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ACID BASE CHEMISTRY

ACID BASE CHEMISTRY. WRITTEN BY JOANNE SWANSON. PART 1, FACTS TO KNOW:. CONCENTRATIONS ARE IN MOLARITY AND ARE DESIGNATED BY SQUARE BRACKETS, [ H 3 O + ] [ H 3 O + ] REPRESENTS MOLAR CONCENTRATION OF ACID [OH - ] REPRESENTS MOLAR CONCENTRATION OF BASE

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ACID BASE CHEMISTRY

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  1. ACID BASE CHEMISTRY WRITTEN BY JOANNE SWANSON

  2. PART 1, FACTS TO KNOW: • CONCENTRATIONS ARE IN MOLARITY AND ARE DESIGNATED BY SQUARE BRACKETS, [ H3O+] • [ H3O+] REPRESENTS MOLAR CONCENTRATION OF ACID • [OH-] REPRESENTS MOLAR CONCENTRATION OF BASE • THE pH SCALE REPRESENTS THE LEVEL OF ACIDITY IN A SOLUTION. IT IS A LOGARITHMIC SCALE.

  3. PART 1, EQUATIONS TO KNOW: • [OH-] x [H3O+] = 1 x 10-14 = Kw • pH + pOH = 14 = Pkw • pH = - LOG [H3O+] • pOH = - LOG [OH-] • [H3O+] = 10-pH • [OH-] = 10-pOH

  4. IMPORTANT DETAILS NEUTRAL WATER CONTAINS EQUAL CONCENTRATIONS OF HYDRONIUM IONS, [H3O+], AND HYDROXIDE IONS, [OH-]. THE pH SCALE IS FROM ZERO TO 14. THE MIDDLE, 7, IS NEUTRAL. BELOW 7 IS ACIDIC AND ABOVE 7 IS BASIC.

  5. ACIDS SOUR TASTING TURN BLUE LITMUS PAPER RED REACT WITH ACTIVE METALS ARE ELECTROLYTES BASES BITTER TASTING FEEL SLIPPERY TURN RED LITMUS PAPER BLUE ARE ELECTROLYTES SOME IMPORTANT PROPERTIES:

  6. ACIDS AND BASES DEFINED: • A BRONSTED-LOWRY ACID IS DEFINED AS A PROTON DONAR. • A BRONSTED-LOWRY BASE IS DEFINED AS A PROTON ACCEPTOR. NOTE: A HYDROGEN ATOM THAT HAS LOST ITS ELECTRON IS REFERRED TO AS A PROTON. Ex. HCl + NaOH HOH + Na+ + Cl- NOTICE THAT HCl DONATED ITS PROTON AND NaOH ACCEPTED THE PROTON.

  7. STRONG ACIDS: • COMPLETELY DISSOCIATE IN WATER, (IONIZE). HCl + HOH  H3O+ + Cl- (aq) • THEREFORE ARE STRONG ELECTROLYTES • THEY HAVE A LARGE Ka VALUE (THIS IS CALLED THE ACID EQUILIBRIUM CONSTANT). SINCE STRONG ACIDS COMPLETELY DISSOCIATE, Ka IS USUALLY USED FOR WEAK ACIDS. • STRONG ACIDS ARE: HCl, HNO3 , H2SO4 HBr, HI, HClO4

  8. STRONG BASES: • THEY ARE SOLUBLE METAL HYDROXIDES. • THEY COMPLETELY DISSOCIATE IN WATER (IONIZE). NaOH + HOH  Na+(aq) + OH- (aq) • THEY ARE STRONG ELECTROLYTES. • SOME STRONG BASES ARE: LiOH, NaOH, KOH, Ca(OH)2 , Sr(OH)2 ,Ba(OH)2

  9. OTHER DEFINITIONS OF ACIDS AND BASES WILL BE ADDRESSED LATER IN THIS LESSON.

  10. SOLVING PROBLEMS: • IF GIVEN EITHER THE ACID or BASE CONCENTRATION OF A SOLUTION, ONE CAN SOLVE FOR THE OTHER CONCENTRATION, THE pH, AND THE pOH. • IF GIVEN THE pH or THE pOH, ONE CAN SOLVE FOR THE OTHER, AND THE CONCENTRATIONS OF ACID AND BASE IN THE SOLUTION.

  11. PH pOH [OH-] pH = - Log [H3o+] pH = - Log (1 x 10-3) pH = 3 pOH + pH = 14 14 – pH = pOH 14 – 3 = pOH = 11 [OH-] x [H3O+] = 1 x 10-14 [OH-] = 1 x 10-14 / [H3O+] [OH -]=1 X 10-14 / 1 X 10-3 =1 X 10-11 GIVEN THAT THE [H3O+] IN A SOLUTION = 1 x 10 –3 CALCULATE:

  12. ANOTHER EXAMPLE: IF THE pH OF A SOLUTION IS = 8, CALCULATE THE pOH, THE [H3O+], AND THE [OH-]. • 14 – pH = pOH, 14 – 8 = 6, pOH = 6 • [H3O+] = 10-pH , [H3O+] = 10-8 , [H3O+]= 10-8 • [OH-] = 1 x 10-14 / [H3O+], • [OH-] = 1 x 10-14 / 10-8 = 1 x 10-6 [OH-]= 1 x 10-6

  13. WEAK ACIDS : THOSE THAT ARE ONLY PARTIALLY DISSOCIATED • THERE ARE THOUSANDS OF WEAK ACIDS. (CHEMICAL INDICATORS ARE ALSO WEAK ACIDS). • IN WATER, A REVERSIBLE REACTION OCCURS. • A SMALL AMOUNT OF HYDROGEN IONS ARE PRODUCED FROM WEAK ACIDS. • THERE ARE BASICALLY 3 TYPES: MOLECULES WITH AN IONIZABLE HYDROGEN (LIKE HF, HClO,HC2H3O2 ,) ANIONS WITH AN IONIZABLE HYDROGEN (LIKE HSO4- ,HPO42-) AND CATIONS WITH AN IONIZABLE HYDROGEN (LIKE NH4+ ).

  14. EXAMPLES OF WEAK ACID REACTIONS IN WATER: MOLECULES WITH AN IONIZABLE HYDROGEN: HF + H2O H+ (aq) + F- (aq) HC2H3O2 + H2OH+ (aq) + C2H3O2- (aq) ANIONS WITH AN IONIZABLE HYDROGEN: HSO4-(aq) H+ (aq) + SO42- (aq) H2PO41-(aq) H+ (aq) + HPO42- (aq) CATIONS WITH AN IONIZABLE HYDROGEN: NH4+(aq) H+ (aq) + NH3 (aq)

  15. WEAK BASES: THOSE THAT DO NOTDIRECTLY FURNISH THE OH- IONS A HYDROXIDE ION IS GENERATED BY THE REACTION OF THE WEAK BASE WITH WATER: NH3(aq) + H2ONH4+ + OH- CH3NH2 (aq) + H2O CH3NH3+ + OH- C2H3O2-(aq) + H2O HC2H3O2 + OH- MANY WEAK BASES ARE ANIONS. ANIONS OF WEAK ACIDS ARE WEAK BASES (LIKE THE ACETATE ION).

  16. ACID BASE PROPERTIES OF SALT SOLUTIONS: • THE ANIONS OF STRONG ACIDS, SUCH AS Cl-, Br-,I-,ClO4-,NO3-, AND SO42- PRODUCE NEUTRAL SALT SOLUTIONS. • THE CATIONS OF STRONG BASES , SUCH AS Li+ , Na+ , K+ Ca2+ , Ba2+ , PRODUCE NEUTRAL SALT SOLUTIONS. • THE ANIONS OF WEAK ACIDS, SUCH AS C2H3O2-, F-, CO32-, PO43-, CN-, PRODUCE BASIC SALT SOLUTIONS. • TRANSITION METAL IONS, Al+3, Mg+2, NH4+1,HSO4-, AND H2PO42- PRODUCE ACIDIC SALT SOLUTIONS.

  17. ACID BASE REACTIONS:STRONG ACID + STRONG BASE THE REACTION OF A STRONG ACID AND A STRONG BASE IN STOICHIOMETRIC AMOUNTS (BALANCED MOLE TO MOLE RATIOS), WILL PRODUCE A SALT AND WATER AS THE PRODUCTS. THE [H+]AND THE [OH-]ARE EQUAL AND THE CONCENTRATION OF THIS PRODUCT IS 1 X 10-14 Ex. HCl + NaOH  H2O + Na+ +Cl-

  18. STRONG ACID + WEAK BASE A STRONG ACID AND A WEAK BASE WILL PRODUCE AN ACIDIC SOLUTION Ex. HCl + NH3 NH4+ + Cl- RECALL THAT HCl COMPLETELY DISSOCIATES TO FORM THE NEUTRAL Cl- ANION. THE WEAK BASE NH3 WILL PICK UP THE PROTON FROM THE STRONG ACID AND FORM NH4+ , A WEAK ACID. THEREFORE THE SOLUTION WILL BE ACIDIC OR HAVE A pH < 7.

  19. WEAK ACID + STRONG BASE: A WEAK ACID AND A STRONG BASE PRODUCE A BASIC SOLUTION. THEREFORE THE pH WILL BE > 7. Ex. HC2H3O2 + NaOH  H2O+ C2H3O2- + Na+ RECALL THAT THE STRONG BASE COMPLETELY DISSOCIATES INTO NEUTRAL Na+ AND STRONG BASE, OH-. THIS HYDROXIDE ION WILL THEN TAKE THE PROTON FROM THE ACETIC ACID TO FORM WATER AND LEAVE BEHIND THE BASIC ACETATE ION.

  20. ACID-BASE TITRATIONS: A TECHNIQUE USED IN THE LAB, TO DETERMINETHECONCENTRATION OF A SPECIES IN A SOLUTION. TO PERFORM A TITRATION OF AN ACID OF UNKNOWN CONCENTRATION, A KNOWN QUANTITYOF A KNOWN CONCENTRATION OF A BASE IS ADDED TO A KNOWN QUANTITYOF AN ACID OF WHICH THE CONCENTRATION IS UNKNOWN UNTIL THE ACID IS COMPLETELY REACTED (THIS IS THE EQUIVALENCE POINT OR ENDPOINT).

  21. WHEN TITRATING, A CHEMICALINDICATOR IS ADDED TO THE ACID BEING TITRATED. A CHEMICAL INDICATOR IS A SUBSTANCE THAT TURNS TO A DIFFERENT COLOR AT A pH WHERE ALL OF THE LIMITING REACTANT IS USED UP (THE EQUIVALENC POINT OF THE REACTION).

  22. WHAT HAPPENS WHEN THE BASE IS ADDED TO THE UNKNOWN ACID WHICH CONTAINS A CHEMICAL INDICATOR? WHEN THE HYDROXIDE ION FROM THE BASE USES UP (OR COMPLETELY REACTS WITH) ALL OF THE ACID, THE EQUIVALENCE POINT WILL BE REACHED. AT THIS POINT THE INDICATOR WILL CHANGE COLOR. THEN A CALCULATION CAN BE USED TO DETERMINE THE CONCENTRATION OF THE ACID, BASED ON THE AMOUNT OF BASE THAT WAS NEEDED TO REACT WITH IT.

  23. TITRATION CALCULATION: DETERMINE THE CONCENTRATION OF A NICOTINIC ACID SOLUTION , HC6H4NO2, TITRATED TO THE ENDPOINT WITH 36.2 mL OF 0.100 M NaOH SOLUTION. THE SAMPLE OF ACID WAS 200. mL. A BALANCED CHEMICAL EQUATION IS NEEDED TO DETERMINE THE MOLE TO MOLE RATIO OF OH- TO ACID. HC6H4NO2 + OH- H2O + C6H4NO2- WE CAN THINK OF THE KNOWN SOLUTION BEING ADDED INTO THE UNKNOWN SOLUTION OR…

  24. KNOWN UNKNOWN = 0.0180 M HC6H4NO2 THE SAME TYPE OF CALCULATION CAN BE PERFORMED TO DETERMINE THE CONCENTRATION OF A BASE USING A KNOWN CONCENTRATION OF A STRONG ACID AS THE TITRANT.

  25. AN APPROPRIATE CHEMICAL INDICATOR MUST BE CHOSEN FOR A TITRATION. WHEN TITRATING A WEAK ACID WITH A STRONGBASE, THE SOLUTION WILL TURN BASIC, THEREFORE THE pH WILL BE GREATER THAN 7. AN INDICATOR THAT CHANGES COLOR AT A pH GREATER THAN 7 WILL BE NEEDED, SUCH AS PHENOLPHTHALEIN WHICH CHANGES AT ABOUT pH=9.

  26. WHEN TITRATING A WEAK BASE WITH A STRONGACID, THE SOLUTION WILL BE ACIDIC AT THE EQUIVILENCE POINT, THEREFORE AN INDICATOR THAT CHANGES COLOR AT A pH BELOW 7 WILL BE NEEDED SUCH AS METHYL RED. IT CHANGES COLOR AT ABOUT PH=5 WHEN TITRATING A STRONG ACID WITH A STRONG BASE, EITHER OF THESE TWO INDICATORS WILL SUFFICE BECAUSE THE pH WILL BE AT 7 BUT THE CHANGE OCCURS OVER A RANGE THAT ENCOMPASSES BOTH OF THESE INDICATOR RANGES.

  27. IDENTIFYING CONJUGATE ACID-BASE PAIRS IN A BRONSTED-LOWRY ACID-BASE REACTION: WHEN AN ACID LOSES A PROTON, ITS PRODUCT IS THE CONJUGATE BASE. WHEN THE BASE THAT REACTS WITH THE ACID GAINS THE PROTON, ITS PRODUCT IS THE CONJUGATE ACID. Ex. HCl + H2O H3O+ + Cl- HERE, THE HCl GIVES THE PROTON TO WATER. THE CONJUGATE BASE IS Cl- (WHAT THE HCl BECAME AFTER LOSING ITS PROTON). THE BASE REACTANT WAS WATER BECAUSE IT WAS THE PROTON ACCEPTOR. IT BECAME H3O+ WHICH IS THE CONJUGATE ACID.

  28. ANOTHER EXAMPLE OF CONJUGATES: HC2H3O2 + OH- HOH + C2H3O2- HERE THE ACETIC ACID BECOMES THE ACETATE ION (WHICH IS THE CONJUGATE BASE) AND THE BASE (OH-) GAINS THE PROTON FROM THE ACID TO BECOME THE CONJUGATE ACID (WATER, HOH).

  29. PART 2, ACID - BASE EQUILIBRIA: WE WILL LOOK MORE IN DEPTH AT THE EQUILIBRIUM CONSTANTS FOR ACIDS AND BASES (Ka AND Kb) IN THIS SECTION. HB (aq) H+(aq) + B-(aq) THE GENERAL FORMAT FOR THE CALCULATION OF Ka FOR A WEAK ACID, AS SHOWN IN THE ABOVE CHEMICAL EQUATION IS: Ka = [H+] x [B-] [HB]

  30. ANOTHER EXAMPLE OF Ka CALCULATION: NH4+(aq) H+(aq) + NH3(aq) Ka = [NH3] x [H+] [NH4+] TEXT BOOKS CONTAIN TABLES OF ACID AND BASE DISSOCIATION CONSTANTS. THEY ARE ARRANGED BY DEACREASING ACID (OR BASE) STRENGTH. Ex. THE Ka FOR HF = 1 x 10-4, FOR HC2H3O2 = 1.8 x 10-5

  31. THE ACID EQUILIBRIUM CONSTANT : Ka • THE EQUILIBRIUM CONSTANT EXPRESSES TO WHAT EXTENT THE ACID DISSOCIATES IN WATER. THIS IN TURN INDICATES THE STRENGTH OF THE ACID. • THE STRONG ACIDS HAVE VERY LARGE Ka VALUES BECAUSE THEY COMPLETELY DISSOCIATE. • pKa IS ANALOGOUS TO pH: pKa= - log Ka

  32. EXAMPLE OF Ka CALCULATION: ASPIRIN IS A WEAK ACID. WE WILL REPRESENT IT BY HAsp. A SOLUTION OF 0.100 MOLAR SOLUTION IS PREPARED. THE CONCENTRATION OF H+ IN THIS SOLUTION IS FOUND TO BE 0.0057 M. CALCULATE THE Ka FOR THE ASPIRIN. HAsp (aq) H+(aq) + Asp-(aq)Ka = [H+] x [Asp-] [Hasp] FROM THE REACTION WE SEE THAT THE FOR EVERY MOLE OF HAsp DISSOCIATED, AN EQUAL NUMBER OF MOLES OF H+ISPRODUCED, AND AN EQUAL NUMBER OF MOLES OF Asp- IS PRODUCED.

  33. CONTINUED: IT HELPS TO MAKE A TABLE OF BEFORE AND AFTER CONCENTRATIONS IN ORDER TO CALCULATE THE Ka ORIGINAL CONC. CHANGE IN EQUIL. CONC M CONC. Hasp 0.100 -0.0057 0.0943 Asp- 0.000 +0.0057 0.0057 H+ 0.000 +0.0057 0.0057 Ka = (0.0057)2 = 3.4 x 10-4 0.0943

  34. WE CAN ALSO CALCULATE EQUILIBRIUM CONCENTRATIONS GIVEN THE EQUILIBRIUM CONSTANT, Ka. Ex. Nicotinic acid, HC6H4NO2 HAS A Ka= 1.4 x 10-5. DETERMINE THE [H+] IN A 0.10 M SOLUTION OF THE ACID. (ASSUMING 1 L). WE WILL REPRESENT NICOTINIC ACID WITH HNic. HNic(aq) H+(aq) + Nic-(aq) 1.4 x 10-5 = [H+] x [Nic-] [HNic] ORIG.CONC CHANGE EQUIL. CONC. HNic 0.10 -x 0.10 - x H+ 0.0 +x x Nic- 0.0 +x x

  35. CONTINUED: 1.4 x 10-5 = (x) (x) 0.10 - x NORMALLY THIS TYPE OF PROBLEM WOULD REQUIRE THE USE OF THE QUADRATIC EQUATION. WE CAN MAKE AN APPROXIMATION TO SIMPLIFY THE MATH. WE ASSUME THAT x << 0.10, AND WE REWRITE THE PROBLEM AS: 1.4 x 10-5 = x2 , x = 1.2 x 10-3 = [H+] = [Nic-] 0.10 ,[HNic] = 0.10 – 1.2 x 10-3 ~ 0.10 CONTINUED

  36. IF (PERCENT DISSOCIATION < 5% THEN WE CAN ASSUME THE APPROXIMATION THAT ‘ x IS SMALL’ IS A VALID ASSUMPTION. OTHERWISE WE HAVE TO SOLVE THE PROBLEM USING THE QUADRADIC EQUATION. 1.2 x 10-3 x 100 = 1.2 % < 5% 0.10 OUR ASSUMPTION WAS CORRECT, THEREFORE THE QUADRATIC EQUATION WAS NOT NEEDED HERE.

  37. Ex. 3 CALCULATE THE [H+] IN A 0.100 M SOLUTION OF NITROUS ACID, FOR WHICH Ka= 4.5 x 10-4. HNO2(aq) H+(aq) + NO2-(aq) 4.5 x 10-4 = x2 0.100 - x WE ASSUME x << 0.100, SOLVING GIVES x = 6.7 x 10-3 NOW CHECK THE ASSUMPION: 6.7 x 10-3 x 100 = 6.7% 0.100 OUR ASSUMPTION WAS NOT VALID, THEREFORE THE PROBLEM HAS TO BE SOLVED USING THE QUADRATIC EQUATION. CONTINUED……

  38. REWRITE THE EQUATION IN THE FORM, ax2 + bx + c = 0 x2 + 4.5 x 10-4 x - 4.5 x 10-5 = 0, where a =1, b = 4.5 x 10-4, c = -4.5 x 10-5 and X = -b + b2 – 4ac , - 4.5 x 10-4 + ( 4.5 x 10-4)2 – 4(-4.5 x 10-5) 2a 2 x = 6.5 x 10-3 and -6.9 x 10-3. [H+] cannot be negative so the correct answer is x = 6.5 x 10-3 ANOTHER METHOD CALLED SUCCESSIVE APPROXIMATIONS CAN ALSO BE USED TO CALCULATE THIS ANSWER, RATHER THAN THE QUADRATIC EQUATION. THIS IS ADDRESSED IN YOUR TEXT.

  39. BUFFER SOLUTIONS: THESE ARE SOLUTIONS THAT RESIST pH change EVEN UPON ADDITION OF A STRONG ACID OR BASE. TO PREPARE A BUFFER, AN AQUEOUS SOLUTION OF A WEAK ACID AND A SALT OF ITS CONJUGATE BASE ARE PREPARED: LIKE ACETIC ACID + SODIUM ACETATE. WE WILL BEGIN BY DETERMING THE [H+] IN A BUFFER SOLUTION. IN SOLVING THIS TYPE OF PROBLEM WE ASSUME THAT AT EQUILIBRIUM, VERY LITTLE OF THE WEAK ACID AND ITS CONJUGATE BASE ARE CONSUMED.

  40. Ex. 1. A BUFFER SOLUTION OF 1.00 MOLE OF LACTIC ACID, HLac + 1.00 MOLE OF SODIUM LACTATE, NaLac ARE DISSOLVED IN ENOUGH WATER TO MAKE ONE LITER OF SOLUTION. CALCULATE [H+] AND THE pH of THIS SOLUTION. THE Ka = 1.4 x 10-4. Ka = [H+] x [Lac-] , [H+] = Ka x [HLac] [Hlac] [Lac-] THE ORIGINAL CONCENTRATION OF HLac = 1.00 M AND THE ORIGINAL CONCENTRATION OF Lac- = 1.00 M BECAUSE SODIUM LACTATE IS A SALT AND COMPLETELY DISSOCIATES.

  41. THEREFORE, [H+] = 1.4 x 10-4 x 1.00 = 1.4 x 10-4 1.00 pH = - log (1.4 X 10-4) = 3.85 IN PREPARING A BUFFER SOLUTION, IT IS BEST TO CHOOSE A WEAK ACID - WEAK BASE PAIR WHERE THE pKa OF THE ACID IS CLOSE TO THE DESIRED pH AND THE ACID BASE PAIR CONCENTRATIONS ARE ROUGHLY EQUAL. WE LOOK AT A TABLE OF Ka AND pKa VALUES TO FIND THE THE BEST PAIRS.

  42. Ex. 2 • IF YOU WANT TO PREPARE A BUFFER SOLUTION OF ACETIC ACID AND ACETATE ION, WITH A pH OF 5.00, AND THE Ka = 1.8 x 10-5, • WHAT IS THE [H+] IN THIS BUFFER? • WHAT IS THE RATIO OF [HC2H3O2 ] / [C2H3O2-] IN THE BUFFER? • HOW MANY MOLES OF SODIIUM ACETATE SHOULD BE ADDED TO ONE LITER OF 0.100 M ACETIC ACID? • CONTINUED……..

  43. [H+] = 10-pH = 1 x 10 –5 • [H+] = Ka x [HC2H3O2] ; • [C2H3O2-] • [H+] = [HC2H3O2] = 1 x 10 –5=0.56 • Ka [C2H3O2-] 1.8 x 10-5 • TAKING THE EQUATION FROM ‘B’ ABOVE, AND SOLVING FOR [C2H3O2-], • [C2H3O2-] = 1.8 x 10-5 x 0.100 = 0.18 M • 1.0 x 10-5 • SO, 0.18 MOLES OF SODIUM ACETATE SHOULD BE ADDED TO THE 1 L OF ACETIC ACID SOLUTION.

  44. FROM THE PREVIOUS PROBLEM, HOW MANY GRAMS OF ACETIC ACID AND SODIUM ACETATE WOULD BE NEEDED? SINCE WE ALREADY DETERMINED THE MOLARITY OF THE ACETATE ION, WE CAN CONVERT THOSE MOLES TO GRAMS VERY SIMPLY. WE ALSO KNOW THE MOLARITY OF THE ACETIC ACID. THIS INFORMATION CAN ALSO BE USED TO CONVERT FROM MOLES TO GRAMS. 0.18 M NaC2H3O2 x 82 g / 1 mol = 14.8 = 15 g 0.10 M HC2H3O2 x 60. g / mol = 6.0 g

  45. CALCULATING CONCENTRATIONS OF A BUFFER SOLUTION AFTER THE ADDITION OF A STRONG ACID OR BASE: FOR THE LACTIC ACID BUFFER USED EARLIER, WHERE HLac AND Lac- = 1.00 M, Ka= 1.4 x 10-4, AND THE pH = 3.85, CALCULATE THE NEW CONCENTRATIONS AND THE pH AFTER ADDITION OF a) 0.10 M HCl AND b) 0.010 M NaOH. a) H+ + Lac- HLac [HLac] = 1.00 M + 0.10 M = 1.10 M [Lac-] = 1.00 M - 0.10 M = 0.90 M [H+] = Ka x [HLac] / [Lac-] = 1.4 x 10-4 x 1.10 / 0.90 = 1.7 x 10-4 pH = 3.77 (notice the very small decrease in pH).

  46. b) • HLac + OH- Lac- + H2O • [HLac] = 1.00 M - 0.10 M =0.90 M • [Lac-] = 1.00 M + 0.10 M = 1.10 M • [H+] = Ka x [HLac] / [Lac-] = 1.4 x 10-4 x 0.90 / 1.10 = 1.2 x 10-4 • pH = 3.92 (notice the very small increase in pH). THERE IS A LIMIT TO HOW MUCH OH- AND H+ A BUFFER CAN ABSORB WITHOUT CHANGING pH. LETS EXAMINE WHAT HAPPENS TO THE pH WHEN 0.50 M H+ IS ADDED………….

  47. Lac- + H+ Hlac [Lac- ] = 1.00 – 0.50 = 0.50 M [HLac] = 1.00 + 0.50 = 1.50 M [H+] = Ka x HLac / Lac- = 1.4 x 10-4 x 1.50 / 0.50 = 4.2 x 10-4 pH = -Log (4.2 x 10-4) = 3.38(NOTICE THIS IS A FAIRLY LARGE CHANGE FROM THE ORIGINAL 3.85) IN SUMMARY, TOCALCULATE THE pH OF A BUFFER SOLUTION AFTER ADDINGSTRONG ACID OR STRONG BASE:IF ADDINGSTRONG ACID, ADD THE MOLES OF H+ TO THE ACID CONCENTRATIONAND SUBTRACT IT FROM THE CONJUGATE BASE CONCENTRATION.IF ADDING STRONG BASE, SUBTRACT THAT # OF MOLES FROM THE ACID AND ADD IT TO THE CONJUGATE BASE.

  48. THE HENDERSON-HASSELBACK EQUATION: THIS EQUATION IS THE SAME EQUATION WE HAVE BEEN USING, TO SOLVE FOR [H+] BUT WITH THE pH PORTION BUILT IN TO THE EQUATION. ITS SIMPLE AND YOU MAY OR MAY NOT CHOOSE TO USE IT. [H+]= Ka x [HB] / [B-] Log [H+]= Log Ka + Log [HB] / [B-] pH = - Log Ka - Log [HB] / [B-] (FLIP RATIO TO MAKE +) pH = pKa + Log [B-] / [HB]

  49. EQUILIBRIUM CONSTANT FOR A WEAK BASE IN WATER: EVERYTHING IS DONE THE SAME WAY, EXCEPT WE LOOK AT A TABLE OF Kb VALUES INSTEAD OF Ka VALUES. WE MUST ALSO BE SURE TO WRITE THE CORRECT EQUATION FOR A WEAK BASE IN WATER IN ORDER TO SET UP THE SOLUTION FOR THE PROBLEM CORRECTLY: NH3 + H2O  NH4+ + OH- (THE CONCENTRATION OF WATER IS INCORPORATED INTO THE Kb VALUE) Kb = [OH-] x [NH4+] / [NH3] THE PROBLEMS ARE ANALOGOUS TO WHAT WE HAVE ALREADY DONE.

  50. Ka , Kb , AND Kw RELATED: SINCE Ka = [H+] x [B-] / [HB] AND Kb = [HB] x [OH-] / [B-] THEN Ka x Kb = [H+] x [B-]/ [HB] x [HB] x [OH-] /[B-] CANCELING LIKE TERMS: Ka x Kb = [H+] x [OH-] = Kw = 1 x 10-14 THEREFORE, IN THE SAME WAY WE CAN SOLVE FOR [H+] GIVEN THE [OH-], (OR VICE VERSA),WE CAN SOLVE FOR Ka OR Kb USING Kw.

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