Advanced Placement Chemistry Acids, Bases, and Aqueous Equilibria

Advanced Placement Chemistry Acids, Bases, and Aqueous Equilibria

AP Chemistry: Chapter 14 Acids, Bases, And Aqueous Equilibria 14.1 THE NATURE OF ACIDS AND BASES ACIDS TASTE SOUR HAVE pH VALUES OF LESS THAN 7 TYPICALLY HAVE A HYDROGEN IN FRONT OF THE FORMULA OR A -COOH GROUP (CARBOXYLIC ACIDS) BASES REFERRED TO AS ALKALI TASTE BITTER AND FEEL

SLIPPERY HAVE pH VALUES OF GREATER THAN 7 TEND OF HAVE A HYDROXIDE (OH-) AT THE BACK OF THE FORMULA THERE ARE THREE GENERAL DEFINITIONS FOR ACIDS AND BASES: ARRHENIUS CONCEPT- ACIDS PRODUCE HYDROGEN IONS IN AQUEOUS SOLUTION WHILE BASES PRODUCE HYDROXIDE IONS. THERE ARE THREE GENERAL DEFINITIONS FOR ACIDS AND BASES: BRONSTED-LOWRY MODEL- ACIDS ARE PROTON (H ) +

DONORS AND BASES ARE PROTON ACCEPTORS. HYDRONIUM ION (H O )- FORMED ON REACTION OF A PROTON 3 + WITH A WATER MOLECULE. H + AND H3O+ ARE USED INTERCHANGEABLY IN MOST SITUATIONS. HA(aq) + H2O(l) H3O+(aq) + A-(aq) CONJUGATE BASE- EVERYTHING THAT REMAINS

OF THE ACID MOLECULE AFTER A PROTON IS LOST CONJUGATE ACID- BASE PLUS A PROTON THERE ARE THREE GENERAL DEFINITIONS FOR ACIDS AND BASES: LEWIS MODEL- ACIDS ARE ELECTRON PAIR ACCEPTORS WHILE BASES ARE ELECTRON PAIR DONORS ACID DISSOCIATION CONSTANT (Ka) ACCORDING TO THESE MODELS, THE ACID HAS A WEAKER HOLD ON THE HYDROGEN ION THAN THE BASES ATTRACTION FOR IT. THEREFORE, THE BASE TAKES THE HYDROGEN ION AND BECOMES THE CONJUGATE ACID.

THE AMOUNT OF CONJUGATE ACID AND BASE FORMED FROM THE EXCHANGE OF THE HYDROGEN ION IN RELATION TO THE AMOUNT OF ACID REMAINING CAN BE DESCRIBED BY THE EQUILIBRIUM Ka = [H3O+][A-] [HA] Ka = [H+][A-] [HA] THE KA VALUE A LARGE K VALUE WOULD INDICATE THAT THE ACID HAD STRONGLY DISSOCIATED AND FORMED A LOT OF CONJUGATE ACID AND BASE. THIS WOULD INDICATE A STRONG ACID. VERY STRONG ACIDS DO NOT HAVE KA VALUES BECAUSE THE AMOUNT OF UNDISSOCIATED REACTANT IS TOO SMALL THE MEASURE ACCURATELY.

A A SMALL K VALUE WOULD INDICATE THAT THE ACID HAD STAYED TOGETHER FOR THE MOST PART AND DID NOT FORM A LOT OF CONJUGATE ACID AND BASE. THIS WOULD INDICATE A WEAK ACID. A Ka = [H+][A-] [HA] When [H+] and [A-] are large and [HA] is small, Ka is large. When [H+] and [A-] are small and [HA] is large, Ka is small. STRONG ACID - ARE MOSTLY DISSOCIATED

EQUILIBRIUM LIES FAR TO THE RIGHT A STRONG ACID YIELDS A WEAK CONJUGATE BASE (MUCH WEAKER THAN H2O) COMMON STRONG ACIDS- ALL AQUEOUS SOLUTIONS (KNOW THESE!) H SO (SULFURIC) HNO (NITRIC) HClO (PERCHLORIC) H CrO (CHROMIC) HMnO (PERMANGANIC) 2 4

3 4 2 4 4 HCl (HYDROCHLORIC) HI (HYDROIODIC) HBr (HYDROBROMIC) WEAK ACID- ARE MOSTLY UNDISSOCIATED EQUILIBRIUM LIES FAR TO THE LEFT

HAS A STRONG CONJUGATE BASE (STRONGER THAN WATER) SULFURIC ACID IS A DIPROTIC ACID THIS MEANS THAT IT HAS TWO ACIDIC PROTONS. THE FIRST (H2SO4) IS STRONG AND THE SECOND (HSO4-) IS WEAK. 14.2 ACID STRENGTH OXYACIDS- FORMED WHEN A COVALENT OXIDE (LIKE CO2) DISSOLVES IN WATER TO MAKE AND ACID (LIKE H2CO3)MOST ACIDS ARE OXYACIDS THE ACIDIC PROTON IS

ATTACHED TO OXYGEN WEAK OXYACIDS: H PO (PHOSPHORIC) HNO (NITROUS) HOCl (HYPOCHLOROUS) 3 4 2 WITHIN A SERIES, ACID STRENGTH INCREASES WITH INCREASING NUMBERS OF OXYGEN ATOMS. FOR EXAMPLE: HClO > HClO > 4

3 HClO2 > HClO AND H SO 2 4 > H2SO3 (ELECTRONEGATIVE O DRAWS ELECTRONS AWAY FROM O-H BOND) ORGANIC ACIDS - HAVE CARBOXYL GROUP - USUALLY WEAK ACIDS CH3COOH ACETIC C6H5COOH BENZOIC

HYDROHALIC ACIDS H IS ATTACHED TO A HALOGEN (HCl, HI, ETC.) HF IS THE ONLY WEAK HYDROHALIC ACID. ALTHOUGH THE H-F BOND IS VERY POLAR, THE BOND IS SO STRONG (DUE TO THE SMALL F ATOM) THAT THE ACID DOES NOT COMPLETELY DISSOCIATE. EFFECTS OF ELECTRONEGATIVITY ON ACID-BASE STRENGTH IN TERMS OF POLARITY: H-F > H-Cl > H-Br > H-I BECAUSE OF THE STRONG HF BOND (SMALL F ATOM), HF IS A WEAK

ACID. IN TERMS OF ELECTRONEGATIVITY AND ACID STRENGTH: HOCl > HOBr > HOI > HOCH3 SOME SUBSTANCES HAVE THE ABILITY TO BOTH GAIN AND LOSE PROTONS. WE CALL THESE AMPHOTERIC SUBSTANCES. AMPHOTERIC SUBSTANCE- SUBSTANCE THAT CAN ACT AS AN ACID OR AS A BASE. EX. H2O Autoionization of water LETS TALK LAB! ION PRODUCT CONSTANT FOR

WATER (KW) H2O + H2O H3O+ + OHKW = [H3O+][OH-] KW = [H+][OH-] AT 25 C, K = 1 X 10 MOL /L BECAUSE [H ] = [OH ] = 1 X 10 M NO MATTER WHAT AN AQUEOUS SOLUTION CONTAINS, AT O W -14 2 2 + 25OC [H+][OH-] = 1 X 10-14 NEUTRAL SOLUTION [H ] = [OH ] ACIDIC SOLUTION [H ] > [OH ]

BASIC SOLUTION [H ] < [OH ] + K W - + - + - VARIES WITH TEMPERATURE -

-7 14.3 THE pH SCALE BECAUSE [H ] IN A SOLUTION IS + TYPICALLY VERY SMALL, WE CAN USE THE pH SCALE, WHICH IS BASED ON THE LOGARITHM OF MOLARITY, TO HELP US RELATE ACIDS AND BASES OF DIFFERING STRENGTHS THE pH SCALE SIGNIFICANT FIGURES IN pH AND OTHER LOG VALUES: THE NUMBER OF DECIMAL PLACES IN THE LOG VALUE SHOULD EQUAL THE NUMBER OF SIGNIFICANT DIGITS IN THE ORIGINAL NUMBER (CONCENTRATION). pH AND pOH ARE LOGARITHMIC FUNCTIONS. THE pH CHANGES

BY 1 FOR EVERY POWER OF 10 CHANGE IN [H+]. pH DECREASES AS [H ] INCREASES. + 14. 4 CALCULATING THE PH OF STRONG ACID SOLUTIONS CALCULATING PH OF STRONG ACID SOLUTIONS IS GENERALLY VERY SIMPLE. THE pH IS SIMPLY CALCULATED BY TAKING THE NEGATIVE LOGARITHM OF CONCENTRATION OF A MONOPROTIC STRONG ACID. FOR EXAMPLE, THE pH OF 0.1 M HCl IS 1.0. HOWEVER, IF THE ACID CONCENTRATION IS LESS THAN 1.0 X 10-7, THE WATER BECOMES THE IMPORTANT SOURCE OF [H+] AND THE pH IS 7.00. THE pH OF AN ACIDIC SOLUTION CAN NOT BE GREATER THAN 7 AT 25OC!

ANOTHER EXCEPTION IS CALCULATING THE pH OF A H2SO4 SOLUTION THAT IS MORE DILUTE THAN 1.0 M. AT THIS CONCENTRATION, THE [H+] OF THE HSO4- MUST ALSO BE CALCULATED. LETS PRACTICE! EX. CALCULATE THE [H ] AND EX. CALCULATE THE pH OF 1.0 + pH IN A 1.0 M SOLUTION OF HCl. THIS IS A STRONG ACID, SO [H+] = [HCl] pH = -LOG (1.0) = 0.00 X 10-10 M HCl.

THIS IS A STRONG ACID, SO [H+] = [HCl] pH = -LOG (1.0 X 10 ) = 10.00 THIS DOES NOT MAKE SENSE -10 FOR AN ACID, SO THE pH IS JUST THE pH OF WATER, OR 7. 14.5 CALCULATING THE pH OF WEAK ACID SOLUTIONS AS WE STATED EARLIER, WEAK ACIDS ARE WEAK BECAUSE THEY ARE MOSTLY UNDISSOCIATED. IN THE SPACE BELOW, DRAW IN THE RESULT OF A STRONG ACID DISSOLVING AND A WEAK ACID DISSOLVING. THE FOLLOWING TABLE CONTAINS EQUILIBRIUM CONSTANTS FOR MANY COMMON WEAK MONOPROTIC ACIDS. NOTE THAT AS THE

VALUE OF K GETS SMALLER, THE FEWER MOLECULES HAVE DISSOCIATED AND THE STRONGER THE CONJUGATE BASE. Formula Name Ka Formula HSO4- Bisulfate ion HClO2 Chlorous acid 1.2 x 10-2 HOCl Name

1.2 x 10-2 HC2H3O2 Acetic acid HC2H2ClO Monochloroac 1.35 x 10-3 HCN etic acid 2 HF Hydrofluoric 7.2 x 10-4 NH4+ acid HNO2 Nitrous acid 4.0 x 10-4 HOC6H5 Hypochlorous acid Hydrocyanic acid Ammonium ion phenol Ka

1.8 x 10-5 3.5 x 10-8 6.2 x 10-10 5.6 x 10-10 1.6 x 10-10 PRACTICE: ARRANGE THE FOLLOWING ACCORDING TO THEIR STRENGTHS AS BASES: HI, H2O, HF, HNO2, HCN HI < HF < HNO2 < H2O < HCN WRITE THE FOLLOWING AS ACIDS AND IN ORDER OF INCREASING PERCENT IONIZATION: OCl-, F-, NO2-, NO3-, NH3 NH 4

+ < HOCl < HNO2 < HF < HNO3 CALCULATING PH OF WEAK ACID SOLUTIONS 1. CALCULATING PH OF WEAK ACIDS INVOLVES SETTING UP AN EQUILIBRIUM. FOLLOW THESE STEPS FOR SUCCESS. 2. ALWAYS START BY WRITING THE EQUATION AND IDENTIFYING THE MAJOR SPECIES INVOLVED. 3. 4. NEXT, SET UP THE ACID EQUILIBRIUM EXPRESSION (K A)

5. FINALLY, SUBSTITUTE VALUES AND VARIABLES INTO THE K A EXPRESSION NOW, DEFINE INITIAL CONCENTRATIONS, CHANGES, AND FINAL CONCENTRATIONS IN TERMS OF X IN A RICE TABLE. AND SOLVE FOR X. EX. CALCULATE THE pH OF A 0.250 M SOLUTION OF HYPOCHLOROUS ACID. HOCl H+ + OClHOCl 3.5 X 10-8 = Ka = [H+][OCl-] = 3.5 X 10-8 [HOCl]

H+ 0.250 -x 0.250 -x 0 +x x [X]2 [0.250-X] 3.5 X 10-8 = X = [H+] = 9.4 X 10-5 + OCl0

+x x [X]2 [0.250] pH = 4.03 3.5 X 10-8 = THE NEGLIGIBLE X [X]2 [0.250-X] 5% of 0.250 = 0.0125 WHEN YOU ASSUME THAT X IS NEGLIGIBLE, YOU MUST CHECK THE VALIDITY OF THIS ASSUMPTION. TO BE VALID, X MUST BE LESS THAN 5% OF THE NUMBER THAT IT WAS TO BE

SUBTRACTED FROM. IN THIS EXAMPLE 9.35 X 10 IS LESS THAN 5% OF 0.250. THIS MEANS THAT THE ASSUMPTION THAT X WAS NEGLIGIBLE WAS VALID. -5 PROPANOIC ACID, HC3H5O2, IONIZES IN WATER AND HAS A KA VALUE OF 1.34 X 10-5. A. WRITE THE EQUATION AND EQUILIBRIUM EXPRESSION FOR THIS REACTION. K = [H+][C H O -] = 1.34 X 10-5 a HC3H5O2 H+ + C3H5O2- 3

5 2 [HC3H5O2] B. WHAT IS THE PH OF A 0.0500 M SAMPLE OF THIS SOLUTION AT EQUILIBRIUM? H C3H5O2 0.0500 -x 0.0500 -x 1.34 X 10-5 = [X]2 [0.0500-X] X = [H+] = 8.18 X 10-4

H+ 0 +x x + C3H5O2- 1.34 X 10-5 = [X]2 [0.0500] pH = 3.087 0 +x x Calculating the pH of Polyprotic Acids

ALL POLYPROTIC ACIDS DISSOCIATE STEPWISE. EACH DISSOCIATION HAS ITS OWN KA VALUE. AS EACH H IS REMOVED, THE REMAINING ACID GETS WEAKER AND THEREFORE HAS A SMALLER KA. AS THE NEGATIVE CHARGE ON THE ACID INCREASES IT BECOMES MORE DIFFICULT TO REMOVE THE POSITIVELY CHARGED PROTON. EXCEPT FOR H SO , POLYPROTIC ACIDS HAVE K 2 4 A2 AND KA3 VALUES SO MUCH WEAKER THAN THEIR KA1 VALUE THAT THE 2ND AND 3RD (IF APPLICABLE) DISSOCIATION CAN

BE IGNORED. THE [H+] OBTAINED FROM THIS 2ND AND 3RD DISSOCIATION IS NEGLIGIBLE COMPARED TO THE [H+] FROM THE 1ST DISSOCIATION. BECAUSE H SO 2 4 IS A STRONG ACID IN ITS FIRST DISSOCIATION AND A WEAK ACID IN ITS SECOND, WE NEED TO CONSIDER BOTH IF THE CONCENTRATION IS MORE DILUTE THAN 1.0 M. THE QUADRATIC EQUATION IS NEEDED TO WORK THIS TYPE OF PROBLEM. EX. CALCULATE THE PH OF A 0.0150 M H2SO4 SOLUTION. SINCE H2SO4 IS STRONG, [H+] = 0.0150 + X, WHERE X IS THE H+ CONTRIBUTED FROM THE SECOND DISSOCIATION. H2SO4 H+ + HSO4HSO4- H+ + SO42- Ka2 = [H+][SO42-] = 1.00 X 10-2

[HSO4-] H2SO4 HSO4- 0.0150 -0.0150 0 0.0150 -x 0.0150 -x H+ + 0

+0.0150 0.0150 H+ HSO4- 0 +0.0150 0.0150 + 0.0150 +x 0.0150 + x SO420 +x x

KEEP GOING! 1.00 X 10-2 = [0.0150 + X][X] [0.0150-X] 1.50 X 10-4 1.00 X 10-2X = 0.0150 X + X2 1.50 x 10-4 2.50 x 10-2x - x2 = 0 Using the quadratic equation, x = 5.81 x 10-3 [H+]= 0.0150 + (.0075) = .0225 pH = 1.648 DETERMINATION OF THE PH OF A MIXTURE OF WEAK ACIDS ONLY THE ACID WITH THE LARGEST K A

VALUE WILL CONTRIBUTE AN APPRECIABLE [H+]. DETERMINE THE PH BASED ON THIS ACID AND IGNORE ANY OTHERS. DETERMINATION OF THE PERCENT DISSOCIATION OF A WEAK ACID % DISSOCIATION = AMOUNT DISSOCIATED (M) X 100 INITIAL CONCENTRATION (MOL/L) FOR A WEAK ACID, PERCENT DISSOCIATION (OR IONIZATION) INCREASES AS THE ACID BECOMES MORE DILUTE. EQUILIBRIUM

SHIFTS TO THE RIGHT. CONCENTRATION, % DISSOCIATION AND THE STRENGTH OF ACIDS MANY THINGS CONTRIBUTE TO THE BEHAVIOR OF AN ACID: AS CONCENTRATION INCREASES, THERE IS LESS WATER TO IONIZE THE ACID, SO PERCENT IONIZATION GOES DOWN. THE MORE DILUTE AN ACID, THE MORE WATER THERE IS TO IONIZE IT, SO THE GREATER THE PERCENT IONIZATION. THE MORE H IONS IN SOLUTION,

THE STRONGER THE ACID. + 14.6 BASES THE HYDROXIDES OF GROUP I AND IIA METALS ARE ALL STRONG BASES. THE GROUP IIA HYDROXIDES ARE NOT VERY SOLUBLE. THIS PROPERTY ALLOWS SOME OF THEM TO BE USED EFFECTIVELY AS STOMACH ANTACIDS. WEAK BASES WEAK BASES (BASES WITHOUT OH ) REACT WITH WATER TO - PRODUCE A HYDROXIDE ION. THIS MEANS THAT WHEN YOU WRITE THESE REACTIONS, YOU MUST ADD WATER TO THE EQUATION!

COMMON EXAMPLES OF WEAK BASES, WHICH YOU NEED TO KNOW, ARE SHOWN IN THE TABLE BELOW. Name Ammonia Methylamine CH3NH2 Formula NH3 1.8 x 10-5 4.38 x 10-4 Kb Structure Ethylamine Aniline Pyridine

C2H5NH2 C6H5NH2 C5H5N 5.6 x 10-4 3.8 x 10-10 1.7 x 10-9 BASES REACT WITH WATER AS SHOWN IN THE EXAMPLES BELOW. B(aq) + H O(l) BH (aq) + OH (aq) CA CB BASE ACID NH + H O NH + OH ACID BASE CA CB

THE LONE PAIR ON N OR THE BASE FORMS A BOND WITH A H . 2 3 2 + - 4 + - + MOST WEAK BASES INVOLVE NITROGEN.

BASE DISSOCIATION CONSTANT + K = [BH ][OH ] B [B] + K = [NH ][OH ] B 4

[NH3] STRONG BASES ARE LIKE STRONG ACIDS EX. CALCULATE THE [OH ], [H ], AND PH OF A 0.0100 M - + SOLUTION OF NaOH. NaOH IS A STRONG BASE. [OH-] = 0.0100 M [H+] = 1 X 10-14/1 X 10-2 = 1.00 X 10-12 M pH = - log 1.00 X 10-12 = 12.000 WEAK BASES DETERMINATION OF THE pH OF A WEAK BASE IS VERY SIMILAR TO THE DETERMINATION OF THE pH OF A WEAK ACID. FOLLOW THE SAME STEPS.

REMEMBER, HOWEVER, THAT X IS THE [OH ] AND TAKING - THE NEGATIVE log OF X WILL GIVE YOU THE pOH AND NOT THE pH! EX. CALCULATE THE [OH-] AND THE pH FOR A 15.0 M NH3 SOLUTION Kb = [NH4 +][OH-] = 1.8 X 10-5 NH3+ H2O NH4+ + OH- [NH3] NH3 + H2O NH4+ + OH15.0 -x 15.0 -x 1.8 X 10-5 =

[X]2 [15.0-X] X = [OH-] = 1.6 X 10-2 --- 0 +x x 0 +x x 1.8 X 10-5 = [X]2 [15.0]

pOH = -log 1.6 x 10-2 = 1.78 pH = 14-1.78 = 12.22 14.8 ACID-BASE PROPERTIES OF SALTS NEUTRAL SALTS- SALTS THAT ARE FORMED FROM THE CATION OF A STRONG BASE AND THE ANION FROM A STRONG ACID FORM NEUTRAL SOLUTIONS WHEN DISSOLVED IN WATER. EX. NaCl, KNO3 ACID SALTS- SALTS THAT ARE FORMED FROM THE CATION OF A WEAK BASE AND THE ANION FROM A STRONG ACID FORM

ACIDIC SOLUTIONS WHEN DISSOLVED IN WATER. EX. NH4Cl THE CATION HYDROLYZES THE WATER MOLECULE TO PRODUCE HYDRONIUM IONS AND THUS AN ACIDIC SOLUTION. NH4+ + H2O H3O+ + NH3 STRONG ACID WEAK BASE BASIC SALTS SALTS THAT ARE FORMED FROM THE CATION OF A STRONG BASE AND THE ANION FROM A WEAK ACID FORM BASIC SOLUTIONS WHEN DISSOLVED IN WATER. EX. NaC2H3O2, KNO2 THE ANION HYDROLYZES THE WATER MOLECULE TO PRODUCE HYDROXIDE IONS AND THUS A BASIC SOLUTION. C H O 2

3 2 - + H2O OH- + HC2H3O2 STRONG BASE WEAK ACID Ka, Kb AND KW Dont forget, Kw varies with temperature! WHEN DETERMINING THE EXACT pH OF SALT SOLUTIONS, WE CAN USE THE Ka OF THE WEAK ACID FORMED TO FIND THE Kb OF THE SALT OR WE CAN USE THE Kb OF THE WEAK BASE FORMED TO FIND THE Ka OF THE SALT. K a X Kb = K W

EX. CALCULATE THE PH OF A 0.15 M SOLUTION OF SODIUM ACETATE. ASK YOURSELF, WHAT TYPE OF SALT IS SODIUM basic ACETATE? ________________ USE THE WEAKEST PART OF THE SALT AND WATER TO C H O + H O HC H O +

OH 2 3 2 2 2 3 2 WRITE AN EQUATION. ___________________________ USE THIS EQUATION AND A RICE TABLE TO SOLVE FOR THE pH. C2H3O2- + H2O HC2H3O2 + OH0.15 -x 0.15 -x --- Kb = [HC2H3O2][OH-] = 5.6 X 10-10 [C2H3O2-]

0 +x x 0 +x x 5.6 X 10-10 = [X]2 [0.15] Kb = 1 X 10-14 1.8 X 10-5 X = [OH-] = 9.2 X 10-6 pH = 8.96 RANK THE FOLLOWING IN ORDER OF MOST

ACIDIC TO LEAST ACIDIC. HCN, H2SO4, CH3NH2, NaCl, KC2H3O2, NH4Cl, Sr(OH)2 Most Acidic H2SO4 HCN Least Acidic NH4Cl NaCl KC2H3O2 CH3NH2 Sr(OH)2 14.10 ACID-BASE PROPERTIES OF

OXIDES WHEN METALLIC (IONIC) OXIDES DISSOLVE IN WATER THEY PRODUCE A METALLIC HYDROXIDE (BASIC SOLUTION). CaO + H O Ca(OH) 2 2 WHEN NONMETALLIC (COVALENT) OXIDES DISSOLVE IN WATER THEY PRODUCE A WEAK ACID (ACIDIC SOLUTION). CO 2 + H2O H2CO3 SALTS OF HIGHLY CHARGED METALS SALTS THAT CONTAIN A HIGHLY CHARGED METAL ION PRODUCE AN ACIDIC

SOLUTION. AlCl + 6H O Al(H O) + 3Cl Al(H O) Al(H O) (OH) + H THE HIGHER THE CHARGE ON THE METAL ION, THE STRONGER THE 3 2 2 6 3+ 2 2 5

6 3+ 2+ - + ACIDITY OF THE HYDRATED ION. THE ELECTRONS ARE PULLED AWAY FROM THE O-H BOND AND TOWARD THE POSITIVELY CHARGED METAL ION, LEAVING THE HYDROGEN ION FREE TO DISSOCIATE. FeCl3 AND Al(NO3)3 ALSO BEHAVE THIS WAY.

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