152 lecture 7 acids bases ii slides = 78
TRANSCRIPT
Lecture 7Professor Hicks
Inorganic Chemistry (CHE152)
Strong acids HCl HNO3 H2SO4
H+ Cl- H+ NO3- 2H+ SO4
2-
hydrochloric acid nitric acid sulfuric acid
Weak acidsHC2H3O2 HF
H+ C2H3O2- H+ F-
acetic acid hydrofluoric acid
metalsform cations
non-metalsform anions
AcidsH+ + anion
H+ -anion
substances that release H+ ions when dissolved
molecularcompounds
all other molecular
compoundsdissolve
Dissolved molecules
ioniccompounds
dissolve
cations(+ ions)
anions(- ions)
separated ionsH+ + F- ~5%(separated ions)
HCl H+ + Cl-
~ 100%
HF ~ 95%(molecules)
acids
dissolve
Acids aremolecular compounds
strong acid
AcidsH+ -anion
• Acids are molecular compound because they candissolve without dissociating into ions
• Ionic compounds must separate into ions to dissolve
• Weak acids have a small percentage of moleculesseparated into H+ and an anion, the rest stay together as one particle
HF ~ 95% H+ and F- 5%
Strong acids separate 100% into H+ and anion in waterHCl ~ 0 % H+ and Cl- ~ 100%
Hydrogen ion • has no electrons!
• to bond other atom must provide both electrons (lone pair)
H+
H+
N
H
H
H
NH3 (aq) + H+ NH4+ (aq)
H N
H
H
H +curved arrows used to showelectron pair movement
1) start at lone pair 2) end at electron acceptor
carboxylic acids
• weak acids
• most common acids in nature C=O
O
R
H
C=O
O
RC=O
O
C=O
O
R
H
R = benzenebenzoic acid
R = C,H
C=O
O
C=O
O
CH3
H
R = CH3acetic acid
food preservative
5% solution = vinegar
curved arrows1) start at lone pair 2) end at electron acceptor
C=O
O
C=O
O
CH3
-
+ H+
hydronium ion H3O+
• H3O+ = H2O + H+
• form H+ takes in water
• reactions of acids in water can be written with H+ or H3O
+
+
0 Q 1010
Gib
bs
Fre
e E
ne
rgy
0 Q 2 x 10-4
Gib
bs
Fre
e E
ne
rgy
acid ionization constant Ka
• equilibrium constant for acid to hydrolyze
• acid + H2O H3O+ + anion
• larger Ka = stronger acid
strong acidsalmost 100%dissociated
weak acidsmuch less
100% dissociated
Q = Ka
Benzoic Acid = C6H5CO2H
C6H5CO2H (aq) + H2O (l) C6H5CO2- (aq) + H3O
+ (aq)
Q =[C6H5CO2
-][H3O+]
[C6H5CO2H]
Ka = 6.5 x 10-5
HClO4 Ka= 1010 !
reactants products reactants products
(weak acid)
Bases
- molecular compounds with lone pairs
+ cation
OH-
any cation
hydroxideion
• Bronsted-Lowry base = proton acceptor
• proton acceptors must have lone pair
- ionic compounds with OH- ion
N
H
H
H
lone pair
both proton acceptors = both bases
3 lone pairs O-H
Examples of baseshydroxide ion
ammonia
C
H
H
H
methane
H
no lone pairsnot a base
conjugate acid/base pairs
Whenever an acid-base reaction occurs:
1) the product that is “ acid minus H+ ” is called the conjugate base of the acid
2) The product that is “base plus H+ “ is called the conjugate acid of the base
HC2H3O2 (aq) + NaOH (aq) NaC2H3O2 (aq) + H2O (l)
acid conjugate acidbase conjugate base
strength conjugate acids/bases
• the stronger a base the weaker itsconjugate acid
0 Q 1010
Gib
bs
Fre
e E
ne
rgy
0 Q 2 x 10-4
Gib
bs
Fre
e E
ne
rgy
Q closer to pure conjugate base
Q closer to pure acid
Benzoic acidKa = 6.5 x 10-5
perchloric acid Ka= 1010 !
acidconjugate
base acidconjugate
base
• the weaker an acid the stronger its conjugate base
Water autoionizes
• Keq for this reaction called Kw
• Kw = 10-14 at room temperature
H2O (l) H+ (aq) + OH- (aq) Q = [H+][OH-]
2H2O (l) H3O+ (aq) + OH- (aq)
or written with hydronium ion
Q = [H3O+][OH-]
• Neutral solution has [H+] =10-7 and [OH-] =10-7
Water is amphoteric(an acid and a base)
water acting as a base
H2O (l) + HCl (aq) H3O+ (aq) + Cl- (aq)
water acting as an acid
H2O (l) + NH3 (aq) OH- (aq) + NH4+ (aq)
Conjugate acid of water H3O + = hydronium ion
Conjugate base of water OH- = hydroxide ion
Le Chateliers principle and Kw
2H2O (l) H3O+ (aq) + OH- (aq)
disturbance = add HNO3increase H3O
+ to 0.10 Msystems response= to decrease H3O
+
what if neutral water is perturbed by adding an acid or base?
10-7 M10-7 M
[H3O+][OH-] = K w = 10-14
Q = 10-7 x 10-7 = 10-14
[H3O+][OH-] = Kw
0.10 * [OH-] = 10-14
[OH-] = 10-14 / 0.10=10-13 M
is system is at equilibrium?
system finds a new equilibrium
+ 0.10 M
Why is [H3O+] ~ 0.10 M?
(about)
If rxn used no OH-
[H3O+] = 0.10 + 10-7
= 0.1000001If rxn used all OH-
[H3O+] = 0.10 + 10-7 -10-7 =0.10
initial M
yes b/c Q = Ka
? ?equilibrium M
Approximately = 0.10Acidic solution has [H3O
+] > 10-7
Basic solution has [OH-] > 10-7
[H3O+]
pH
• scale of [H+] concentration
• more convenient than scientific notation
• pH = -log [H3O+]
still not sure? take the log of 10 it should be 1
pOH• same idea as pH
• scale of OH- concentration
• pOH = -log [OH-]
[H3O+]
pH + pOH = 14
Why?
[H3O+][OH-] = 10-14
pH + pOH = -log(10-14) = 14
pH, pOH, and pKw
1) take –log both sides2) log(A*B) = log(A) + log(B)
pH is what most people think in terms of some problems we get a result [OH-] or pOH
use this equation to express it as a pH
pH = 14 - pOH
pH fun facts!
• more bacteria that are not harmful grow in acidic conditions (acidophilus strains)
• more bacteria that are harmful grow in basic conditions
• blood pH about 7.4• stomach pH 1.5 !• H+ and OH- are catalysts for the
reactions that hold together, fats, carbohydrates, and proteins!!!!
control of pH important for life
The Strong Acids
• molarity of a monoprotic strong acid = molarity of [H3O+]
• b/c strong acids completely, 100% dissociate
For example a 1.0 M solution of HCl has a [H3O+] = 1.0 M
HCl, HBr, HI, HNO3, HClO4, H2SO4
Dissociation of weak acids
HF (aq) + H2O (l) H3O+ (aq) + F- (aq)0.55 - 0 0-x - +x +x0.55-x - +x +x
initial (M)
Ka =[H3O+][F-]
[HF]
change (M)
equilibrium (M)
x2
0.55-x3.5 x 10-5 = x = ???
a quicker alternative is to look for an approximationif x is very small compared to 0.55
x2
0.553.5 x 10-5 =
x = sqrt{ 0.55* 3.5 x 10-5 } = 0.004387 Mx = change in [HF] to reach equilibrium,
and final equilibrium molarities [H3O+], [F-]
Calculate [H3O+] at equilibrium for a 0.55 M solution of HF in water.
look up Ka for HF
write hydrolysis reactionAcid + H2OH3O
+ + conj base
to solve for x you could use the quadratic equation
HF beforedissociation
HF (aq)H+ (aq)F- (aq)
dissolve
% dissociated =
- x
initial molarity
initial molarity x
“x”
initial molarity
x 100%PercentDissociated
at equilibrium
Dissociation of weak acidsHF (aq) + H2O (l) H3O+ (aq) + F- (aq)0.55 - 0 0-x +x +x0.55-x +x +x
initial (M)
Ka =[H3O+][F-]
[HF]
change (M)
equilibrium (M)
x2
0.55-x3.5 x 10-5 = x = ???
to solve for x you could use the quadratic equation
a quicker alternative is to look for an approximation
if x is very small compared to 0.55x2
0.553.5 x 10-5 =
x = sqrt{ 0.55* 3.5 x 10-5 } = 0.004387 M
is 0.004387 is small compared to 0.55 M? 5% error is widely accepted in science
0.004387
0.55 x 100% = 0.79%
approximation is valid b/c %dissociated is less than 5% so
error could not be larger than 5%
% dissociated
The method of successive
approximations
CO32- (aq) + H2O (l) HCO3
- (aq) + OH- (aq)0.010 - 0 0-x +x +x0.01 - x +x +x
initial (M)change (M)
equilibrium (M)
x = sqrt{ 0.01* 1.8 x 10-4 } = 0.00134 M
% dissociated = 0.00134
0.010x100% = 13.4%
more than 5%!approximation not good!
x2
0.010 -x1.8 x 10-4 =
1) substitute this approximationof x back into equation
2) calculate new approximate value of x
x2
0.010 - 0.00134
x = sqrt { 1.8x10-4 (0.01 - 0.00134 } = 0.00128
repeat steps 1, 2x = sqrt { 1.8x10-4 (0.01 - 0.00128 }
=0.00125
repeat steps 1, 2 again
x = sqrt { 1.8x10-4 (0.01 - 0.00125 } = 0.00125
the approximations have stopped changing in the 5th decimal place so the
approximation good to this precision
an improved approximation of x
use the method of successive approximations if % dissociation
is more than 5%
a more improved approximation of x
1) Break into groups of 2-3 - each group will be assigned an acid2) Determine the pH and % dissociated.
acid Ka
assigned M
1) HClO2 1.10 x 10 -02 5.0
2) HCHO2 1.80 x10 -04 9.0E-02
3) C7H6O2 6.50 x10 -05 3.0E-02
4) HC2H3O2 1.80 x10 -05 1.0E-02
5) HClO 2.90 x10 -08 2.0E-05
6) HCN 4.90 x10 -10 2.5E-07
7) HC6H5O 1.30 x10 -10 6.0E-08
8) HF 3.50 x10 -04 1.8E-01
9) HNO2 4.60 x10 -04 2.0E-01
Determine the pH and % dissociated for a 1.5 x 10-4 M solution of acetic acid using the method of successive approximations
Le Chateliers Principle and % dissociated
HF + H2O F + H3O+ Ka = 3.5 x 10-4
disturbance = increase H2O (dilute acid)
response = decrease [H2O]
1%10%30%40%50%60%70%80%90%~100%
percent dissociated
at “infinite” dilution
1.0 M 0.10 M
0.00010 M
0.001 M
etc
molarity declines with dilution
Polyprotic Acids• often acid molecules have more than one ionizable H –
these are called polyprotic acids
– 1 H = monoprotic, 2 H = diprotic, 3 H = triprotic• HCl = monoprotic, H2SO4 = diprotic, H3PO4 = triprotic
• ionizable H ’s have different Ka’s
• polyprotic acids ionize in steps– each ionizable H removed sequentially
• removing of the first H+ makes removal of the second H+
harder– H2SO4 is a stronger acid than HSO4
– H3PO4 is a stronger acid than H2PO4-
– H2PO4- is a stronger acid than HPO4
2-
Sulfuric acid is a diprotic acid
H2SO4 (aq) H2O (l) H3O+ (aq) + HSO4
-
~ 100%
HSO4- (aq) H2O (l) H3O
+ (aq) + SO42-
weak acidmuch less 100%
a 1.0 M solution of H2SO4 has [H3O+] = 1.0 M (first step)
+ a little more H3O+ from second dissociation
strong acid
• Estimate the concentration of sulfate ion in a 5.0 M solution of sulfuric acid.
Hints:
• In the first hydrolyses of sulfuric acid it acts as a strong acid
• For the second hydrolyses Ka = 1.0 x10-2
Strong bases
• calculate pOH from molarity [OH-]
• calculate pH
pH = 14 - pOH
Example: A 0.0010 M solution of NaOH has a [OH-] = 0.0010 M
pOH = -log(0.0010) = 3.0
pH = 14 - 3 = 11
ionic compounds with OH- ionKOH, NaOH, LiOH.
Weak bases
• react with water to produce OH-
Base + H2O Base-H+ + OH-
NH3 (aq) + H2O (l) NH4+ (aq) + OH- (aq)
• do not react completely
• equilibrium constant Kb
hydrolysis reaction
Dissociation of weak bases
CO32- (aq) + H2O (l) HCO3
- (aq) + OH- (aq)0.10 - 0 0-x +x +x
0.10 - x x x
initial (M)
Kb =[OH-][HCO3
-]
[CO32-]
change (M)
equilibrium (M)
x2
0.10-x1.8 x 10-4 =
x2
0.101.8 x 10-4 =
x = sqrt{ 0.10* 1.8 x 10-4 } = 0.00424 M
Calculate pH at equilibrium for a 0.10 M solution of Na2CO3 in water.
look up Kb for CO32-
write hydrolysis reactionbase + H2O conj acid + OH-
% dissociated = 0.00424
0.10x100% = 4.2 %
less than 5%!approximation is OKpOH = -log[OH-]=-log(0.00424)=2.37
pH =14-pOH =14-2.37=11.62
dissociation of weak bases
CO32- (aq) + H2O (l) HCO3
- (aq) + OH- (aq)0.010 - 0 0-x +x +x
0.010 - x x x
initial (M)
Kb =[OH-][HCO3
-]
[CO32-]
change (M)
equilibrium (M)
x2
0.010-x1.8 x 10-4 =
x2
0.0101.8 x 10-4 =
x = sqrt{ 0.010* 1.8 x 10-4 } = 0.00134 M
Calculate pH at equilibrium for a 0.010 M solution of Na2CO3 in water.
look up Kb for CO32-
write hydrolysis reactionbase + H2O conj acid + OH-
% dissociated = 0.00134
0.010x100% =13.4%
more than 5%!approximation not good enough!
Need to use the quadratic equationor successive approximations method
% dissociated increases asthe acid/base is more dilute
acidsH+ + anion
H+ -anion
+ anion
Na+ -anion
salts of acids
K+ -anion
Strong acids HCl HNO3 H2SO4
H+ Cl- H+ NO3- 2H+ SO4
2-
hydrochloric acid nitric acid sulfuric acid
Weak acidsHC2H3O2 HF
H+ C2H3O2- H+ F-
acetic acid hydrofluoric acid
Salts of Strong acids LiCl NaNO3 K2SO4
Li+ Cl- Na+ NO3- 2K+ SO4
2-
lithium chloride sodium nitrate potassium sulfate
Salts of Weak acidsMg(C2H3O2)2 CsF
Mg2+ 2C2H3O2- Cs+ F-
magnesium acetate cesium fluoride
often from group Isince all group I salts are soluble
replace H+
any cation
if an acid is unchargedits conjugate base is
negatively charged
conjugate bases of acids exist as ionic compoundsaka salts
0 % dissociated 100
Gib
bs
Fre
e E
ne
rgy
0 % dissociated 100
Gib
bs
Fre
e E
ne
rgya weak acid
conjugate base acetate ion
is a weak base
acetate ion acetic acidacetic acid acetate ion
conjugate bases of weak acids• most conjugate acid/base pairs are both weak• exception: conjugates of the strong acids and
strong bases (OH-) are weak bases/acids
HCl, HBr, HI, HNO3, HClO4, H2SO4
Weak acids
Increasing Acid Strength
“Strong Acids”weaker
weak acidsstronger
weak acids
Increasing Base Strength
Cl-, Br-, I-, NO3-, ClO4
-,
Conjugate bases of “Strong Acids”
Conjugate bases of weak acids
weakerweak bases
strongerweak bases
OH-
alcoholsC2H5OH
“Strong bases”
H2O
C2H5O-
HClO2HCN
ClO2-CN-
(so weak they do not affect pH)
(so weak they donot affect pH)
weak base salt of itsconjugate acid
compound with lone pairsoften a N containing compound
N
H
H
H
ammonia amines
N
H
H
H
R
N
H
H
H
H + N
H
H
H
R
H +
-anion
-anion
and
when they act as bases gaining H+ they become positively charged
examples NH4Cl CH3NH3(ClO4)
weak bases and the salts of their conjugate acidsif a base is unchargedits conjugate acid is
positively charged
conjugate acids of bases exist as ionic compoundsaka salts
Ka Kb Kw
HC2H3O2(aq) + H2O(l) C2H3O2-(aq) + H3O+(aq) Ka=1.76 x10-5
for acetic acid the hydrolysis reaction is
for acetate ion the hydrolysis reaction is
C2H3O2- (aq) + H2O HC2H3O2 (aq) + OH- (aq) Kb=5.68 x10-10
notice if you add them the conjugate acid and base cancel
2H2O(l) H3O+(aq) + OH- (aq) K = ?Kw = 10-14
when reactions are added the overall Keq is the product of the Keq’s
KaKb = 1.76 x10-5 x 5.68 x10-10 = 10-14 !!!!!!!!!
overall reaction becomes
=
0 %ionized 100
Gib
bs
Fre
e E
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rgy
0 %ionized 100
Gib
bs
Fre
e E
ne
rgy
NH3 a weak baseNH4
+ is a weak acid
NH4+
NH3NH3 NH4
+
conjugate acids of weak bases
• most weak bases conjugate acids are weak
• the conjugate acid of the strongbase OH- = H2O is a weak acid
Kb = 1.76 x 10-5
Ka = 5.68 x 10-10
same free energy bowl looked at from other side
not on table Ka’s
Ka = 10-14
Kb
use
salts of weak acids/bases
• if soluble fully dissociate into ions
• initial molarity calculated from chemical formula
Example
0.33 M NaC2H3O2 a solution 0.33 M in C2H3O2-
0.24 M Ca(C2H3O2)2 a solution 0.48 M in C2H3O2-
2 C2H3O2- per 1 Ca(C2H3O2)2
ICE tables for salts of weak acidsWhat is the pH of a 0.66 M solution of sodium acetate?
C2H3O2- (aq) + H2O HC2H3O2 (aq) + OH- (aq)
Kb=
initial
change
equil
0.66 0 0
-x +x +x
0.66-x +x +x
by the usual approximation x = square root (0.66*5.68 x 10 -10) = 1.936 x 10-5
= 1.936 x 10-5[OH-]
pOH = 4.71
pH = 14 - 4.71 = 9.29a basic solution b/c we added
the conjugate base of acetic acid
acetate ion
this problem is setup like other weak base problems but you will not find acetate in table of bases you must recognize
it as the conjugate base of a weak acid and calculate its Kb
= 5.68 x10-10
10-14
Ka (acetic acid) 10-14
1.76 x10-5=
trends in strengths of acids
2 factors
• electronegativity of conjugate base
• bond strength
weaker bonds break more easily stronger acidA H
A H - +
higher electronegativity of Amakes H closer to H+1 stronger acid
bond more polarized
Strengths of Binary Acids
• binary acid strength increases to the right across a period
example H-C < H-N < H-O < H-F
• binary acid strength increases down the column
example H-F < H-Cl < H-Br < H-I
1) the more + H-X - polarized the bond, the more acidic the bond
2) the weaker the H-X bond, the stronger the acid
Strengths of Oxoacids
• more oxygen atoms stronger acid– helps polarize the H-O bond
• more oxygen atoms in the chemical formula, like adding a single atom of greater electronegativity– used to compare similar acids
Example: H2SO4 is a strong acid but H2SO3 is a weak acid
Lewis Acids• Any substance that can accept an electron pair
to form a new bond is called a Lewis acid• H+ qualifies as a Lewis acid because it has no
electrons – it must accept electron pairs to form bonds
• Many metal ions accept electron pairs to form Coordinate Covalent Bonds
• Electron deficient species in general - such as boron compounds that violate the octet rule are Lewis acids
H
B
H
H
H
N
H
H
H
B
H
H
H
N
H
H
complex ions
• metal ion + base new complex• structure of base stays intact• metal ion is acting as a Lewis Acid• bases are called ligands when they bind to
metals
Al3+H-O-H repeat 6x
Al(H2O)63+
ligand metal ion complex ion
complex ions
• metal ion + base new complex
• water becomes more acidic when bound to metal
(H2O)5Al O H
H
3+
+ H2O
Al(H2O)5(OH)2+
+ H3O+
Ka = 1.4 x 10-5Al(H2O)63+
acidity of complex ions• increases as metal ion becomes smaller
and/or more highly charged
size decreases
charge increases
both increase acidity
not acidic
metal ion + water complex ion
Classifying Salt Solutions asAcidic, Basic, or Neutral
• cations of group 1 (Li+, Na+, K+, etc) will not change the pH
• anions that are conjugate bases of strong acids are such weak bases they will not change the pHNaCl LiNO3 KBr
group 1 ions conjugate bases of strong acids
neutral solutions
Cl– NO3– Br –
Classifying Salt Solutions asAcidic, Basic, or Neutral
• if the anion is the conjugate base of a weak acid, it will form a basic solution
NaF KNO2
group I ions
Na+ K+
conjugate bases of weak acids
F– NO2-
basicneutral
solution will be basic
Classifying Salt Solutions asAcidic, Basic, or Neutral
• if the salt cation is the conjugate acid of a weak base and the anion is the conjugate base of a strong acid, it will form an acidic solution
NH4Cl
weak acid conjugate base of a strong acid
acidic
neutral
Classifying Salt Solutions asAcidic, Basic, or Neutral
• if the salt cation is a small / highly charged metal ion and the anion is the conjugate base of a strong acid, it will form an acidic solution
Al(NO3)3
weak acid conjugate base of a strong acid
acidic
neutral
Classifying Salt Solutions asAcidic, Basic, or Neutral
• if the salt cation is the conjugate acid of a weak base and the anion is the conjugate base of a weak acid, the pH of the solution depends on the relative strengths of the acid and base
NH4F
is larger than Kb of the F−;
solution will be acidic
5.68 x 10-102.86 x 10-11
Ka of NH4+
Example: Determine whether a solution of the following salts is acidic, basic, or neutral
a) SrCl2
Sr2+ slightly acidic
Cl− is the conjugate base of a strong acid
pH neutral
solution will be slightly acidic
b) AlBr3
Al3+ is a small, highly charged metal ion
weak acid
Br− is the conjugate base of a strong acid,
pH neutral
solution will be acidic
Example: Determine whether a solution of the following salts is acidic, basic, or neutral
c) CH3NH3NO3
CH3NH3+ conjugate acid of a weak base
acidic
NO3− is the conjugate base of a strong acid, pH
neutral
solution will be acidic
Example: Determine whether a solution of thefollowing salts is acidic, basic, or neutral
Example: Determine whether a solution of the following salts is acidic, basic, or neutral
d) NaCHO2
Na+ is in group I,
neutral
CHO2− base of a weak acid
basic
solution will be basic
Example: Determine whether a solution of the following salts is acidic, basic, or neutral
e) NH4(HCO2)
NH4+ conjugate acid of a weak base
acidic
HCO2− conjugate base of a weak acid
basic
Ka(NH4+) > Kb(F
−); solution will be acidic(5.68 x 10 -10) (2.8 x 10 -11)
Classifying Salt Solutions asAcidic, Basic, or Neutral
NH4OH
weak acid strong base
Basic
Estimate the pH of a 0.10 M NH4OH solution(when bad approximations go good)
Forms NH4
+ and OH-
conjugate acid of NH3
base
will the solution be basic or acidic?
Write down an equilibrium reaction that includes OH- and NH4+
NH3 (aq) + H2O (l) NH4+ (aq) + OH- (aq) Kb = 1.76 x 10-5
initial (M) 0 0.10 0.10
change (M) x -x -x
equilibrium (M) x 0.10 - x 0.10 - x
(0.10 – x)2
x1.76 x 10-5 =
(0.10 – x)(0.10 – x)
x1.76 x 10-5 = (0.10 )2
1.76 x 10-5x =
= 568 !!!!!!!!! Very bad approximation !!!!!!!!
0 100%
Gib
bs
Fre
e E
ne
rgy
Estimate the pH of a 0.10 M NH4OH solution
NH3 (aq) + H2O (l) NH4+ (aq) + OH- (aq) Kb = 1.76 x 10-5
initial (M) 0.10 0 0
change (M) -x +x +x
equil (M) 0.10 -x x x
x2
0.10 - x1.76 x 10-5 =
x = sqrt {0.10 x 1.76 x10-5}
x = 0.001326
If approximation is very bad then x is large
x small0 100%
Gib
bs
Fre
e E
ne
rgy
x large
placing the ball in a different initial position does not change where it will end
up at equilibrium
reactants and products can beimagined to react into any set of
concentrations and then “allowed”to move to equilibriumlike the ball in the bowl
% dissociated = (0.001326/0.010) x100% = 1.3%