Chapter 16 Aqueous Ionic Equilibrium 2008, Prentice Hall Chemistry: A Molecular Approach, 1 st Ed....

Chapter 16Aqueous Ionic

Equilibrium

2008, Prentice Hall

Chemistry: A Molecular Approach, 1st Ed.Nivaldo Tro

2

The Danger of Antifreeze• each year, thousands of pets and wildlife die

from consuming antifreeze• most brands of antifreeze contain ethylene

glycol– sweet taste– initial effect drunkenness

• metabolized in the liver to glycolic acid– HOCH2COOH

• if present in high enough concentration in the bloodstream, it overwhelms the buffering ability of HCO3

−, causing the blood pH to drop• when the blood pH is low, it ability to carry O2 is

compromised– acidosis

• the treatment is to give the patient ethyl alcohol, which has a higher affinity for the enzyme that catalyzes the metabolism of ethylene glycol

ethylene glycol(aka 1,2–ethandiol)

Tro, Chemistry: A Molecular Approach 3

Buffers

• buffers are solutions that resist changes in pH when an acid or base is added

• they act by neutralizing the added acid or base

• but just like everything else, there is a limit to what they can do, eventually the pH changes

• many buffers are made by mixing a solution of a weak acid with a solution of soluble salt containing its conjugate base anion


Making an Acid Buffer


How Acid Buffers WorkHA(aq) + H2O(l) A−

(aq) + H3O+(aq)

• buffers work by applying Le Châtelier’s Principle to weak acid equilibrium

• buffer solutions contain significant amounts of the weak acid molecules, HA – these molecules react with added base to neutralize it– you can also think of the H3O+ combining with the OH− to

make H2O; the H3O+ is then replaced by the shifting equilibrium

• the buffer solutions also contain significant amounts of the conjugate base anion, A− - these ions combine with added acid to make more HA and keep the H3O+ constant


H2O

How Buffers Work

HA + H3O+A−A−

AddedH3O+

newHA

HA


H2O

HA

How Buffers Work

HA + H3O+

A−

AddedHO−

newA−

A−


Common Ion Effect HA(aq) + H2O(l) A−

(aq) + H3O+(aq)

• adding a salt containing the anion, NaA, that is the conjugate base of the acid (the common ion) shifts the position of equilibrium to the left

• this causes the pH to be higher than the pH of the acid solution– lowering the H3O+ ion concentration


Common Ion Effect


Ex 16.1 - What is the pH of a buffer that is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2?

Write the reaction for the acid with water

Construct an ICE table for the reaction

Enter the initial concentrations – assuming the [H3O+] from water is ≈ 0

HC2H3O2 + H2O C2H3O2 + H3O+

[HA] [A-] [H3O+]

initial 0.100 0.100 ≈ 0

change

equilibrium


[HA] [A-] [H3O+]

initial 0.100 0.100 0

change

equilibrium


represent the change in the concentrations in terms of x

sum the columns to find the equilibrium concentrations in terms of x

substitute into the equilibrium constant expression

+x+xx0.100 x 0.100 + x x

x

xxKa

100.0

100.0

OHHC

]OH][OH[C

232

3-232



x

xxKa

100.0

100.0

OHHC

]OH][OH[C

232

3-232

determine the value of Ka

since Ka is very small, approximate the [HA]eq = [HA]init and [A−]eq

= [A−]init solve for x

x 5108.1

[HA] [A-] [H3O+]

initial 0.100 0.100 ≈ 0

change -x +x +x

equilibrium 0.100 0.100 x0.100 x


0.100 +x

100.0

100.0

OHHC

]OH][OH[C

232

3-232 x

Ka

Ka for HC2H3O2 = 1.8 x 10-5



Ka for HC2H3O2 = 1.8 x 10-5

check if the approximation is valid by seeing if x < 5% of [HC2H3O2]init

%5%018.0%1001000.1

108.11

5

the approximation is valid

x = 1.8 x 10-5

[HA] [A-] [H3O+]

initial 0.100 0.100 ≈ 0

change -x +x +x

equilibrium 0.100 0.100 x



x = 1.8 x 10-5

substitute x into the equilibrium concentration definitions and solve

M 100.0108.1100.0100.0OHHC 5232 x

M 108.1]OH[ 53

x

[HA] [A-] [H3O+]

initial 0.100 0.100 ≈ 0

change -x +x +x

equilibrium 0.100 0.100 1.8E-5

M 100.0108.1100.0100.0]OHC[ 5232 x

0.100 + x x 0.100 x



substitute [H3O+] into the formula for pH and solve

74.4108.1log

OH-logpH5

3

[HA] [A-] [H3O+]

initial 0.100 0.100 ≈ 0

change -x +x +x




check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka

the values match

[HA] [A-] [H3O+]

initial 0.100 0.100 ≈ 0

change -x +x +x


5

5232

3-232

108.1100.0

108.1100.0

OHHC

]OH][OH[C

aK

Ka for HC2H3O2 = 1.8 x 10-5


Practice - What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF?





Enter the initial concentrations – assuming the [H3O+] from water is ≈ 0

HF + H2O F + H3O+

[HA] [A-] [H3O+]

initial 0.14 0.071 ≈ 0

change

equilibrium


[HA] [A-] [H3O+]

initial 0.14 0.071 0

change

equilibrium





+x+xx0.14 x 0.071 + x x

x

xxKa

14.0

071.0

HF

]OH][[F 3-

HF + H2O F + H3O+


x

xxKa

14.0

071.0

HF

]OH][[F 3-




x 3104.1

[HA] [A-] [H3O+]

initial 0.14 0.071 ≈ 0

change -x +x +x



0.071 +x

14.0

071.0100.7 4 x

Ka

Ka for HF = 7.0 x 10-4

4

15.3

100.7

1010

a

pKa

K

K a



Ka for HF = 7.0 x 10-4


%5%1%100104.1

104.11

3


x = 1.4 x 10-3

[HA] [A2-] [H3O+]

initial 0.14 0.071 ≈ 0

change -x +x +x




x = 1.4 x 10-3


M 14.0104.114.014.0HF 3 x

M 104.1]OH[ 33

x

[HA] [A2-] [H3O+]

initial 0.14 0.071 ≈ 0

change -x +x +x


M 072.0104.1071.0071.0]OHC[ 3232 x

0.071 + x x 0.14 x




85.2104.1log

OH-logpH3

3

[HA] [A-] [H3O+]

initial 0.14 0.071 ≈ 0

change -x +x +x





the values are close enough

[HA] [A-] [H3O+]

initial 0.14 0.071 ≈ 0

change -x +x +x


4

3

3-

102.714.0

104.1072.0

HF

]OH][[F

aK

Ka for HF = 7.0 x 10-4


Henderson-Hasselbalch Equation• calculating the pH of a buffer solution can be

simplified by using an equation derived from the Ka expression called the Henderson-Hasselbalch Equation

• the equation calculates the pH of a buffer from the Ka and initial concentrations of the weak acid and salt of the conjugate base– as long as the “x is small” approximation is valid

initial

initiala acid][weak

anion] base conjugate[logp pH K


Deriving the Henderson-Hasselbalch Equation

][A

[HA]]OH[

HA

]OH][[A

-3

3-

a

a

K

K

][A

[HA]log]OHlog[

-3 aK

]Olog[H- pH 3

][A

[HA]loglog]OHlog[

-3 aK

][A

[HA]loglogpH

-aK

aK log- pKa

][A

[HA]logppH

-aK

[HA]

][Alog

][A

[HA]log

[HA]

][AlogppH

-

aK



Assume the [HA] and [A-] equilibrium concentrations are the same as the initial

Substitute into the Henderson-Hasselbalch Equation

Check the “x is small” approximation


][HA

][AlogppH

-

aK

050.0

0.150log781.4pH

Ka for HC7H5O2 = 6.5 x 10-5

781.4105.6log

logp5

aa KK

4.66pH

54.66-3

-pH3

102.210]OH[

10]OH[

%5%044.0%100050.0

102.2 5





find the pKa from the given Ka




HF + H2O F + H3O+

][HA

][AlogppH

-

aK

86.2

14.0

0.071log15.3pH

32.86-3

-pH3

104.110]OH[

10]OH[

%5%1%10014.0

104.1 3


Do I Use the Full Equilibrium Analysis or the Henderson-Hasselbalch Equation?• the Henderson-Hasselbalch equation is generally

good enough when the “x is small” approximation is applicable

• generally, the “x is small” approximation will work when both of the following are true:

a) the initial concentrations of acid and salt are not very dilute

b) the Ka is fairly small• for most problems, this means that the initial acid

and salt concentrations should be over 1000x larger than the value of Ka


How Much Does the pH of a Buffer Change When an Acid or Base Is Added?• though buffers do resist change in pH when acid or

base are added to them, their pH does change• calculating the new pH after adding acid or base

requires breaking the problem into 2 parts1. a stoichiometry calculation for the reaction of the

added chemical with one of the ingredients of the buffer to reduce its initial concentration and increase the concentration of the other– added acid reacts with the A− to make more HA– added base reacts with the HA to make more A−

2. an equilibrium calculation of [H3O+] using the new initial values of [HA] and [A−]


Ex 16.3 - What is the pH of a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L that has

0.010 mol NaOH added to it?

If the added chemical is a base, write a reaction for OH− with HA. If the added chemical is an acid, write a reaction for it with A−.

Construct a stoichiometry table for the reaction

HC2H3O2 + OH− C2H3O2 + H2O

HA A- OH−

mols Before 0.100 0.100 0

mols added - - 0.010

mols After




Fill in the table – tracking the changes in the number of moles for each component

HC2H3O2 + OH− C2H3O2 + H2O

HA A- OH−

mols Before 0.100 0.100 ≈ 0


mols After 0.090 0.110 ≈ 0






Enter the initial concentrations – assuming the [H3O+] from water is ≈ 0, and using the new molarities of the [HA] and [A−]


[HA] [A-] [H3O+]

initial 0.090 0.110 ≈ 0

change

equilibrium


[HA] [A-] [H3O+]

initial 0.090 0.110 0

change

equilibrium






+x+xx0.090 x 0.110 + x x

x

xxKa

090.0

110.0

OHHC

]OH][OH[C

232

3-232



x

xxKa

090.0

110.0

OHHC

]OH][OH[C

232

3-232




x 51074.1

[HA] [A-] [H3O+]

initial 0.100 0.100 ≈ 0

change -x +x +x




0.110 +x

090.0

110.0

OHHC

]OH][OH[C

232

3-232 x

Ka

Ka for HC2H3O2 = 1.8 x 10-5

090.0

110.0108.1 5 x



0.010 mol NaOH added to it?Ka for HC2H3O2 = 1.8 x 10-5


%5%016.0%100100.9

1074.12

5


x = 1.47 x 10-5

[HA] [A-] [H3O+]

initial 0.090 0.110 ≈ 0

change -x +x +x





x = 1.47 x 10-5


M 090.01074.1090.0090.0OHHC 5232 x

M 1074.1]OH[ 53

x

[HA] [A-] [H3O+]

initial 0.090 0.110 ≈ 0

change -x +x +x


M 110.01074.1110.0110.0]OHC[ 5232 x

0.110 + x x 0.090 x





83.41074.1log

OH-logpH5

3

[HA] [A-] [H3O+]

initial 0.090 0.110 ≈ 0

change -x +x +x






the values match

[HA] [A-] [H3O+]

initial 0.090 0.110 ≈ 0

change -x +x +x


5

5

232

3-232

108.1090.0

1074.1110.0

OHHC

]OH][OH[C

aK

Ka for HC2H3O2 = 1.8 x 10-5


Ex 16.3 - What is the pH of a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L that

has 0.010 mol NaOH added to it?

find the pKa from the given Ka



547.4

108.1log

logp5

aa KK

Ka for HC2H3O2 = 1.8 x 10-5

[HA] [A-] [H3O+]

initial 0.090 0.110 ≈ 0

change -x +x +x



Ex 16.3 - What is the pH of a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L that has 0.010 mol NaOH added to it?




][HA

][AlogppH

-

aK

83.4

090.0

0.110log547.4pH

%5%016.0%100090.0

1074.1 5

pKa for HC2H3O2 = 4.745

54.83-3

-pH3

1074.110]OH[

10]OH[


Ex 16.3 – Compare the effect on pH of adding 0.010 mol NaOH to a buffer that has 0.100 mol HC2H3O2 and 0.100

mol NaC2H3O2 in 1.00 L to adding 0.010 mol NaOH to 1.00 L of pure water?


][HA

][AlogppH

-

aK

83.4

090.0

0.110log547.4pH

pKa for HC2H3O2 = 4.745

M 010.0L 1.00

mol 010.0]OH[

00.2100.1log

]OHlog[pOH2

12.00

2.00-14.00

pOH -14.00pH

00.14pOHpH


Basic BuffersB:(aq) + H2O(l) H:B+

(aq) + OH−(aq)

• buffers can also be made by mixing a weak base, (B:), with a soluble salt of its conjugate acid, H:B+Cl−

H2O(l) + NH3 (aq) NH4+

(aq) + OH−(aq)


Ex 16.4 - What is the pH of a buffer that is 0.50 M NH3 (pKb = 4.75) and 0.20 M NH4Cl?

find the pKa of the conjugate acid (NH4

+) from the given Kb

Assume the [B] and [HB+] equilibrium concentrations are the same as the initial



NH3 + H2O NH4+ + OH−

][HB

[B]logppH aK

65.9

20.0

0.50log25.9pH

109.65-3

-pH3

1032.210]OH[

10]OH[

%5%10020.0

1032.2 10

9.254.75-14p -14p

14pp

ba

ba

KK

KK


Buffering Effectiveness• a good buffer should be able to neutralize moderate

amounts of added acid or base• however, there is a limit to how much can be added

before the pH changes significantly• the buffering capacity is the amount of acid or base a

buffer can neutralize• the buffering range is the pH range the buffer can be

effective• the effectiveness of a buffer depends on two factors

(1) the relative amounts of acid and base, and (2) the absolute concentrations of acid and base

HA A- OH−



mols After 0.17 0.030 ≈ 0

Effect of Relative Amounts of Acid and Conjugate Base

Buffer 10.100 mol HA & 0.100 mol A-

Initial pH = 5.00


Initial pH = 4.05pKa (HA) = 5.00

][HA

][AlogppH

-

aK

09.5

090.0

0.110log00.5pH

after adding 0.010 mol NaOHpH = 5.09

HA + OH− A + H2O

HA A- OH−



mols After 0.090 0.110 ≈ 0

25.4

17.0

0.030log00.5pH


%8.1

%1005.00

5.00-5.09

Change %

%0.5

%1004.05

4.05-4.25

Change %

a buffer is most effective with equal concentrations of acid and base

HA A- OH−



mols After 0.49 0.51 ≈ 0

%6.3

%1005.00

5.00-5.18

Change %

HA A- OH−



mols After 0.040 0.060 ≈ 0

Effect of Absolute Concentrations of Acid and Conjugate Base


Initial pH = 5.00


Initial pH = 5.00pKa (HA) = 5.00

][HA

][AlogppH

-

aK

02.5

49.0

0.51log00.5pH


HA + OH− A + H2O

18.5

040.0

0.060log00.5pH


%4.0

%1005.00

5.00-5.02

Change %

a buffer is most effective when the concentrations of acid and base are largest


Effectiveness of Buffers

• a buffer will be most effective when the [base]:[acid] = 1– equal concentrations of acid and base

• effective when 0.1 < [base]:[acid] < 10• a buffer will be most effective when the [acid]

and the [base] are large

53

Buffering Range• we have said that a buffer will be effective when

0.1 < [base]:[acid] < 10• substituting into the Henderson-Hasselbalch we can

calculate the maximum and minimum pH at which the buffer will be effective

][HA

][AlogppH

-

aKLowest pH

1ppH

10.0logppH

a

a

K

K

Highest pH

1ppH

10logppH

a

a

K

K

therefore, the effective pH range of a buffer is pKa ± 1when choosing an acid to make a buffer, choose one whose is pKa is closest to the pH of the buffer


Ex. 16.5a – Which of the following acids would be the best choice to combine with its sodium

salt to make a buffer with pH 4.25?

Chlorous Acid, HClO2 pKa = 1.95Nitrous Acid, HNO2 pKa = 3.34Formic Acid, HCHO2 pKa = 3.74Hypochlorous Acid, HClO pKa = 7.54


Ex. 16.5a – Which of the following acids would be the best choice to combine with its sodium

salt to make a buffer with pH 4.25?

Chlorous Acid, HClO2 pKa = 1.95Nitrous Acid, HNO2 pKa = 3.34Formic Acid, HCHO2 pKa = 3.74Hypochlorous Acid, HClO pKa = 7.54

The pKa of HCHO2 is closest to the desired pH of the buffer, so it would give the most effective buffering range.


Ex. 16.5b – What ratio of NaCHO2 : HCHO2 would be required to make a buffer with pH 4.25?Formic Acid, HCHO2, pKa = 3.74

][HA

][AlogppH

-

aK

][HCHO

][CHOlog51.0

][HCHO

][CHOlog74.325.4

2

2

2

2 24.3][HCHO

][CHO

1010

2

2

51.0][HCHO

][CHOlog

2

2

to make the buffer with pH 4.25, you would use 3.24 times as much NaCHO2 as HCHO2


Buffering Capacity• buffering capacity is the amount of acid or base that can

be added to a buffer without destroying its effectiveness• the buffering capacity increases with increasing

absolute concentration of the buffer components• as the [base]:[acid] ratio approaches 1, the ability of the

buffer to neutralize both added acid and base improves• buffers that need to work mainly with added acid

generally have [base] > [acid]• buffers that need to work mainly with added base

generally have [acid] > [base]


Buffering Capacity

a concentrated buffer can neutralize more added acid or base than a dilute buffer


Titration• in an acid-base titration, a solution of unknown

concentration (titrant) is slowly added to a solution of known concentration from a burette until the reaction is complete– when the reaction is complete we have reached the

endpoint of the titration• an indicator may be added to determine the endpoint– an indicator is a chemical that changes color when the pH

changes

• when the moles of H3O+ = moles of OH−, the titration has reached its equivalence point


Titration


Titration Curve• a plot of pH vs. amount of added titrant• the inflection point of the curve is the equivalence

point of the titration• prior to the equivalence point, the known solution in

the flask is in excess, so the pH is closest to its pH• the pH of the equivalence point depends on the pH of

the salt solution– equivalence point of neutral salt, pH = 7 – equivalence point of acidic salt, pH < 7 – equivalence point of basic salt, pH > 7

• beyond the equivalence point, the unknown solution in the burette is in excess, so the pH approaches its pH


Titration Curve:Unknown Strong Base Added to

Strong Acid



Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH

• HCl(aq) + NaOH(aq) NaCl(aq) + H2O(aq)

• initial pH = -log(0.100) = 1.00• initial mol of HCl = 0.0250 L x 0.100 mol/L = 2.50 x 10-3

• before equivalence point

NaOH added molesL 1

NaOH mol 0.100NaOH added of L

added 5.0 mL NaOH

NaOH mol 100.5L 1

NaOH mol 0.100NaOH L 0.0050

4

used HCl moles NaOH mol 1

HCl mol 1NaOH mole

5.0 x 10-4 mol NaOH

used HCl mol 100.5NaOH mol 1

HCl mol 1NaOH mol 100.5

4

4

excess HCl mol used HCl mol -HCl mol initial

excess HCl mol102.00

used HCl mol105.0 -HCl mol 102.503-

-4-3

]O[HHCl M NaOH LHCl L

excess HCl mol

3

]O[HHCl M 0.0667 NaOH L 0050.0HCl L 0.0250

HCl mol102.00

3

-3

2.00 x 10-3 mol HCl

]O-log[HpH 3 18.10667.0-logpH


excess NaOH mol105.0

used NaOH mol102.50 -NaOH mol 103.004-

-3-3

NaOH mol 1000.3

L 1


3

][OHNaOH M 0.00909 NaOH L 0300.0HCl L 0.0250

NaOH mol100.5 -4

123

14

3

1001.11009.9

101

]OH[]OH[

wK

][OHNaOH M NaOH LHCl L

excess NaOH mol


• HCl(aq) + NaOH(aq) NaCl(aq) + H2O(aq)

• at equivalence, 0.00 mol HCl and 0.00 mol NaOH• pH at equivalence = 7.00• after equivalence point

NaOH added molesL 1


added 30.0 mL NaOH

5.0 x 10-4 mol NaOH xs

excess NaOH mol HCl mol initial -added NaOH mol

wK ]][OHOH[ 3

96.111001.1log- pH 9- ]Olog[H- pH 3


added 30.0 mL NaOH0.00050 mol NaOHpH = 11.96


Adding NaOH to HCl

25.0 mL 0.100 M HCl0.00250 mol HClpH = 1.00

added 5.0 mL NaOH0.00200 mol HClpH = 1.18




added 25.0 mL NaOHequivalence pointpH = 7.00




Titration of 25 mL of 0.100 M HCHO2 with 0.100 M NaOH

• HCHO2(aq) + NaOH(aq) NaCHO2(aq) + H2O(aq)

• Initial pH:[HCHO2] [CHO2

-] [H3O+]

initial 0.100 0.000 ≈ 0

change -x +x +x

equilibrium 0.100 - x x x

Ka = 1.8 x 10-4

M 1042.4]O[H

100.0100.0108.1

]HCHO[

]O][H[CHO

33

24

2

32

x

x

x

xx

Ka

37.210424.-log

]Olog[H- pH3-

3

%5%2.4%100100.0

102.4 3



• HCHO2(aq) + NaOH(aq) NaCHO2 (aq) + H2O(aq)

• initial mol of HCHO2 = 0.0250 L x 0.100 mol/L = 2.50 x 10-3

• before equivalence added 5.0 mL NaOH

NaOH mol 100.5L 1


4

HA A- OH−

mols Before 2.50E-3 0 0

mols added - - 5.0E-4

mols After 2.00E-3 5.0E-4 ≈ 0

2

2

HCHO mol

CHO mollogppH aK

14.3pH10.002

10.05log74.3pH

5-

4-

74.3108.1-log

log- p4-

aa KK

72

M 107.1][OH

0500.00500.0106.5

]CHO[

]][OH[HCHO

6

211

2

2

x

x

x

xx

Kb

96

14

3

109.5107.1

101

]OH[]OH[

wK

114

14

CHO ,

106.5108.1

101

2

a

wb K

KK

22-

2-2

2-2

-3

CHO M105.00

NaOH L102.50HCHO L102.50

CHO mol102.50

2

2

2

CHO M

NaOH LHCHO L

CHO mol



• initial mol of HCHO2 = 0.0250 L x 0.100 mol/L = 2.50 x 10-3

• at equivalence added 25.0 mL NaOH

NaOH mol 1050.2L 1


3

HA A- OH−

mols Before 2.50E-3 0 0

mols added - - 2.50E-3

mols After 0 2.50E-3 ≈ 0

[HCHO2] [CHO2-] [OH−]

initial 0 0.0500 ≈ 0

change +x -x +x

equilibrium x 5.00E-2-x x

CHO2−

(aq) + H2O(l) HCHO2(aq) + OH−(aq)

Kb = 5.6 x 10-11

23.8109.5-log

]Olog[H- pH9-

3

[OH-] = 1.7 x 10-6 M


excess NaOH mol105.0

used NaOH mol102.50 -NaOH mol 103.004-

-3-3

NaOH mol 1000.3L 1


3

][OHNaOH M NaOH LHCl L

excess NaOH mol

][OHNaOH M 0.0091 NaOH L 0300.0HCl L 0.0250

NaOH mol100.5 -4



• after equivalence point

NaOH added molesL 1


added 30.0 mL NaOH

5.0 x 10-4 mol NaOH xs

excess NaOH mol HCHO mol initial -added NaOH mol 2

12

3

14

3

1001.1101.9

101

]OH[]OH[

wK

96.111001.1log- pH 9- ]Olog[H- pH 3

wK ]][OHOH[ 3

Date post:	03-Jan-2016
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Chapter 16 Aqueous Ionic Equilibrium 2008, Prentice Hall Chemistry: A Molecular Approach, 1 st Ed....

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