8



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Titration:

A laboratory method for determining the

concentration of an unknown acid or base

using a neutralization reaction.

A standard solution,(a solution of known

concentration-titrant), is used.


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Titration

A known concentration of base(or acid) is slowly added to asolution of acid (or base).

After titration- we have a knownvolume and concentration fromtitrant

(M1)(V1). The unknown has aknown volume(V2) so we can

calculate (M2).

(M1)(V1)= (M2)(V2)

Titrant-known ofconcentration

Unknownconcentration with aknown volume


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TitrationA pH meter or indicatorsare used to determine

when the solution has

reached the equivalence

point,at which the

stoichiometric amount of

acid equals that of base.


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pH Titration Curve

pH titration curve, a

graph of pH as a

function of volume

of the added titrant.

The pH curve can:*help determine

equivalence point

*determine the pH

indicators needed for

Ka or Kb determination.


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pH Titration Curve

Can you identify the

titrant?

Is it an acid or base.

Base

*pH is increasing

*it levels off with lots of

base.

Strong Acid Strong Base Titration


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Strong Acid- Strong Base Titration

4 regions of a titration curve1st- Initial pH-

its really low- probably

a strong acid.

2nd- Between initial

and equivalence pt.rises slowly then rapidly

around the ~SAME~

volume as the

unknown.

Strong Acid Strong Base Titration


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Strong Acid- Strong Base Titration

4 regions of a titration curve3rdEquivalence pt

[H+] = [OH-] = pH 7

4thAfter equivalence

pt.

Has plateaued with

excess base

Titration of a Strong Acid with a


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Strong Base

From the start of

the titration to near

the equivalencepoint, the pH goes

up slowly.



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Strong Base

Just before and

after the

equivalence point,

the pH increases

rapidly.



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Strong BaseAt the equivalence

point, moles acid =moles base, and the

solution contains

only water and the

salt from the cationof the base and the

anion of the acid.



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Strong Base

As more base is

added, the increase in

pH again levels off.


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We will learn how to CALCULATE the

equivalence point later.


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Titration of a Weak Acid with a

Strong Base

Which way will the equilibrium shift in the case

of weak acid and strong base?

f k d h


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Titration of a Weak Acid with a Strong Base

1st- Initial pH-

~3 or 4 is a stronger weakacid

2nd- Between initial andequivalence pt.2 things to consider

1. neutralization of weakacid by strong base

2. Strong base acts as abuffer so it resists thetitration

pH of your weak acid is theamount for neutralization ofbase- of 50 mls (25 mls)and pH of ~4.8

f k d h


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3rdAt equivalence pt

~The pH here is above 7

which is what we

expected

4thAfter equivalence ptThe curve looks very

similar to a strong

acid/strong base curve.


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Titration of a Weak Base with a

Strong AcidWhich way will the equilibrium shift in the case of weak base

and strong acid?


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Titration of a Weak base with a Strong acid

1st- Initial pH-

its really high probably a

strong base.

2nd- Between initial and

equivalence pt.2 things to consider

1. neutralization of weak

base by strong acid

2. Strong acid acts as a

buffer so it resists the

titration

Strong base


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Titration of a Weak base with a Strong acid

3rdAt equivalence pt

~The pH here below 7

which is what we

expected

4thAfter equivalence pt

The curve looks very

similar to a strong

acid/strong basecurve.

Strong base


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With weaker acids,

the initial pH is higher

and pH changes nearthe equivalence point

are more subtle.


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Titrations of Polyprotic Acids

In these cases there is

an equivalence point

for each dissociation.


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03b_Buffer Ws Answers and Titration Notes

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