Transcript ACIDS, BASES & SALTS

Copyright Sautter 2003
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?
PROPERTIES OF ACIDS
• CONTRARY TO COMMON BELIEF ACIDS DO NOT
ATTACK ALL SUBSTANCES. MANY ARE VITAL TO
OUR VERY EXISTENCE !
I’VE GOT
• ALL ACIDS DO HOWEVER HAVE SEVERAL TOO MUCH
HCl !
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
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
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 TRADITATIONAL 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 !
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(aq)
(acid)
(base)
(salt)
(water)
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
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)
A CLOSER LOOK AT BRONSTED –
LOWRY ACID – BASE REACTIONS
HCl(aq) + H2O(aq)  H3O+(aq
(aq)
acid
(1)
(2)
(3)
(4)
base
conjugate acid
)
+ Clconjugate acid
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 SITUATON.
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.
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.
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)
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)
COMMON STRONG ACIDS
• STRONG ACIDS
•
HCLO4
HI
HBr
HCl
HNO3
H2SO4
PERCHLORIC ACID
HYDROIODIC ACID
HYDROBROMIC ACID
HYDROCHLORIC ACID
NITRIC ACID
SULFURIC ACID
COMMON STRONG BASES
• STRONG BASES
•
•
•
•
•
•
•
•
LiOH
NaOH
KOH
RbOH
CsOH
Ca(OH)2
Sr(OH)2
Ba(OH)2
LITHIUM HYDROXIDE
SODIUM HYDROXIDE
POTASSIUM HYDROXIDE
RUBIDIUM HYDROXIDE
CESIUM HYDROXIDE
CALCIUM HYDROXIDE
STRONTIUM HYDROXIDE
BARIUM HYDROXIDE
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 .0001 OR 10-4. THE LOG OF .0001 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 .00345 IS –2.46 AS DETERMINED
BY THE CALCULATOR.
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+] = 0.000001 M ?
SOLUTION:
0.000001 = 10-6
PH = - LOG (10-6) = - ( -6.00) = 6.00
THE PH SCALE
0 1 2
3
4
5
6
7 8
9 10 11 12 13 14
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+]
PH , [H+]
AND [OH-]
AND [OH-]
POH OF SOLUTIONS
• POH = - LOG [OH-]
•
WHERE THE BRACKETS AROUND OHMEANS “CONCENTRATION OF OH-”
• SAMPLE PROBLEM:
• WHAT IS THE POH OF A SOLUTION WHEN
ITS [OH-] = 0.00001 M ?
• SOLUTION:
• 0.00001 = 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
FORMATION OF A HYDROGEN ION
(AN AQUEOUS PROTON)
Atomic number
1
Atomic mass
1
A HYDROGEN ATOM
LOSES ITS ELECTRON
TO FORM A
HYDROGEN ION
1e-
1 P+
0 N0
STRONG & WEAK ACID
DISSOCIATION
STONG ACIDS DISSOCIATE READILY
(NITRIC ACID HNO3)
FREE MOBILE IONS
READILY FORM
ALL MOLECULES
DISSOCIATE
WEAK ACIDS DISSOCIATE POORLY
(HYDROFLOURIC ACID HF)
FREE MOBILE IONS
FORM BUT WITH
DIFFICULTLY
FEW MOLECULES
DISSOCIATE
COMPARISON OF ACID AND BASE
STRENGTHS FOR SEVERAL ACIDS
STRONGEST
ACID
ACID
GETS
S
T
R
O
N
G
E
R
WEAKEST
CONJUGATE
BASE
HClO4  H+ + ClO4HCl  H+ + ClHF  H+ + FHCOOH  H+ + HCOOHC2H3O2  H+ + C2H3O2NH4+  H+ + NH3
WEAKEST
CONJUGATE
ACID
STRONGEST
BASE
BASE
GETS
S
T
R
O
N
G
E
R
ACID- BASE NEUTRALIZATION
H+(aq) + OH-(aq)  H2O(l)
ACID
BASE
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
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”
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”
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”
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.
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)
ADVANCED ACID BASE CHEMISTRY
ACID CONSTANTS (CONT’D)
• ACID CONSTANT FOR SOME COMMON ACIDS
ACID
STRONG
ACIDS
WEAK
ACIDS
HYDROCHLORIC
NITRIC
HYDROFLOURIC
ACETIC
FORMIC
BENZOIC
Ka
UNDEFINED
UNDEFINED
6.7 x 10-4 WEAKEST
1.8 x 10-5
ACID
1.8 x 10-4
6.0 x 10-5
AS Ka , ACID GETS WEAKER
UNDEFINED INDICATES A VERY LARGE VALUE
& ACID IS A STRONG ACID (100% DISSOCIATION)
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
PH OF A STRONG ACID
• WHAT IS THE PH OF 2.0 LITERS OF NITRIC ACID
SOLUTION WHICH CONTAINS 15.75 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 = 15.75 GRAMS / 63.0 = 0.25 MOLES
• [H+] = 0.25 MOLES / 2.0 LITERS = 0.125 M
• PH = - LOG [H+] = - LOG (0.125) = 0.903
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 10-10 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.
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] = 0.10 - X
ADDITIONALLY IF Ka IS VERY SMALL THEN THE ACID IS
VERY WEAK AND [H+] IS VERY SMALL THEREFORE
0.10
–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
CALCULATIONS INVOLVING
WEAK BASES
• THE PH OF A 0.10 M AMMONIA SOLUTION IS 11.37.
WHAT IS THE Kb FOR NH3 ?
• NH3(aq) + H2O  NH4+(aq) + OH-(aq)
• PH + POH =14.0, POH = 14.0 – 11.37 = 2.87
Kb = [NH4+] x [OH-] AND POH – LOG [OH-]
[NH3]
[OH-] = 10-2.87 = 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
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.
TITRATION (CONT’D)
STRONG
ACID (HCl)
END POINT
SOLUTIONS
STRONG
BASE (KOH)
NEUTRAL
PH = 7.00
WEAK
ACID (HF)
ACIDIC
PH < 7.00
BASIC
PH > 7.00
ACIDIC
BASIC OR
NEUTRAL
WEAK
BASE (NH3)
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
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
~ 1 NORMAL HCl
• 1 MOLAR H2SO4 ~ 2 NORMAL H2SO4
• 1 MOLAR H3PO4 ~ 3 NORMAL H3PO4
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
~
• 1 MOLAR Ca(OH)2 ~
• 1MOLAR Al(OH)3 ~
1 NORMAL NaOH
2 NORMAL Ca(OH)2
3 NORMAL Al(OH)3
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)
0.10
x 0.100
= 0.0010 (H+)
• BASE (NaOH) 0.10
x 0.050
= - 0.0005 (OH-)
• H+(aq) + OH-(aq)  H2O(l)
0.005 EXTRA H+
• [H+] REMAINING = 0.005MOLES / (0.100 + 0.050) L
• [H+] = .033 M, PH = -LOG(0.033) = 1.48
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) 0.10
x 0.100
= 0.0010 (H+)
• BASE (NaOH) 0.10
x 0.100
= - 0.0010 (OH-)
• H+(aq) + OH-(aq)  H2O(l)
0.00 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.
13
12
11
10
9
pH 8
7
6
5
4
3
2
1
0
0.10 M HCl + 0.10 M NaOH
Equivalent point
moles acid = moles base
Strong acid – strong base
Titration equivalent point at
pH = 7.00
Volume of Base Added
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)
0.10
x 0.100 = 0.0100 (HAc)
• BASE (NaOH) 0.10
x 0.025
= - 0.0025 (OH-)
• H+(aq) + OH-(aq)  H2O(l)
0.0075 EXTRA HAc
• [HAc] REMAINING = 0.0075 MOLES / (0.100 + 0.025) L
• [HAc] = 0.006 M (A WEAK ACID)
• MOLES SALT FORMED = MOLES OF BASE ADDED
• [Ac-] = MOLES SALT / LITER
• [Ac-] = 0.0025 MOLES / 0.125 L = 0.020 M
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-]
0.020
PH = - LOG [H+] = - LOG (5.4 x10-6 ) = 5.26
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)
0.10
x 0.100 = 0.0100 (HA
• BASE (NaOH) 0.10
x 0.100
= - 0.0100 (OH-)
• H+(aq) + OH-(aq)  H2O(l)
0.00 EXTRA HAc
• MOLES SALT FORMED = MOLES OF BASE ADDED
• [Ac-] = MOLES SALT / LITER
• [Ac-] = 0.0100 MOLES / 0.125 L = 0.080 M (ONLY A SALT
SOLUTION REMAINS)
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 0.080 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.
TITRATION CALCULATIONS (CONT’D)
(WEAK ACID – STRONG BASE)
• EQUATIONS:
• (1) HAc + NaOH  NaAc + H2O
(2) NaAc  Na+ + Ac(3) Ac- + H2O  HAc + OHKw = Ka x Kb, Kb = Kw / Ka , Kb = 1.0 x 10-14 / 1.8 x 10-5
Kb = 5.56 x 10-10 = [HAc] x [OH-] = X x X = X2
[Ac-]
0.080
0.080
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)
13
12
11
10
9
pH 8
7
6
5
4
3
2
1
0
0.10 M HAc + 0.10 M NaOH
Equivalent point
moles acid = moles base
Weak acid – strong base
Titration equivalent point at
pH > 7.00
Buffer
region
Volume of Base Added
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
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.
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+
+
Acweak 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 AcIONS AND THEREBY CONSUME THE WEAK ACID (HAc)
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).
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-]
0.05
PH = - LOG (3.6 x10-5) = 4.44
ADDING ACID TO AN UNBUFFERED SYSTEM
• HOW DOES THE PH OF A LITER OF WATER
CHANGE WHEN 0.0001 MOLES OF HCl ARE
ADDED?
• SOLUTION:
• PURE WATER HAS A PH = 7.00
• HCl IS A STRONG ACID (100% DISSOCIATION)
• HCl  H+ + Cl- , [H+] = 0.0001 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)
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 0.0001 MOLES
OF HCl ARE ADDED?
• SOLUTION:
• HAc  H+ + Ac-, ADDING AN ACID SHIFTS THE SYSTEM
TO THE LEFT.
• MOLARITY x LITERS = MOLE (ORIGINAL MOLES)
0.10
x 1.0
= 0.10 MOLES OF HAc
0.05
x
1.0
= 0.05 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 =0.10 + 0.0001 = 0.1001
MOLES OF Ac- = 0.05 – 0.0001 = 0.0499
ADDING ACID TO AN UNBUFFERED SYSTEM
• [HAc]eq = 0.1001 MOLES / 1 LITER = 0.1001 M
• [Ac-]eq = 0.0499 MOLES / 1 LITER = 0.0499 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.