1. Acids, Bases
&
Salts
Copyright Sautter 2003

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4. ACIDS, BASES & SALTS WHAT IS AN ACID ? WHAT IS A BASE ?
WHAT ARE THE PROPERTIES OF ACIDS AND BASES ?
WHAT ARE THE DIFFERENT KINDS OF ACIDS AND BASES ?
HOW ARE ACIDS AND BASES NAMED?

5. PROPERTIES OF ACIDS
CONTRARY TO COMMON BELIEF ACIDS DO NOT ATTACK ALL SUBSTANCES. MANY ARE VITAL TO OUR VERY EXISTENCE !
ALL ACIDS DO HOWEVER HAVE SEVERAL COMMON CHARACTERISTICS.
(1) ACIDS TASTE SOUR
(2) ACIDS TURN LITMUS RED (LITMUS IS A DYE THAT CHANGES COLOR DEPENDING ON ACIDITY)
(3) ACIDS REACT WITH ACTIVE METALS TO FORM HYDROGEN GAS
(4) ACIDS REACT WITH BASES TO FORM SALTS AND WATER
I’VE GOT
TOO MUCH
HCl !

6. PROPERTIES OF BASES
(1) BASES TASTE BITTER (MEDICINES ARE OFTEN BASES THUS THE TERM “BITTER MEDICINE”)
(2) BASES TURN LITMUS BLUE
(3) BASES FEEL SLIPPERY
(4) BASES REACT WITH ACIDS TO FORM SALTS AND WATER

7. DEFINITION OFACIDS AND BASES
ACIDS AND BASES AND THE REACTIONS WHICH RESULT CAN BE DESCRIBED USING SEVERAL DIFFERENT THEORIES.
THE THREE MOST COMMON THEORIES ARE:
(1) THE ARRENHIUS OR TRADITIONAL THEORY
(2) THE BRONSTED – LOWRY THEORY
(3) THE LEWIS THEORY
EACH OF THE THREE THEORIES VIEW ACIDS AND BASES SLIGHTLY DIFFERENTLY BUT THEY DO NOT CONTRADICT EACHOTHER IN ANY WAY. ONE MERELY EXPANDS ON THE OTHER !

8. THE ARRENHIUS OR TRADITIONAL ACID – BASE THEORY
AN ACID IS A SUBSTANCE WHICH RELEASES HYDROGEN IONS (H+) IN SOLUTION.
HNO3(aq)  H+(aq) + NO3-(aq)
A BASE IS A SUBSTANCE WHICH RELEASES HYDROXIDE IONS (OH-) IN SOLUTION.
NaOH(S)  Na+(aq) + OH-(aq)
WHEN AN ACID AND BASE REACT (A REACTION CALLED NEUTRALIZATION), A SALT AND WATER ARE FORMED.
HNO3(aq) + NaOH(aq)  NaNO3(aq) + H2O(l)
(acid) (base) (salt) (water)

9. COMMON ACIDS & BASES HYDROCHLORIC ACID (STOMACH ACID) – HCl
ACETIC ACID (VINEGAR) – HC2H3O2
CARBONIC ACID (SODA WATER) – H2CO3
SODIUM HYDROXIDE (DRAINO) – NaOH
AMMONIA WATER (CLEANING AGENT) – NH4OH
ALUMINUM HYDROXIDE (ROLAIDS) – Al(OH)3
MAGNESIUM HYDROXIDE (TUMS) – Mg(OH)2

10. THE BRONSTED – LOWRY ACID AND BASE THEORY
AN ACID IS A PROTON DONOR. A PROTON IN SOLUTION CONSISTS OF A HYDROGEN ION (H+). (HYDROGEN WITH AN ATOMIC NUMBER OF ONE AND A MASS NUMBER OF ONE HAS ONE PROTON, NO NEUTRONS AND AFTER LOSING ONE ELECTRON TO FORM AN ION, HAS NO ELECTRONS.)
A BASE IS A PROTON ACCEPTOR AND IT NEED NOT CONTAIN HYDROXIDE IONS.
AN ACID – BASE REACTION CONSISTS OF A PROTON TRANSFER FROM AN ACID TO A BASE. WHEN THIS OCCURS A NEW ACID AND BASE ARE FORMED. THIS IS BRONSTED- LOWRY NEUTRALIZATION.
HCl(aq) + H2O(aq)  H3O+(aq) + Cl-(aq)
(acid) (base) (new acid) (new base)

11. A CLOSER LOOK AT BRONSTED – LOWRY ACID – BASE REACTIONSH+ H+
HCl(aq) + H2O(aq)  H3O+(aq ) Cl-(aq)
acid
base
conjugate acid
conjugate acid
(1) WATER CAN ACT AS A BASE. AT TIMES IT CAN EVEN ACT AS A ACID.. THE TERM IS AMPHIPROTIC MEANS THAT IT CAN BE EITHER DEPENDING ON THE SITUATION.
WHEN WATER ACTS AS A BASE H3O+ ION IS FORMED. THIS CALLED HYDRONIUM ION.
THE ORIGINAL BASE (H2O) AFTER RECEIVING THE PROTON CAN NOW FUNCTION AS AN ACID IN THE REVERSE REACTION. HYDRONIUM ION IS CALLED THE CONJUGATE ACID OF THE BASE WATER IN THIS REACTION.
THE ORIGINAL ACID (HCl)AFTER LOSING THE PROTON CAN NOW FUNCTION AS AN BASE IN THE REVERSE REACTION. CHLORIDE ION IS CALLED THE CONJUGATE BASE OF THE ACID HYDROCHLORIC ACID IN THIS REACTION.

12. LEWIS ACID – BASE THEORY
THE LEWIS ACID – BASE THEORY EXPANDS THE ARRENHIUS AND BRONSTED LOWRY THEORIES TO INCLUDE EVEN MORE SUBSTANCES WHICH HAVE BEEN FOUND EXPERIMENTALLY TO BE ACIDIC OR BASIC BUT NOT COMPLETELY EXPLAINED BY EITHER.
THE LEWIS THEORY DESCRIBES ACIDS AS ELECTRON PAIR ACCEPTORS AND BASES AS ELECTRON PAIR DONORS. AS A RESULT THE OBSERVED ACIDIC PROPERTIES OF METAL IONS IN SOLUTION CAN BE EXPLAINED.
ADDITIONALLY, THE BASIC PROPERTIES OF SUBSTANCES SUCH AS AMMONIA CAN AS BE EXPLAINED AS ELECTRON PAIR DONORS EVEN THOUGH AMMONIA CONTAINS NO HYDROXIDE IONS.

13. WHERE DO ACIDS & BASES COME FROM?
ACIDS RESULT FROM THE ADDITION OF NONMETAL OXIDES TO WATER. THESE OXIDES ARE CALLED ACID ANHYDRIDES (ACIDS WITHOUT WATER). EVEN CARBON DIOXIDE WHEN ADDED TO WATER WILL MAKE THE SOLUTION MILDLY ACIDIC.
CO2(g) H2O(l)  H2CO3(aq) (CARBONIC ACID)
SO2(g) H2O(l)  H2SO3(aq) (SULFUROUS ACID)
BASES ARE FORMED BY METALLIC OXIDES AND WATER. THEY ARE CALLED BASIC ANHYDRIDES.
CaO(s) H2O(l)  Ca(OH)2(s) (CALCIUM HYDROXIDE)
Na2O(s) + H2O(l)  2 NaOH(s) (SODIUM HYDROXIDE)

14. ACID & BASE STRENGTH
WHEN DISSOLVED SUBSTANCES SEPARATE INTO FREE MOBILE IONS THIS IS CALLED DISSOCIATION.
THE STRENGTH OF ACIDS AND BASES DEPENDS ON THEIR ABILITY TO DISSOCIATE IN SOLUTION.
CONCENTRATION REFERS TO THE MOLARITY OF THE SOLUTION.
CONCENTRATION AND STRENGTH DO NOT MEAN THE SAME THING BUT ARE RELATED.
THERE ARE SEVERAL STRONG ACIDS AND BASES. THESE DISSOCIATE WELL (~ 100%). ALL OTHER ACIDS AND BASES ARE WEAK (DISSOCIATE POORLY)

15. COMMON STRONG ACIDS STRONG ACIDS HCLO4 PERCHLORIC ACID
HI HYDROIODIC ACID
HBr HYDROBROMIC ACID
HCl HYDROCHLORIC ACID
HNO NITRIC ACID
H2SO SULFURIC ACID

16. COMMON STRONG BASES STRONG BASES LiOH LITHIUM HYDROXIDE
NaOH SODIUM HYDROXIDE
KOH POTASSIUM HYDROXIDE
RbOH RUBIDIUM HYDROXIDE
CsOH CESIUM HYDROXIDE
Ca(OH)2 CALCIUM HYDROXIDE
Sr(OH)2 STRONTIUM HYDROXIDE
Ba(OH)2 BARIUM HYDROXIDE

17. PH OF SOLUTIONS
PH IS A CONVENIENT SYSTEM FOR THE MEASURING THE ACIDITY OF A SOLUTION.
PH IS DEFINED AS THE NEGATIVE LOGARITHM OF THE HYDROGEN ION CONCENTRATION IN A SOLUTION.
A LOGARITHM (LOG) IS A POWER OF 10. IF A NUMBER IS WRITTEN AS 10X THEN ITS LOG IS X.
FOR EXAMPLE 100 COULD BE WRITTEN AS 102 THEREFORE THE LOG OF 100 IS 2.
IN CHEMISTRY CALCULATIONS OFTEN SMALL NUMBERS ARE USED LIKE OR THE LOG OF IS THEREFORE –4.
FOR NUMBERS THAT ARE NOT NICE EVEN POWERS OF 10 A CALCULATOR IS USED TO FIND THE LOG VALUE. FOR EXAMPLE THE LOG OF IS –2.46 AS DETERMINED BY THE CALCULATOR.

18. PH OF SOLUTIONS (CONT’D)
PH = - LOG [H+]
P MEANS NEGATIVE LOG AND THE BRACKETS AROUND H+ MEANS “CONCENTRATION OF H+”
SAMPLE PROBLEM:
WHAT IS THE PH OF A SOLUTION WHEN ITS [H+] = M ?
SOLUTION:
= 10-6
PH = - LOG (10-6) = - ( -6.00) = 6.00

19. THE PH SCALE 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 PH , [H+] AND [OH-]
ACID RANGE
BASE RANGE
NEUTRAL
[H+] = 1.00 x 10-7 M IN PURE WATER
PH = - LOG (1.00 x 10-7) = 7.00
LOW PH VALUES INDICATE HIGH ACIDITY
HIGH PH VALUE INDICATE HIGH BASITY
7.00 IS THE PH OF PURE WATER
PH , [H+] AND [OH-]
PH , [H+] AND [OH-]

20. * POH OF SOLUTIONS POH = - LOG [OH-]
WHERE THE BRACKETS AROUND OH- MEANS “CONCENTRATION OF OH-”
SAMPLE PROBLEM:
WHAT IS THE POH OF A SOLUTION WHEN ITS [OH-] = M ?
SOLUTION:
= 10-5
POH = - LOG (10-5) = - ( -5.00) = 5.00
PH + POH = PKw = 14.0
PH = 14.0 –5.0 = 9.0, THE SOLUTION IS BASIC

21. FORMATION OF A HYDROGEN ION (AN AQUEOUS PROTON)
Atomic number 1
Atomic mass + H
A HYDROGEN ATOM
LOSES ITS ELECTRON
TO FORM A
HYDROGEN ION
1e-
ONLY A PROTON
REMAINS
1 P+
0 N0

22. STRONG & WEAK ACID DISSOCIATION
STONG ACIDS DISSOCIATE READILY
(NITRIC ACID HNO3) +
FREE MOBILE IONS
READILY FORM
ALL MOLECULES
DISSOCIATE - H NO NO
(aq) 3
(aq) 3
WEAK ACIDS DISSOCIATE POORLY
(HYDROFLOURIC ACID HF)
FREE MOBILE IONS
FORM BUT WITH
DIFFICULTLY
FEW MOLECULES
DISSOCIATE + - F H F F F
(aq)
(aq)

23. COMPARISON OF ACID AND BASE STRENGTHS FOR SEVERAL ACIDS
HClO4  H+ + ClO4-
HCl  H+ + Cl-
HF  H+ + F-
HCOOH  H+ + HCOO-
HC2H3O2  H+ + C2H3O2-
NH4+  H+ + NH3
STRONGEST
ACID
WEAKEST
CONJUGATE
BASE
BASE
GETS S T R O N G E
ACID
GETS S T R O N G E
WEAKEST
CONJUGATE
ACID
STRONGEST
BASE

24. ACID- BASE NEUTRALIZATION H+(aq) + OH-(aq)  H2O(l)

25. NAMING OF ACIDS (NOMENCLATURE)
HELLO
MY NAME IS:
NAMING OF ACIDS (NOMENCLATURE)
ACIDS ARE OF TWO TYPES FOR NAMING PURPOSES
(1) BINARY ACIDS – CONSIST OF TWO ELEMENTS ONE OF WHICH IS HYDROGEN
(2) TERNARY ACIDS – CONSIST OF THREE ELEMENTS ONE OF WHICH IS HYDROGEN AND ANOTHER OXYGEN. THESE ARE ALSO CALLED OXYACIDS

26. NAMING BINARY ACIDS
BINARY ACIDS ARE COMPOSED OF TWO ELEMENTS NOT NECESSARILY JUST TWO ATOMS. FOR EXAMPLE, H2S CONSISTS OF JUST HYDROGEN AND SULFUR AND IS THEREFORE BINARY EVEN THOUGH IT HAS THREE ATOMS !
ALL BINARY ACID NAMES BEGIN WITH “HYDRO” AND END IN “IC”. THE NAME OF THE NON HYDROGEN ELEMENT LIES INBETWEEN THE PREFIX AND SUFFIX OF THE NAME.
FOR EXAMPLE, H2S IS NAMED “HYDROSULFURIC ACID”
HBr IS NAMED “HYDROBROMIC ACID”

27. NAMING TERNARY (OXYACIDS)
TERNARY ACIDS ARE COMPOSED OF THREE ELEMENTS NOT NECESSARILY JUST THREE ATOMS. FOR EXAMPLE, H2SO4 IS COMPOSED OF HYDROGEN, SULFUR AND OXYGEN EVEN THOUGH IT CONTAINS SEVEN ATOMS.
NAMING TERNARY ACIDS REQUIRES THAT YOU KNOW HOW TO NAME THE ANION THAT IS CONTAINED IN THE ACID. IN H2SO4 THE NEGATIVE ION IS SO4-2, SULFATE ION.
WHEN THE ANION NAME ENDS IN “ATE” THE ACID NAME ENDS IN “IC”.
THE NAME FOR H2SO4 IS THEREFORE “SULFURIC ACID”
THE NAME FOR HNO3,WHICH CONTAINS THE NITRATE ION, NO3- , IS “NITRIC ACID”

28. NAMING TERNARY (OXYACIDS) (CONT’D)
FOR EXAMPLE H2SO3 CONTAINS THE ANION SO3-2 , SULFITE ION. THE ACID NAME THEREFORE IS “SULFUROUS ACID”
WHEN THE NAME OF THE ANION CONTAINED IN THE ACID ENDS IN “ITE” THE ACID NAME ENDS IN “OUS”.
THE NAME FOR HNO2, WHICH CONTAINS NO2-, THE NITRITE ION IS “NITROUS ACID”

29. ADVANCED ACID - BASE
CONCEPTS

30. ADVANCED ACID BASE CHEMISTRY (ACID CONSTANTS)*
ADVANCED ACID BASE CHEMISTRY (ACID CONSTANTS)
A MEASURE OF ACID STRENGTH IS THE EQUILIBRIUM CONSTANT (Ka). LIKE ALL EQUILIBRIUM CONSTANTS IT IS CALCULATED BY DIVIDING EQUILIBRIUM PRODUCT CONCENTRATIONS BY REACTANT CONCENTRATIONS.
THIS MEANS THAT THE SIZE OF Ka WILL INDICATE RELATIVE ACID STRENGTH. SINCE PRODUCT CONCENTRATIONS ARE PLACED IN THE NUMERATOR OF THE CALCULATION, AS THEY INCREASE, Ka ALSO INCREASES. IF Ka IS SMALL, FEW PRODUCTS ARE FORMED.

31. ADVANCED ACID BASE CHEMISTRY ACID CONSTANTS (CONT’D)*
THE STRENGTH OF AN ACID DEPENDS ON THE DEGREE OF DISSOCIATION OF THE ACID (CONCENTRATIONS OF PRODUCT IONS FORMED)
AS Ka INCREASES, ACID STRENGTH INCREASES
SINCE THE STRENGTH OF AN ACID IS INVERSELY RELATED TO THE STRENGTH OF ITS CONJUGATE BASE, AS Ka FOR AN ACID INCREASES, THE STRENGTH OF ITS CONJUGATE DECREASES.
BASES STRENGTHS ARE OFTEN MEASURED BY Kb VALUES.
Kw = Ka x Kb
(REMEMBER Kw ALWAYS EQUALS 1.00 x 10-14)

32. ADVANCED ACID BASE CHEMISTRY ACID CONSTANTS (CONT’D)*
ACID CONSTANT FOR SOME COMMON ACIDS
ACID Ka
HYDROCHLORIC UNDEFINED
NITRIC UNDEFINED
HYDROFLOURIC x 10-4
ACETIC x 10-5
FORMIC x 10-4
BENZOIC x 10-5
STRONG
ACIDS
WEAKEST
ACID
WEAK
ACIDS
AS Ka , ACID GETS WEAKER
UNDEFINED INDICATES A VERY LARGE VALUE
& ACID IS A STRONG ACID (100% DISSOCIATION)

33. CALCULATING HYDROGEN AND HYDROXIDE CONCENTRATIONS*
PURE WATER CONTAINS VERY SMALL CONCENTRATIONS OF BOTH HYDROGEN AND HYDROXIDE IONS. IN PURE WATER THE [H+] = [OH-] AND BOTH EQUAL 1.00 X 10-7 MOLAR
USING THIS FACT THE EQUILIBRIUM CONSTANT (Kw) FOR THE DISSOCIATION OF PURE WATER INTO HYDROGEN AND HYDROXIDE ION CAN BE CALCULATED AS 1.00 x 10-14
H2O(l)  H+(aq) + OH-(aq)
Kw = [H+] x [OH-]
(1.00 x 10-7)(1.00 x 10-7) = 1.00 x 10-14

34. *
PH OF A STRONG ACID
WHAT IS THE PH OF 2.0 LITERS OF NITRIC ACID SOLUTION WHICH CONTAINS GRAMS OF THE ACID?
SOLUTION: NITRIC ACID (HNO3) IS A STRONG AND DISSOCIATES COMPLETELY.
HNO3  H+ + NO3-
THE MOLAR MASS OF HNO3 IS 63.0 GRAMS
MOLES OF ACID = GRAMS / 63.0 = 0.25 MOLES
[H+] = 0.25 MOLES / 2.0 LITERS = M
PH = - LOG [H+] = - LOG (0.125) = 0.903

35. CALCULATING HYDROGEN AND HYDROXIDE CONCENTRATIONS (CONT’D)*
USING THE Kw CONSTANT FOR THE DISSOCIATION OF PURE WATER WE CAN CALCULATE THE [H+] IF THE [OH-] IS KNOWN OR VISE VERSA.
FOR EXAMPLE:
WHAT IS THE CONCENTRATION OF HYDROGEN ION IN A SOLUTION WITH THE [OH-] = 2.0 x 10-5 M ?
Kw = [H+] x [OH-] = 1.00 x 10-14
[H+](2.00 x 10-5) = 1.00 x 10-14, [H+] = 5.00 x M
THE SOLUTION IS ACIDIC.
SOLUTIONS WITH [H+] > 1.00 x 10-7 M AND [OH-] < 1.00 x 10-7 M ARE ACIDIC.
SOLUTIONS WITH [H+] < 1.00 x 10-7 M AND [OH-] > 1.00 x 10-7 M ARE BASIC.

36. PH OF WEAK ACIDS (CALCULATIONS USING Ka)*
FOR A WEAK ACID HX
HX(aq)  H+(aq) X-(aq)
K a = [H+] x [X-] AND PH = -LOG [H+]
[HX]
WHAT IS THE PH OF A 0.10 M SOLUTION OF ACETIC ACID (Ka = 1.8 x 10-5 ) ?
BASED ON THE BALANCED EQUATION FOR EVERY H+ FORMED AN X- IS ALSO FORMED.
IF [H+] = X THEN [X-] = X AND [HX] = X
ADDITIONALLY IF Ka IS VERY SMALL THEN THE ACID IS VERY WEAK AND [H+] IS VERY SMALL THEREFORE –X ~ 0.10 M
1.8 x 10-5 = ( X2 / 0.10)
[H+] = X = ((1.8 x 10-5 )(0.10))1/2 = 1.3 x 10-3 M
PH = - LOG (1.3 x 10-3 ) = 2.87

37. CALCULATIONS INVOLVING WEAK BASES*
THE PH OF A 0.10 M AMMONIA SOLUTION IS WHAT IS THE Kb FOR NH3 ?
NH3(aq) H2O  NH4+(aq) + OH-(aq)
PH + POH =14.0, POH = 14.0 – = 2.87
Kb = [NH4+] x [OH-] AND POH – LOG [OH-]
[NH3]
[OH-] = = 1.35 x 10-3 M, FROM THE EQUATION FOR EACH OH- ONE NH4+ ALSO FORMS SO [NH4+] = 1.35 x 10-3 M AND [NH3] = 0.10 – 1.35 x 10-3 = .0986
Kb = ( 1.35 x 10-3) 2 / (.0986) = 1.8 x 10-5

38. TITRATION
TITRATION REFERS TO THE ADDITION OF AN ACID AND BASE IN MEASURED QUANTITIES. OFTEN THE TITRATION IS CARRIED OUT TO AN END POINT. THE END POINT IS THE POINT WHERE THE MOLES OF ADDED ACID AND BASE ARE EQUAL. THE END POINT IS NOT ALWAYS THE NEUTRAL POINT
WHEN STRONG ACIDS ARE TITRATED WITH STRONG BASES TO THE END POINT A NEUTRAL SOLUTION RESULTS (PH =7.00)
WHEN STRONG ACIDS ARE TITRATED WITH WEAK BASES TO THE END POINT AN ACIDIC SOLUTION RESULTS (PH < 7.00)
WHEN STRONG BASES ARE TITRATED WITH STRONG BASES TO THE END POINT A BASIC SOLUTION RESULTS (PH > 7.00)
WHEN WEAK ACIDS AND BASES ARE TITRATED TO THE END POINT THE RELATIVE STRENGTHS OF EACH DETERMINES THE ACIDITY OF THE RESULTING SOLUTION.

39. TITRATION (CONT’D) END POINT SOLUTIONS STRONG ACID (HCl) STRONG
BASE (KOH)
NEUTRAL
PH = 7.00
ACIDIC
PH < 7.00
WEAK
BASE (NH3)
WEAK
ACID (HF)
BASIC
PH > 7.00
ACIDIC
BASIC OR
NEUTRAL

40. NORMALITY AND TITRATION*
NORMALITY IS A SYSTEM OF MEASURING THE CONCENTRATION OF SOLUTIONS WHICH IS OFTEN USED IN TITRATIONS.
THE NORMALITY OF AN ACID IS EQUAL TO THE MOLES OF HYDROGEN IONS AVAILABLE PER LITER OF SOLUTION.
THE NORMALITY OF A BASE IS EQUAL TO THE MOLES OF HYDROXIDE IONS AVAILABLE PER LITER OF SOLUTION.
NACID = MOLES OF H+ IONS / LITER
NBASE = MOLES OF OH- IONS / LITER

41. NORMALITY AND TITRATION (CONT’D)*
NORMALITY AND TITRATION (CONT’D)
A MOLE OF H+ IONS OR OH- IONS IS CALLED AN EQUIVALENT. THEREFORE NORMALITY MAY BE DEFINED AS EQUIVALENTS PER LITER.
THE NORMALITY OF AN ACID CAN BE RELATED TO ITS MOLARITY BY THE NUMBER OF REPLACEABLE H+ IONS CONTAINED IN THE ACID.
1 MOLAR HCl ~ NORMAL HCl
1 MOLAR H2SO4 ~ NORMAL H2SO4
1 MOLAR H3PO4 ~ NORMAL H3PO4

42. NORMALITY AND TITRATION (CONT’D)*
THE NORMALITY OF AN BASE CAN BE RELATED TO ITS MOLARITY BY THE NUMBER OF REPLACEABLE OH- IONS CONTAINED IN THE BASE
1 MOLAR NaOH ~ NORMAL NaOH
1 MOLAR Ca(OH)2 ~ NORMAL Ca(OH)2
1MOLAR Al(OH)3 ~ NORMAL Al(OH)3

43. TITRATION CALCULATIONS (STRONG ACID – STRONG BASE)
FIND THE PH OF A SOLUTION OBTAINED BY MIXTURE 100 MLS OF HCl 0.10 M WITH 50 MLS OF NaOH 0.10 M.
SOLUTION:
MOLARITY x LITERS = MOLES
ACID (HCl) x = (H+)
BASE (NaOH) x = (OH-)
H+(aq) + OH-(aq)  H2O(l) EXTRA H+
[H+] REMAINING = 0.005MOLES / ( ) L
[H+] = .033 M, PH = -LOG(0.033) = 1.48

44. TITRATION CALCULATIONS (CONT’D) (STRONG ACID – STRONG BASE)
FIND THE PH OF A SOLUTION OBTAINED BY MIXTURE 100 MLS OF HCl 0.10 M WITH 100 MLS OF NaOH 0.10 M.
SOLUTION:
MOLARITY x LITERS = MOLES
ACID (HCl) x = (H+)
BASE (NaOH) x = (OH-)
H+(aq) + OH-(aq)  H2O(l) EXTRA H+
WHEN MOLES OF H+ = MOLES OF OH- THEN SOLUTION IS NEUTRAL AND THE PH MUST BE 7.00
THIS IS TRUE FOR ALL STRONG ACID – STRONG BASE TITRATION END POINTS.

45. Titration of Strong Acid vs Strong Base13 12 11 10 9 8 7 6 5 4 3 2 1
0.10 M HCl M NaOH pH
Equivalent point
moles acid = moles base
Strong acid – strong base
Titration equivalent point at
pH = 7.00
Volume of Base Added

46. TITRATION CALCULATIONS (WEAK ACID – STRONG BASE)*
FIND THE PH OF A SOLUTION OBTAINED BY MIXTURE 100 MLS OF HAc 0.10 M WITH 25 MLS OF NaOH 0.10 M.(Hac = HC2H3O2 ACETIC ACID)
SOLUTION: EACH MOLE OF ADDED BASE CONSUMES A MOLE OF ACID AND FORMS A MOLE OF SALT (AC- ION)
MOLARITY x LITERS = MOLES
ACID (HAc) x = (HAc)
BASE (NaOH) x = (OH-)
H+(aq) + OH-(aq)  H2O(l) EXTRA HAc
[HAc] REMAINING = MOLES / ( ) L
[HAc] = M (A WEAK ACID)
MOLES SALT FORMED = MOLES OF BASE ADDED
[Ac-] = MOLES SALT / LITER
[Ac-] = MOLES / L = M

47. TITRATION CALCULATIONS (CONT’D) (WEAK ACID – STRONG BASE)*
TITRATION CALCULATIONS (CONT’D) (WEAK ACID – STRONG BASE)
Ka = 1.8 x 10-5, [H+] = X
HAc  H+ + Ac-, Ka = ([H+] x [Ac-]) / [HAc]
[H+] = Ka x [HAc] = (1.8 x 10-5)(0.006) = 5.4 x10-6 M
[Ac-]
PH = - LOG [H+] = - LOG (5.4 x10-6 ) = 5.26

48. TITRATION CALCULATIONS (CONT’D) (WEAK ACID – STRONG BASE)*
FIND THE PH OF A SOLUTION OBTAINED BY MIXTURE 100 MLS OF HAc 0.10 M WITH 100 MLS OF NaOH 0.10 M.(Hac = HC2H3O2 ACETIC ACID)
SOLUTION: EACH MOLE OF ADDED BASE CONSUMES A MOLE OF ACID AND FORMS A MOLE OF SALT (AC- ION)
MOLARITY x LITERS = MOLES
ACID (HAc) x = (HA
BASE (NaOH) x = (OH-)
H+(aq) + OH-(aq)  H2O(l) EXTRA HAc
MOLES SALT FORMED = MOLES OF BASE ADDED
[Ac-] = MOLES SALT / LITER
[Ac-] = MOLES / L = M (ONLY A SALT SOLUTION REMAINS)

49. TITRATION CALCULATIONS (CONT’D) (WEAK ACID – STRONG BASE)*
SINCE ONLY A SALT SOLUTION IS PRESENT THE QUESTION NOW BECOMES, “WHAT IS THE PH OF A M SOLUTION OF NaAc ?”
NaAc  Na+ + Ac- (ALKALI SALTS DISSOCIATE COMPLETELY)
Na+ CAN ACT AS NEITHER ACID NOR BASE. IT CAN’T ACCEPT PROTONS (BOTH IT AND A PROTON ARE POSITIVE) AND IT HAS NO H+ IONS TO LOSE.
Ac- CAN’T ACT AS AN ACID (IT HAS NO H+ IONS) BUT BEING THE CONJUGATE BASE OF A WEAK ACID (ACETIC ACID) IT CAN ACCEPT PROTONS AND ACT AS A BASE.
Ac- + H2O  HAc + OH- (THE FORMATION OF HYDROXIDE ION MAKES THE SOLUTION BASIC)
THE SALT OF A WEAK ACID ANION AND A STRONG BASE CATION FORMS A BASIC SOLUTION.

50. TITRATION CALCULATIONS (CONT’D) (WEAK ACID – STRONG BASE)*
EQUATIONS:
(1) HAc + NaOH  NaAc H2O
(2) NaAc  Na+ + Ac-
(3) Ac- + H2O  HAc + OH-
Kw = Ka x Kb, Kb = Kw / Ka , Kb = 1.0 x / 1.8 x 10-5
Kb = 5.56 x = [HAc] x [OH-] = X x X = X2
[Ac-]
X = [OH-] = 6.67 x 10-6 M,
POH = - LOG (6.67 x 10-6 ) = 5.17, PH = 14.0 – POH
PH = 8.82 (BASIC SOLUTION)

51. Titration of Weak Acid vs Strong Base *13 12 11 10 9 8 7 6 5 4 3 2 1
0.10 M HAc M NaOH
Equivalent point
moles acid = moles base pH
Weak acid – strong base
Titration equivalent point at
pH > 7.00
Buffer
region
Volume of Base Added

52. NORMALITY AND TITRATION (CONT’D)*
AT THE END POINT OF ACID – BASE TITRATION (ALSO CALLED EQUIVALENCE POINT), MOLES OF H+ IONS ADDED FROM THE ACID EQUAL MOLES OF OH- IONS ADDED FROM THE BASE.
NACID = MOLES H+ / LITERS
MOLES H+ = NACID x LITERS
NBASE = MOLES OH- / LITERS
MOLES OH- = NBASE x LITERS
AT ENDPOINT MOLES H+ = MOLES OH-
THEREFORE: NACID x VOLACID = NBASE x VOLBASE

53. BUFFERS WHAT IS A BUFFER?
A WEAK ACID AND ITS SALT OR A WEAK BASE AND ITS SALT.
WHAT DOES A BUFFER DO?
A BUFFER SOLUTION RESISTS PH CHANGES WHEN SMALL QUANTITIES OF ACID OR BASE ARE ADDED.
HOW DO BUFFERS WORK?
THEY USE THE EQUILIBRIUM CONCEPTS DESCRIBED BY LE CHATELIER’S PRINCIPLE.

54. BUFFERS (CONT’D)
LE CHATELIER’S PRINCIPLE STATED THAT A SYSTEM AT EQUILIBRIUM CONSUME ADDED REACTANTS OR PRODUCTS BY SHIFTING AWAY FROM THE ADDED COMPONENT AND WILL REPLACE REMOVED REACTANT OR PRODUCT BY SHIFTING TOWARDS THE REMOVED COMPONENT.
IN A BUFFER AN EQUILIBRIUM EXISTS BETWEEN A WEAK ACID , IT ANION AND THE HYDROGEN ION.
HAc  H Ac-
weak acid hydrogen ion acid anion (conjugate base)
ADDING ACID (H+) WILL SHIFT THE SYSTEM LEFT THEREBY CONSUMING ADDED ACID ALONG WITH THE ANION (Ac-) AND FORMING MORE WEAK ACID (HAc)
ADDING BASE (OH-) WILL DECREASE H+ ION CONCENTRATION AND SHIFT THE SYSTEM RIGHT CREATING REPLACEMENT H+ IONS ALONG WITH Ac- IONS AND THEREBY CONSUME THE WEAK ACID (HAc)

55. BUFFERS (CONT’D)
SUMMARY OF EFFECTS OF ADDING AN ACID TO AN ACIDIC BUFFER:
[WEAK ACID] , [CONJUGATE BASE] , SYSTEM SHIFTS LEFT (TOWARDS REACTANTS)
SUMMARY OF EFFECTS OF ADDING A BASE TO AN ACIDIC BUFFER:
[WEAK ACID] , [CONJUGATE BASE] , SYSTEM SHIFTS RIGHT (TOWARDS PRODUCTS)
ADDING ACID OR BASE TO A BASIC BUFFER (WEAK BASE AND ITS SALT) WILL HAVE EXACTLY THE OPPOSITE EFFECT ON THE WEAK BASE AND ITS CONJUGATE ACID (CATION).

56. BUFFERS (CONT’D) *
WHAT IS THE PH OF A BUFFER THAT CONSISTS OF 0.10 M HAc AND 0.05 M NaAc ?
HAc  H+ + Ac-, Ka = [H+] x [Ac-]
[HAc]
[SALT] = [Ac-] = 0.05 M, [ACID WEAK] = [HAc] =0.10 M
[H+] = Ka x [HAc] = (1.8 x 10-5) (0.10) = 3.6 x10-5 M
[Ac-]
PH = - LOG (3.6 x10-5) = 4.44

57. ADDING ACID TO AN UNBUFFERED SYSTEM*
HOW DOES THE PH OF A LITER OF WATER CHANGE WHEN MOLES OF HCl ARE ADDED?
SOLUTION:
PURE WATER HAS A PH = 7.00
HCl IS A STRONG ACID (100% DISSOCIATION)
HCl  H+ + Cl- , [H+] = MOLES / 1.0 LITER
PH = - LOG (0.0001) = 4.00
THE CHANGE IN PH IS FROM 7.00 TO 4.00 OR 3.00 PH UNITS. THIS CHANGE MEANS THE SYSTEM HAS BECOME 1000 TIMES MORE ACIDIC (103 = 1000)

58. ADDING ACID TO AN BUFFERED SYSTEM*
HOW DOES THE PH OF A LITER OF A BUFFER COMPOSED OF 0.10 M HAc AND 0.05 M NaAc CHANGE WHEN MOLES OF HCl ARE ADDED?
SOLUTION:
HAc  H+ + Ac-, ADDING AN ACID SHIFTS THE SYSTEM TO THE LEFT.
MOLARITY x LITERS = MOLE (ORIGINAL MOLES)
x = MOLES OF HAc
x = MOLES NaAc (Ac-)
WHEN EQUILIBRIUM SHIFTS THE MOLES OF HAc WILL INCREASE BY THE NUMBER OF MOLES OF ADDED ACID AND MOLES OF Ac- WILL DECREASE BY THE NUMBER OF MOLES OF ADDED ACID.
THEREFORE AT EQUILIBRIUM:
MOLES OF HAc = =
MOLES OF Ac- = 0.05 – =

59. ADDING ACID TO AN UNBUFFERED SYSTEM*
[HAc]eq = MOLES / 1 LITER = M
[Ac-]eq = MOLES / 1 LITER = M
[H+]eq = X
Ka = [H+] x [Ac-] , [H+] = Ka x [HAc]
[HAc] [Ac-]
[H+] = (1.8 x 10-5) (0.1001) = 3.61 x10-5 M
0.0499
PH = 4.442, PH OF THE ORGINAL BUFFER PREVIOUSLY CALCULATED = 4.444
THE PH OF THE BUFFERED SOLUTION HARDLY CHANGES WITH THE ADDED ACID WHILE WHEN THE SAME QUANTITY OF ACID IS ADDED TO PLAIN WATER THE PH CHANGES DRAMATICALLY. A BUFFER STABILIZES PH.

60. THE END