Contributed by:
The highlights are:
1. Arrhenius Definition
2. Bronsted-Lowry Definitions
3. What happens when an acid dissolves in water?
4. Conjugate acid and bases
5. Acid and base strength
6. Autoionization of water
7. pH and pOH
8. Organic lesson
9. Titrations
10. Lewis acids and bases
2.
Arrhenius Definition
Acids are substances that produce
hydrogen ions when dissolved in water.
HCl → H+ + Cl-
Bases are substances that produce
hydroxide ions when dissolved in water.
NaOH → Na+ + OH-
Problem: NH3 (ammonia) when dissolved in water
forms NH4OH, a weak base, but NH3 could not be an
Arrhenius base based on traditional definition because
NH3 does not have a hydroxide to donate.
3.
Brønsted-Lowry Definitions
A Brønsted–Lowry acid…
…must have a removable (acidic)
proton.
A Brønsted–Lowry base…
…must have a pair of nonbonding
electrons.
4.
Brønsted-Lowry Definitions
According to this theory, an acid is a
proton (hydrogen ion, H+) donor
and a base is a proton (hydrogen
ion, H+) acceptor.
…Consider introducing HCl(g) into
EOS
5.
What Happens When an Acid
Dissolves in Water?
Water acts as a
Brønsted–Lowry base
and extracts a proton
(H+) from the acid,
becoming a proton
acceptor.
As a result, the
conjugate base of the
acid and a hydronium
ion (H3O+) are formed.
6.
NH3 works under this definition as a
base. Let’s see how.
Notice that water now behaves as an
7.
If it can be either an acid or base…
...it is amphiprotic
(amphoteric).
HCO3−
HSO4−
H2O
8.
Conjugate Acids and Bases:
From the Latin word conjugare, meaning “to
join together.”
Reactions between acids and bases always
yield their conjugate bases and acids.
Conjugate acid and
base pairs are related
by a hydrogen ion on
either side of the
If the reaction proceeds in the forward direction, then HNO2 acts as
an acid by donating a hydrogen ion (proton). However, if the
reaction were to then go backwards, then NO2-1 would act as a
base by accepting a hydrogen ion (proton)
9.
The conjugate acid of a base is the base
PLUS the attached proton and the
conjugate base of an acid is the acid
MINUS the proton
10.
Identify the acid, base, conjugate
acid, conjugate base and pairs
HCN + NH3 CN- +NH4+
Acid – HCN
Base – NH3
Conjugate Base - CN-
Conjugate Acid - NH4+
11.
Acid and Base Strength
A strong acid/base undergoes complete
ionization
Reaction goes to completion
Full dissociation. Equilibrium lies
completely to the product side
LARGE value of Kc
Ex: HCl(aq) H3O+(aq) + Cl-(aq)
Ex: NaOH Na+(aq) + OH-(aq)
(aq)
The conjugate base/acid therefore is
extremely weak
Ex: Cl- has VERY poor ability to attract
protons to itself
12.
Acid and Base Strength
Weak acids/bases have
very little ionization.
Very little (partial)
dissociation in water
Equilibria lies mostly to
the LEFT. SMALL value
of Kc
Their conjugate
bases/acidss are weak to
exceedingly strong
As acid/base strength
decreases, the
conjugate base/acid
strength increases
13.
Acid and Base Strength
Substances with
negligible acidity do
not dissociate in
water.
Their conjugate bases
are exceedingly
strong.
CH4 + H2O CH3- + H3O+
CH3- is a VERY strong base due to its
+
14.
Acid and Base Strength
In any acid-base reaction, the
equilibrium will favor the reaction
that moves the proton to the
stronger base.
HCl(aq) + H2O(l) H3O+(aq) + Cl−(aq)
H2O is a much stronger base than Cl−, so
the equilibrium lies so far to the right K is
not measured (K>>1).
ASSUMPTION: 100% dissociation. [H+] or
[H3O+] = [ACID given] ** important when
calculating pH
15.
Acid and Base Strength
HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2−(aq)
Acetate ion is a stronger base than H2O,
Predict K>>>1 K>1 K<1 K<<<1
equilibrium favors the left side (K<<<1).
16.
Consider the four diagrams where in each
case, HA is an acid, H+ is a hydrogen ion,
A- is an anion, and water molecules
HA H+
A- H2O
1. Which diagram represents a relatively concentrated weak acid?
Explain
2. Which diagram represents a relatively concentrated strong acid?
Explain
3. Assign relative strengths and concentrations to the two remaining
17.
Carefully note that the concentration of
H+ and OH- are dependent upon TWO,
separate factors.
Strength of acid or base, i.e, degree of
dissociation/ionization
Amount of water present, i.e.
concentration of sample solution
It is possible to have a dilute, strong
acid and a weak, concentrated acid
with the SAME hydronium ion
concentration
Therefore, the same pH value
18.
Examples of Acids and Bases
Weak Acids Organic (carboxylic)
acids, such as
butanoic, propanoic,
ethanoic and
methanoic
Strong Acids HCl, HBr, HI, HClO4,
HNO3, H2SO4
Weak Bases Ammonia, and organic
bases such as amines
and pyridines
Strong Bases Group 1 and Group 2
hydroxides
19.
Autoionization of Water
As we have seen, water is
amphoteric.
In pure water, a few molecules act as
bases and a few act as acids.
H2O(l) + H2O(l) H3O+(aq) + OH−(aq)
This is referred to as autoionization.
20.
pH (power of Hydrogen)
Many of the concentration measurements
in acid-base problems are given to us in
terms of pH (power of H+) and pOH (power
of OH-).
p (anything) = -log (anything)
pH = -log [H+]
pOH = -log [OH-]
pKa = -log Ka
21.
pH (power of H)
pH scale is used to indicate the
strength of an acid or base
SAMPLE.
Traditional range 0-14
Negative pH is possible
0- <7 Acidic
7> - 14 Basic
7 Neutral
22.
Recall: For a strong acid/base, ionization is
considered to be 100%. Therefore
H+ or OH- concentration can be determined directly
from the stoichiometric ratio in the balanced
equation and the concentration of acid or base.
Ex: HNO3 is a strong acid. If 0.10 M HNO3
dissociates, [H+] is ALSO 0.10 M
HNO3 H+ + NO3-
0.10 M 0.10 M 0.10M
Calculate pH of above sample
pH = -log[H+]
pH = -log [0.10]
pH = 1.00
23.
pH
pH= -log[H+]
Used because [H+] is usually very
small
As pH decreases, [H+] increases
exponentially
24.
Practice
Calculate pH of each of the following
solutions
0.0030 M HCl
0.030 M HCl
0.30 M HCl
3.0 M HCl
Carefully note that since pH is based on a
log scale, there is a 10 fold change in
concentration associated with a pH change
of 1 unit (EXPONENTIAL)
25.
pH – cont.
pH of 3 = 10 times more
concentrated than pH of ???, and ???
times more concentrated than pH of
5
pH of 3 = 10 times more
concentrated than pH of 4, and 100
times more concentrated than pH of
5
26.
Determining [conc] from
pH/pOH
pH is a measure of the pOH is a measure of the
strength of an acid strength of a base
sample; sample;
low pH = stronger acid low pOH = stronger
pH = –log[H3O+] and base
pOH = –log[OH–]
[H3O+] = 10(–pH) (also means HIGH pH
and [OH ] = 10
– (–pOH)
EOS
27.
Practice: Predict the following from
highest to lowest [H+]. Then
determine [H+] if pH is
1.58
9.5
14
-0.87
4.2
28.
Additionally, at 298 K,
14 = pH + pOH
Therefore, ALL values (pH, pOH, OH- or
H+ ) can be determined based on ANY
one known value
29.
Practice
Calculate the pH of a 0.010 M
solution of Lithium hydroxide
Calculate the hydrogen ion
concentration in a solution with a pH
of 4.32
Calculate the pH of a solution made
by dissolving 2.00g KOH in water to a
total volume of 250. mL
30.
pH of weak acids
•What is the definition of a weak acid?
•Incomplete dissociation
•What does the pH and/or strength of sample
depend on?
• amount of acid dissociated [H+] AND
amount of water present.
Weak acids do not completely
dissociate in water.
Equilibrate!!
Problems solved similar to
equilibrium problems.
31.
HA(aq) + H2O(l) H3O+(aq) + A-(aq)
I X 0 0
C -x +x +x
E X-x 0+x 0+x
• We do NOT know how much acid actually
•Therefore, CANNOT use pH = - log [H+] because
CANNOT assume the [H+] = [acid].
•Carefully note that since ionization is minimal, value
of x (dissociation) is small.
•Considered NEGLIGIBLE in terms of (X – x)
32.
• Additionally
• [H+] = [A-]
• H2O (pure liquid) does not appear in Ka (Acid
dissociation constant) expression
• Therefore, the following Ka expression can be
derived
[H+][A-] [H ]
+ 2
Ka =
[HA] [HA]
Ka = OR
Mathematically
And equal. MUST NOT
pKa = -log Ka use as the
EQUATION on ap
frq. SHOW the
ACTUAL equation.
33.
Practice Problems
If ethanoic acid has a pKa 0f 4.74, what is its Ka?
If propanoic acid has a Ka of 1.38 X 10-5, what is
it’s pKa?
Which is a stronger acid, propanoic or
ethanoic?
What is the pH of a 1.00 M solution of ethanoic
acid?
Calculate the Ka value of a 0.100 M solution of a
weak acid with a [H+] of 1.75 X 10-3 M
Calculate the Ka of a solution of 0.250 M of a
weak acid with a pH of 5.11
34.
ORGANIC LESSON
Carboxylic acid (organic acid)
General makeup
R-COOH
Example: ethanoic
Ethane (C2H6): remove methyl (CH3) and add –COOH
Formula: CH3COOH
Example: propanoic
propane: (C3H8): remove methyl (CH3) and add –
COOH
Formula: C2H5COOH
Determine formula for methanoic and butanoic
acids. Draw Structure.
35.
Strength of carboxylic acid depends upon the
stability of the anion (called: ___oate) formed
when labile (mobile) proton is lost.
All carboxylic acids that dissociates, give the following
equilibrium, where R (hydrocarbon chain) can vary
RCOOH RCOO- + H+
Anions may form, releasing H+ to form acidic solution
Stability of anion depends on size of R group.
Smaller R group = FEWER electrons “pumped” to COO-
anion = HIGHER the stability of anion = STRONGER acid
= LARGER Ka = SMALLER pKa
Explain in terms of polarity:
Higher polarity of a smaller particle (methanoate) will
cause greater attraction to water, therefore higher
chance of H+ ionizing. Lower polarity of a larger non-
polar R group (butanoate) will have far less attraction to
water, therefore smaller chance of H+ ionizing.
RECALL: carboxylic acids up to 4 carbons are
soluble/acidic
36.
Carefully note: as carboxylic acid R
group increases in size, acid strength
decreases.
Ka decreases, pKa increases.
37.
Other related
molecules
Similar effects are
observed when
comparing relative
strenghts of halogen
substituted carboxylic
acids.
As the number of
chlorine atoms present
increases, so does the
acid strength.
The high
electronegativity of Cl
atoms attract electron
density away from
COO-
38.
High Halogen Electronegativity = high Ka =
Stronger Acid = Low pKa
Why?
Greater attraction for electrons, so flourine pulls
electrons away from the COO-, allowing easier
removal of H+
39.
Kb: Equilibrium Constant for Weak Bases
Similar to weak acid equilibria, weak base
ionization is incomplete
Cannot directly determine amount of base
dissociated based on Molarity of solution
Cannot use pOH = -log [Base]
Must determine and use ACTUAL [OH-]
Same assumptions as weak acid equilibria
calculations
Assume x (dissociated ion) is negligible relative to
X (base molarity)
Assume [cation] = [OH-]
40.
NH3 + H2O NH4+ + OH-
Kb = [OH -
][NH 4
+
]
[NH3]
DO NOT USE ON AP FRQ.
Often written as Use ACTUAL Kb expression.
CLEARLY SPECIFY that you
Kb = [OH-]2 are ASSUMING [OH-] =
[Cation]
[NH3] CLEARLY SPECIFY that you
are ASSUMING [OH-] to be
And negligible, therefore using
[Base] as given
pKb = -log Kb
42.
Autoionization of Water
Although pure water is essentially
covalent, a small amount of self-
ionization occurs
Water autoionizes, by donating a
proton to another water molecule:
H2O + H2O H3O+ + OH-
or
H2O H+ + OH-
No individual ion remains ionized for long
At 298 K, only about 2 out of every 109
molecules are ionized at any given
moment
43.
Autoionization of Water
Equilibrium constant, Kw, written as
Kw = [H+][OH-]
At 298 K, Kw = 1.0 X 10-14
Since [H+] = [OH-], therefore
1.0 X 10-14 = [H+][OH-]
√1.0
and [H ] = [OH ] =
+ - X 10-14
= 1.0 X 10-7
M
Therefore, pH = -log [1.0 X 10-7 M] = 7
44.
Determine pKw
pKw = -log Kw
= -log [1.0 X 10-14]
= 14
Since we know that
Kw = [H+][OH-] therefore
K w = K a · Kb therefore
pKw = pKa + pKb therefore
14 = pH + pOH
*** KNOW THIS
45.
Temperature dependent
Carefully note that at temperatures
other than 298 K, Kw value may be
different, HOWEVER,
[H+] = [OH-]
pH may be different from 7
HOWEVER, still considered NEUTRAL due to
[H+] = [OH-] (no excess H+ or OH-)
46.
Autoionization of Water
H2O H+ + OH-
Since this is specifically for the
autoionization of water we use Kw as
the equilibrium constant.
At 25oC:
Kw = [H+][OH-] = 1.0 x 10-14
[H+] > [OH-] acid
[H+] < [OH-] base
[H+] = [OH-] neutral
47.
Autoionization of Water
H2O H+ + OH-
Kw = [H+][OH-] = 1.0 x 10-14
At this point it is very useful to
remember that pH = - log [H+]
Since Kw is a constant, if we are given
[OH-] it is a very simple matter to
calculate pH
Since pH, pOH, [H+], [OH-], pKa, pKb,
pKw and Kw are ALL related, knowing
ANY of these values is sufficient to
calculate ALL others.
48.
Titrations
Titration: experimental technique to
perform a stoichiometrically
balanced neutralization reaction.
Accurately graduated glassware
(volumetric flasks, graduated glass
pipets and burets) is used in
quantitative manner
Helps determine unknown concentration of
an acid or a base
49.
Titrations Terms to Know:
Titrant:solution added
FROM the buret
Titrate: solution that
titrant is added to.
Equivalence point:
100% neutralization.
[H+] = [OH-]
End point: Change in
color of indicator
Half Equivalence point:
1/2 total amount of
titrant added.
Buffer solution pH =
pKa
50.
Titrations
As acid or base is added, there is
Very little change in pH
pH change of less than 1.5 units
expected upto the point where 90% of
acid/base is neutralized
At equivalence point, when 100% of acid
or base is neutralized, RAPID change in
pH is observed
Summarized using titration curve plots
51.
Strong acid-Strong base
Titration
At equivalence point,
pH = 7
Carefully note: IF and
ONLY IF concentrations
of monoprotic acid and
base are equal, then
VOLUME of acid and
base will also be equal.
Generally, expect
questions where
EITHER concentration
OR volume of acid OR DUMB QUESTION: In the above
base is UNKNOWN, and example, which is the titrate and which
you are asked to solve is the titrant? How would you know if
labels were not given?
52.
Weak acid-Strong
Base Titration
PREDICT: How will a
weak acid-strong
base curve differ
from strong acid-
strong base curve?
Starts at a HIGHER
pH
Plateaus at relatively
HIGH basic pH Weak acids doesn’t ionize fully. Removal
Equivalence point of dissociated H +
due to neutralization
FORCES equilibrium shift towards
ABOVE 7. Why? dissociation of acid. Reaction rate SLOWS
down as 100% neutralization approaches.
Need EXCESS base to FULLY ionize the
weak acid for complete neutralization
53.
Strong acid-Weak base Titration
PREDICT: How will a
strong acid-weak
base curve differ
from strong acid-
strong base curve?
Starts at LOW pH
Equivalence point
BELOW 7
Plateaus at relatively
LOW basic pH
54.
Weak acid-Weak base Titration
Starts at relatively
HIGH acid pH
Plateaus at relatively
LOW basic pH
No sharp change in pH
at equivalence point
Equivalence point likely
at pH = 7
May be variable based
on relative strength of
acid and base.
55.
Polyprotic acid neutralization
Multiple
equivalence
points
Recall: Acid
strength
decreases with
EACH hydrogen
ion removal
H3PO4 > H2PO4-1
> HPO4-2
56.
Indicators
Most acids and bases form colorless
solutions
Driving force of neutralization reaction
(formation of water) does not produce an
observable change.
Need a method of determining when
equivalence point has been reached.
Indicator: chemical that changes color at
various pH’s
Often weak acids, where ionized and unionized
form have different colors.
Ideal indicators change color over a small, given
57.
Indicators
Choose an indicator that changes color at a pH value as close to the
equivalence point as possible
End point should correspond closely to equivalence point.
• Suitable Indicators:
• Strong acid-Strong base most
• Weak Acid-Strong base phenolphthalein
• Strong acid-Weak base methyl orange
• Weak acid-Weak base none due to lack of
sharp change in pH (use
pH meter)
58.
Buffer Solutions
A buffer solution is one that resists
changes in pH, when either a small
amount of acid or base is added to it.
General composition
Weak acid and one of its salts (i.e. its
conjugate base)
Weak base and one of its salts (i.e its
conjugate acid)
Example: ethanoic acid and sodium
ethanoate
59.
Ethanoic acid acts as the acid in the
buffer solution. Absorbs base
CH3COOH + OH- CH3COO- + H2O
Ethanoate ion acts as the base.
Absorbs acid
CH3COO- + H3O+ CH3COOH + H2O
Process of “mopping up” of added
acids and bases allows pH to remain
relatively unchanged.
60.
Capacity of buffers
Capacity defined as the ability to
continue reacting as extra acid/base
is added. Depends upon
Concentration of components
Higher the concentration, the more acid or
base it can absorb, the higher its capacity
61.
pH of Buffers
pH of buffer is determined by the
pKa or pKb of the weak acid or base, and
Ratio of concentrations of each component
To calculate pH of buffer solution, use
Henderson-Hasselbach equation
pH = pKafor [salt]
+ acidic buffer
[acid]
log
pOH = pKfor+basic [salt]
buffer
b
[base]
62.
Carefully note a crucial application of buffer
solutions (expect ** AP ** question)
Weak acid-strong base or weak base-strong acid
titrations produce a buffer region on the graph
Gradual addition of Strong base to weak
acid results in
Some weak acid neutralization =
BU
Production of the salt of weak acid FF
S ER
Production of water OL
U
TIO
N
64.
When 1/2 the acid is neutralized, what
is the ratio of acid to salt?
1:1
Plugging into Henderson-Hasselbach,
we get log (1) = 0
Therefore when acid to salt ratio 1:1, then
pH = pKa
True at ½ equivalence point
65.
Practice
Ethanoic acid has a pKa of 4.75. Find
the pH of the solution that results
from the addition of 40.0 mL of 0.040
M NaOH to 50.0 mL of 0.075 M
ethanoic acid
Methanoic acid has a Ka = 1.60 X 10-
4. Calculate the pH of the final
solution when 23.90 mL of 0.100 M
sodium h
66.
Acid dissociation constant = Ka
Acid dissociation constant is the
equilibrium constant for an acid.
HA(aq) + H2O(l) H3O+(aq) + A-(aq)
Ka = [H3O+][A-]
[HA]
H3O+ is often written H+ , ignore the
water in equation.
67.
Acid dissociation constant Ka
HA(aq) H+(aq) + A-(aq)
Ka = [H+][A-] [HA]
We can write the expression for any
acid.
Strong acids dissociate completely.
Equilibrium far to right.
Conjugate base is weak.
68.
Write the acid dissociation/ionization
reaction and its Ka (omit water)
HC2H3O2 for Acetic Acid
69.
Kb = Base dissociation constant
B(aq) + H2O(l) BH+(aq) + OH-(aq)
Kb =[BH+][OH-]
[B]
70.
Strong Acids
Equilibrium lies far to the right
Almost 100% dissociation
HA H+ + A-
1% 99%
Gives a weak conjugate base
(weaker than water)
Large Ka
71.
Weak acids
Equilibrium lies to the left
Very little H+
HA H+ + A-
99% 1%
Strong conjugate base (stronger than
water)
Small Ka
72.
Back to Pairs
Strong acids Weak acids
Ka is large Ka is small
[H+] is >>> to [HA] [H+] <<< [HA]
A- is a weaker base A- is a stronger
than water base than water
73.
Weak Acids and Bases
For weak acids and bases, equations can be
written to describe equilibrium conditions
EOS
74.
Acid–Base Strength (cont’d)
Ka values are used to compare the strengths
of weak acids; K, strength
For strong acids, water has a leveling effect; that
is, when the strong acids are dissolved in water,
they all completely ionize to the hydronium ion
EOS
75.
Strong Bases
Large Kb
Almost completely dissociated
Conjugate acid is weak
Alkali metal hydroxides NaOH, LiOH,
KOH
Ba(OH)2
Ca(OH)2 or Sr(OH)2
76.
Kw, Ka, Kb, pH, Oh my!
How does this all relate and tie
together???
The symbols are actually very basic:
K is the equilibrium constant for water
w
K is the acid-dissociation constant
a
Kb is the base-dissociation constant
pH (power of hydrogen) is the –log of
the concentration of hydrogen
77.
Kw, Ka, Kb, Oh my!
Given an acid reaction:
HA(aq) + H2O(l) H+(aq) + A-(aq)
Ka = [H+] [A-]
[HA]
Given a base reaction:
B(aq) + H2O(l) HB+(aq) + OH-(aq)
Kb = [HB+] [OH-]
[B]
78.
Kw, Ka, Kb, Oh my!
Here we have a conjugate acid-base pair:
NH4+(aq) NH3 (aq) + H+ (aq)
NH3(aq) + H2O(l) NH4+ (aq) + OH- (aq)
Ka = [NH3] [H+]
[NH4+]
Kb = [NH4+] [OH-]
[NH3]
79.
Mathematical Approach
We can add the two equations together:
NH4+(aq) NH3 (aq) + H+ (aq)
NH3(aq) + H2O(l) NH4+ (aq) + OH- (aq)
NH3(aq) + H2O(l) + NH4+(aq) NH3 (aq) + H+(aq) + NH4+(aq) + OH- (aq)
Cancel out the duplicate terms on
opposite sides of the arrows…
NH3(aq) + H2O(l) + NH4+(aq) NH3 (aq) + H+(aq) + NH4+(aq) + OH- (aq)
You are now left with:
H2O H+ + OH-
80.
Mathematical Approach
If we try to multiply Ka and Kb. this is
what we get:
Ka x Kb
[NH3] [H+] x [NH4+] [OH-] =
[NH4+] [NH3]
Ka x Kb = [H+] x [OH-]
Ka x Kb = 1.0 x 10-14 = Kw
81.
Watch This!
[H3O+] [OH−] = Kw = 1.0 10−14,
we know that
−log [H3O+] + −log [OH−] = −log Kw =
14.00
or, in other words,
pH + pOH = pKw = 14.00
82.
Mathematical Approach
KW = [H+][OH-] = 1.0 x 10-14
Ka × Kb = Kw
In many books you will find a table
for Ka they rarely are given for Kb
since it is such an easy calculation
83.
Relationships
Kw = [H+][OH-]
-log K = -log([H+][OH-])
w
-log K = -log[H+]+ -log[OH-]
w
pK = pH + pOH
w
K = 1.0 x10-14
w
14.00 = pH + pOH
Given any one of these we can find the
other three:
[H+], [OH-], pH and pOH
84.
Lewis Acids
Lewis acids are defined as electron-pair acceptors.
Includes cations and incomplete octets.
Atoms with an empty valence orbital can be Lewis
acids.
85.
Lewis Bases
Lewis bases are defined as electron-pair donors.
Should have a lone pair of electrons.
Anything that could be a Brønsted–Lowry base is a
Lewis base.
Lewis bases can interact with things other than
protons, however.
86.
Lewis Acids and Bases
Boron triflouride wants more
electrons, because it has an
incomplete octet.
F F H
H
B F :N H F B N H
F H F H
87.
pH of strong acids
Finding the pH of strong acids is easy
since there is almost complete
dissociation
Common strong acids include:
HCl, HBr, HI, HNO3, H2SO4, HClO4
ASSUMPTION: [H+] = [acid given]
88.
Example
ASSUMPTION: [H+] = [acid given]
HCl (aq) + H2O(l) H3O+(aq) + Cl-
(aq)
0.0004% at 99.996% at
equilibrium
equilibrium
89.
Example
Determine the pH of a 0.5 M HBr solution
First,write the equilibrium reaction
For strong acids, we assume 100%
dissociation, therefore
[H+] = [HBr] = 0.5 M
Determine pH
pH = - log [H+]
pH = - log [0.5]
pH = 0.3
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