Upload
others
View
3
Download
0
Embed Size (px)
Citation preview
~ 1 ~
Acids and bases are considered to be strong when
they dissociate or ionize fully (100%) in solution.
This allows them to act as strong electrolytes in
water and have a signficant impact on pH.
Weak acids and bases dissociate, but to a much
lesser extent. This limits their impact on pH and
allows them to act only as weak electrolytes.
Strong Acids: HCl, HBr, HI, HNO3, HClO3, HClO4
H2SO4 (first proton ONLY, 2nd
is weak)
Strong Bases: All Group 1 metal hydroxides (MOH)
Ca(OH)2, Sr(OH)2, and Ba(OH)2
Ex.1) How would the pH of a 0.100 M HCl sol'n compare
with that of a 0.100 M acetic acid (HAc) sol'n?
23.1.1 Acid/Base Strength and Major Species
~ 2 ~
Knowing this, we can predict major species (MS)
present in a solution containing acids or bases. For
example:
HCl is a strong acid and would fully dissociate:
Major Species: H+ Cl
- H2O
Being comparatively weak, HAc partially dissociates.
Not enough H+ or Ac
- is produced to be significant:
Major Species: HAc H2O
List the major species present in each of the
following solutions:
Ex.2) 0.010 M NaOH MS:
Ex.3) 0.90 M HF MS:
~ 3 ~
Ex.4) 1.40 M H2SO4 MS:
Ex.5) 0.175 M Ba(OH)2 MS:
Ex.6) 1.0e-4 M Al(OH)3 MS:
Ex.7) 2.5 M NH4Cl MS:
Ex.8) 1.00 M HClO4 MS:
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
SC.1) Give the MS present in a nitric acid sol'n:
SC.2) Dissolved NaOH is reacted with an excess of of HCl.
Predict the surviving MS:
Answers: SC.1 / H= NO3
- H2O SC.2 / H
= Cl
- H2O Na
+
~ 4 ~
In water, acids dissociate or ionize in water to form
a proton and conjugate base. Weak acids exist in
equilibrium with their conjugates ( ↔ ), while strong
acids dissociate fully ( → ):
HA(aq) ↔ H+(aq) + A
-(aq)
Show the following acids dissociating water:
Ex.1) Acetic acid, HAc
Ex.2) Hydrochloric acid, HCl
Ex.3) Ammonium ion, NH4+
23.1.2 Acid Dissociation Equations in Water
~ 5 ~
Organic acids often show the protons being donated
at the end of the formula. The proton being
donated is bold in the examples below:
Ex.4) Acetic acid, CH3COOH
Ex.5) Pyridinium ion, C5H5NH+
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
SC.1) Anilinium ion, C6H5NH3+
SC.2) Hydrocyanic acid, HCN
Answers: SC.1 / C6H5NH3+(aq) ↔ H
+(aq) + C6H5NH2(aq)
SC.2 / HCN(aq) ↔ CN-(aq) + H
+(aq)
~ 6 ~
Strong bases are typically group 1 and 2 metal
hydroxides which dissociate completely in water to
give hydroxide ions:
MOH(aq) → OH-(aq) + M
+(aq)
Show the following bases dissociating in water:
Ex.1) NaOH
Ex.2) Ba(OH)2
Ex.3) KOH
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
SC.1) Show Ca(OH)2 dissociating in water:
23.1.3 Strong Base Equations in Water
Answers: SC.1 / Ca(OH)2(aq) → Ca2+
(aq) + 2 OH-(aq)
~ 7 ~
Weak bases typically react with water to produce a
hydroxide ion and conjugate weak acid:
A-(aq) + H2O(l) ↔ OH
-(aq) + HA(aq)
Show the following acting as a base in water:
Ex.1) Fluoride ion, F-
Ex.2) Ammonia, NH3
Ex.3) Hypochlorite ion, OCl-
23.1.4 Weak Base Equations in Water
~ 8 ~
Organic bases often contain a highly electronegative
element which accepts the proton in reactions. I've
made these bold in the examples below:
Ex.4) Methylamine, CH3NH2
Ex.5) Acetate ion, CH3COO-
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
SC.1) Aniline, C6H5NH2
SC.2) Phosphate ion, PO43-
Answers: SC.1 / C6H5NH2(aq) + H2O(l) ↔ OH-(aq) + C6H5NH3
+(aq)
SC.2 / PO43-
(aq) + H2O(l) ↔ HPO42-
(aq) + OH-(aq)
~ 9 ~
Within an acid-base equation, acids and bases exist
in pairs known as "conjugates". By donating a
proton, what remains of the acid, typically an anion,
is now known as a conjugate base:
HA(aq) ↔ H+(aq) + A
-(aq)
By accepting a proton, bases become conjugate
acids. This also changes the charge:
A-(aq) + H
+(aq) ↔ HA(aq)
The equations below contain two conjugate pairs.
Identify the pairs and label the CA and CB in each.
Ex.1) HAc(aq) + HCO3-(aq) ↔ H2CO3(aq) + Ac
-(aq)
Ex.2) H2O(l) + H2O(l) ↔ H3O+(aq) + OH
-(aq)
23.1.5 Identifying Conjugate Acid/Base Pairs
~ 10 ~
Ex.3) NH3(aq) + H2O(l) ↔ NH4+(aq) + OH
-(aq)
In a conjugate pair, HA/A- , the acid will always have
exactly one more proton than its conjugate base.
Amphiprotic substances will form 2 pairs, depending
on whether they initially act as an acid or as a base.
Give the conjugate(s) for each of the following:
Ex.4) NH3 ↔
Ex.5) H2CO3 ↔
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
SC.1) Write and label the conjugates of the hydrogen
carbonate ion, HCO3- :
SC.2) Given that HCl is a very strong acid, what can we
infer about the nature / strength of its conjugate?
Answers: SC.1 / H2CO3 is the CA, CO32-
is the CB SC.2 / chloride is an incredibly weak base.
~ 11 ~
Ionic compounds, or salts, often have acidic/basic
properties according to the following conditions:
Neutral: cation is from a SB (ex. group 1), and anion
is a conjugate of a SA.
Acidic: cation is CA of a WB, anion is neutral.
Basic: cation is neutral, anion is CB of a WA.
Classify each of the following salts as a WA, WB, or
neutral. Indicate which part of the salt, if any, has
acidic or basic tendencies:
Ex.1) NaF Ex.2) NH4Cl
Ex.3) KCl Ex.4) NaAc
23.1.6 Acidity and Basicity of Salts
~ 12 ~
Occasionally, a salt may contain both acidic and
basic components. In these instances, compare K
values to determine whether the sol'n will be acidic,
basic, or neutral (table 21 in your data booklet).
Ex.5) NH4Ac NH4+ (Ka = 5.6e-10) / Ac
- (Kb = 5.6e-10)
Ex.6) NH4NO2 NO2- (Kb = 1.4e-11)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
SC.1) NH4NO3 SC.2) KNO3
SC.3) NH4F (HF has Ka = 6.6e-4)
Answers: SC.1 / acidic SC.2 / neutral SC.3 / acidic
~ 13 ~
Covalent oxides consist of a non-metal paired with
oxygen. In a reaction with water, many of these
non-metals oxides (NMOs) produce an oxoacid.
Oxoacids are polys and typically contain 3+ oxygens.
NMO(g) + H2O(l) → Oxoacid(aq)
CO2(g) + H2O(l) → H2CO3(aq)
When an NMO combines with water, "tack on" an
extra oxygen and only keep the hydrogen atoms
necessary to satisfy the charge. For example:
NO2 → NO3-
→ HNO3
Ex.1) SO2 → →
Ex.2) SO3 → →
Ex.3) ClO2 → →
23.1.7 Behavior of Covalent Oxides in Water
~ 14 ~
For larger, heavily-oxidized NMOs, assume the acid
produced will have the same oxidation numbers:
+5 -2 +1 -2 +1 +5 -2 +1 +3 -2
N2O5 + H2O → HNO3 not HNO2
Ex.4) P4O10 + H2O →
Ex.5) Cl2O7 + H2O →
Ex.6) N2O3 + H2O →
NMOs in lower oxidation states tend to have little
or no acidic nature. If you cannot "strip" a poly acid
of hydrogens and one oxygen to produce an NMO,
it is unlikely to have acidic behavior:
Ex.7) Which carbon oxides are acidic, which are not?
~ 15 ~
Ex.8) Which sulfur oxides are acidic, which are not?
Ex.9) What is the lowest oxidation number sulfur can
have in a sulfur oxide and still be acidic?
Ex.10) Would N2O be acidic in water? Why or why not?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
SC.1) Fossil fuels contain nitrogen and sulfur compounds.
Why would the combustion of these substances be
an environmental concern?
SC.2) Which sulfur oxide, SO2 or SO3, is most acidic?
Answers: SC.1 / the nitrogen oxides and sulfur oxides which are produced as a by-product of
combustion can both combine with atmospheric water to form acid rain. SC.2 / SO3 is most
highly oxidized and will produce the stronger sulfuric acid, as opposed to sulfurous.
~ 16 ~
Ionic oxides are compounds consisting of a metal
and oxygen and are often also known as metal
oxides. In a reaction with water, metal oxides form
metal hydroxides:
MO(s) + H2O(l) → MOH(aq)
Show the following ionic oxides reacting with water:
Ex.1) MgO:
Ex.2) Fe2O:
Ex.3) CaO:
Ex.4) Based on the equation above, how does the
addition of a metal oxide influence the pH of pure
water:
23.1.8 Behavior of Ionic Oxides in Water
~ 17 ~
In additional being called ionic oxides, most metal
oxides are also known as basic oxides due to their
common behavior in water.
Which basic oxide will produce the following
hydroxides in water?
Ex.5) Sr(OH)2 Ex.6) KOH
Ex.7) LiOH Ex.8) Ba(OH)2
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
SC.1) Show Na2O reacting with water:
SC.2) Which ionic oxide produces cesium hydroxide in
water?
Answers: SC.1 / Na2O(s) + H2O(l) → 2 NaOH(aq) SC.2 / Cs2O(s)
~ 18 ~
"p" is a mathematical function commonly found
when working with very large or small numbers,
such as concentrations (pH, pOH) and equilibrium
constants (Ka, Kb):
pX = - log10X 10 -pX
= X
Convert each of the following into their respective
p-values: Ka → pKa , [H+] → pH, etc. Only numbers
past the decimal are significant in "log" answers:
Ex.1) Ka = 1.3 x 106 Ex.2) [OH
-] = 1.3e-6
Ex.3) [H+] = 4.5e-2 Ex.4) Kb = 1.8 x 10
-5
Reverse the following p-calcs. Remember sigfigs!
Ex.5) pH = 10.23 Ex.6) pKa = -8.552
23.1.9 p - Calculations
~ 19 ~
Ex.7) pOH = 0.54 Ex.8) pKb = 13.29
Ex.9) What is the relationship between the size of the
original number and the corresponding p-value?
Larger numbers produce....
Smaller numbers produce...
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
SC.1) Estimate the pKa of an acid with a Ka of 1.2e-3:
SC.2) Calculate the actual pKa value from the SC.1:
SC.3) Give the [OH-] in a solution with pOH = 4.57
Answers: SC.1 / a little less than 3 SC.2 / 2.92 SC.3 / 2.69e-5
~ 20 ~
Due to the fact that they only partially dissociate or
ionize in water, weak acids and bases exist in a state
of equilibrium with their products. As a result, we
can use RICE to predict the extent to which
dissociation will occur as well as the amount of H+
or OH- which will be produced. Specialized versions
of K, the equilibrium constant, are used:
Ka or Kb = acid or base dissociation constants
Equation → RICE → K expression → solve for x
Ex.1) Calculate the [F-] in a 0.200 M HF solution once it
reaches equilibrium (Ka for HF = 6.6e-4) :
23.2.1 Weak Acid/Base Equilibria
~ 21 ~
Ex.2) What is the hydroxide concentration present in a
0.900 M solution of NH3 (Kb = 1.8e-5) :
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
SC.1) Calculate the [H+] present in a 0.100 M solution of
benzoic acid, C6H5COOH (Ka = 6.3e-5) :
Answers: SC.1 / [H+] = 2.51e-3 M
~ 22 ~
The percent dissociation for an acid or base in
solution is calculated by comparing the initial
molarity of the substance against the amount which
is consumed as the reaction reaches equilibrium. In
most cases, the value of "x" in the RICE expression is
the amount consumed. For a weak acid:
% dissociation = x • 100%
[HA]0
A similar equation for a weak base would simply
substitute "A" for "HA". In general, strong acids and
bases dissociate 100%, while weak acids and bases
dissociate partially, typically less than 5%.
Ex.1) A 0.0800 M solution of HAc contains [Ac-] = 1.2e-3
M at equilibrium. Determine the % dissociation:
23.2.2 Calculating Percent Dissociation
~ 23 ~
Ex.2) Calculate the % dissociation of a 0.500 M HAc
solution (Ka = 1.8e-5):
Ex.3) % dissociation of 1.0 M HCl (Ka = 1.3e6):
Ex.4) How is the value of "K" related to % dissociation?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
SC.1) Determine the % dissociation in a 0.800 M
methylamine solution (CH3NH2 , Kb = 4.5e-4):
Answers: SC.1 / 2.37% dissociation
~ 24 ~
Unlike strong acids, weak acids dissociate partially
and establish an equilibrium with their conjugate
bases. To determine pH, we often have to create a
RICE table and calculate the amount of H+ produced
via dissociation. For example, if you wanted to find
the pH of 0.100 M HAc (Ka = 1.8e-5):
Major Species: HAc H+ Ac
- H2O
R HAc(aq) ↔ H+(aq) + Ac
-(aq)
I 0.100 0 0
C -x +x +x
E 0.100 x x
1.8e-5 = x2
x = 1.3e-3 M
0.1
The [H+] produced by HAc is much higher than that
produced by water (1e-7), thus:
pH = -log(1.3e-3) = 2.89
23.2.3 pH of Weak Acid Solutions
~ 25 ~
Ex.1) Determine the pH of 2.500 M HAc:
Ex.2) Calculate the pH of 1.0e-5 M HCN (Ka = 6.1e-10)
~ 26 ~
SC.1) Lemons contain citric acid (HC6H7O7 , Ka = 7.4e-4)
at a concentration of ~0.30 M. Estimate the pH of
lemon juice from this data:
Answers: SC.1 / pH = 1.82
~ 27 ~
Within a conjugate pair, the acid will have a Ka value
and the base will have a Kb value. You can convert
between values using the equation shown below:
Ka • Kb = 1 x 10-14
Given each of the following substances and their K
values, determine the following: the conjugate, its
K value, and the matching equation.
Ex.1) Ammonia, NH3, Kb = 1.8e-5 :
Ex.2) Pyridine, C5H5N, has Kb = 5.9e-6 :
23.2.4 Converting Between Ka and Kb
~ 28 ~
Ex.3) Carbonic acid, H2CO3, has Ka = 4.4e-7
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
SC.1) Nitrous acid, HNO2, has Ka = 4.5e-4
SC.2) Phosphate ion, PO43-
, has Kb = 0.024
Answers: SC.1 / CB = NO2- Kb = 2.2e-11 / NO2
-(aq) + H2O(l) ↔ HNO2(aq) + OH
-(aq)
SC.2 / CA = HPO42-
Ka = 4.2e-13 / HPO42-
(aq) ↔ PO43-
(aq) + H+(aq)
~ 29 ~
Unlike strong bases, weak bases dissociate partially
and establish an equilibrium with their conjugate
acids. To determine pH, we often have to create a
RICE table and calculate the [OH-] produced via
dissociation. For example, if you wanted to find
the pH of 3.0 M NH3 (Kb = 1.8e-5):
Major Species: HAc H+ Ac
- H2O
R NH3(aq) + H2O(l) ↔ NH4+(aq) + OH
-(aq)
I 3.0 0 0
C -x +x +x
E 3.0 x x
1.8e-5 = x2
x = 7.3e-3 M
3.0
The [OH-] produced is much higher than that
produced naturally by water (1e-7), thus:
pOH = -log(3.0e-3) = 2.14 pH = 14 - 2.14 = 11.86
23.2.5 pH of Weak Base Solutions
~ 30 ~
Ex.1) Household ammonia has a [NH3] of 0.500 M. What
is the pH of this ammonia solution?
Ex.2) Methylamine, CH3NH2, has a Kb value of 4.4 x 10-4
.
Find the pH of a 0.65 M solution of methylamine:
~ 31 ~
SC.1) Pyridine, C5H5N has a Ka value of 5.9e-6. Determine
the pH of a 0.0500 M solution of pyridine:
Answers: SC.1 / pH = 10.732 (depending on how you rounded, your answer may be different.)
~ 32 ~
An acid is considered strong if it dissociates fully
in water. This is responsible for the incredibly high
Ka values for strong acids and allows us to make
several assumptions. For example, if you wanted to
find the pH of a 0.015 M solution of HCl:
Major Species: H+ Cl
- H2O
HCl fully dissociates, none remains as HCl(aq)
The H+
produced here is far greater than is normally
present in water (0.015 M >>> 1e-7 M) and chloride
is a neutral ion. We determine pH directly from the
[H+] produced by the dissociation of HCl:
pH = - log (0.015) = 1.82
Ex.1) Determine the pH of a 4.1e-5 mol dm-3
HNO3 sol'n:
Major Species: pH =
23.2.6 pH of Strong Acid Solutions
~ 33 ~
Ex.2) Calculate the pH of 9x10-2
M HClO4:
Major Species: pH =
Be wary - if the concentation of the acid is too low,
water becomes the greatest contributor to pH:
Ex.3) Determine the pH of a 9.05e-9 M HBr:
Major Species: pH =
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
SC.1) pH of 0.10 M HNO3?
SC.2) pH of 6.90e-10 mol dm-3
HCl ?
SC.3) Estimate the [ ] level at which a strong acid no
longer contributes significantly to pH:
Answers: SC.1 / pH = 1.000 SC.2 / 7.0 SC.3 / any [ ] less than e-9, typically.
~ 34 ~
Like strong acids, bases must dissociate or ionize
fully to be considered strong. Assuming a high
enough concentration, we can typically determine
the pH indirecly from the [OH-] which is produced.
In a 2.8e-3 mol dm-3
NaOH solution:
Major Species: Na+ OH
- H2O
NaOH fully dissociates, none remains as NaOH(aq)
The OH- produced is far greater than is normally
present in water (2.8e-3 M >>> 1e-7 M) and sodium
is a neutral ion. We determine pH indirectly using
the [OH-]. (Find pOH, subtract from 14)
pH = 14 - (- log (2.8e-3)) = 11.45
Ex.1) Determine the pH of a 0.80 M KOH solution:
Major Species: pH =
23.2.7 pH of Strong Base Solutions
~ 35 ~
Ex.2) Calculate the pH of 3.34e-5 M Ba(OH)2:
Major Species: pH =
Be wary - if the concentation of the base is too low,
water becomes the greatest contributor to pH:
Ex.3) Determine the pH of 8.8x10-10
M LiOH:
Major Species: pH =
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
SC.1) pH of 2.3e-11 mol dm-3
Ca(OH)2?
SC.2) pH of 0.78 M NaOH:
SC.3) At what concentration does an SB no longer
contribute significantly to pH?
Answers: SC.1 / pH = 7.00 SC.2 / 13.89 SC.3 / any [ ] less than e-9, typically.
~ 36 ~
A buffer is a solution which contains a weak acid
and its conjugate base. A buffered solution is
extremely resistant to changes in pH and can be
found in a variety of biological (homeostasis!) and
industrial systems. Buffers work by converting SA
and SB into WA and WB which have a substantially
smaller impact of pH:
A buffer made from acetic acid and sodium acetate
would produce an equlibrium between the
following two substances:
HAc(aq) ↔ Ac-(aq)
Ex.1) If added to this buffer, a strong base such as NaOH
would react with __________ and increase the
concentration of __________.
Ex.2) If added to this buffer, a strong acid such as HCl
would react with __________ and increase the
concentration of __________.
23.3.1 Buffers and How They Work
~ 37 ~
A mixture of ammonia and ammonium chloride are
dissolved in water. Describe how the buffer will
respond to each of the following:
Ex.3) The addition of a strong acid will convert _________
into _________ and slightly _________ the pH of
the solution.
Ex.4) The addition of a strong base will convert ________
into _________ and slightly _________ the pH of
the solution.
Buffer capacity is a rough measure of how long a
buffer can continue to neutralize SA and SB before
running out of either the WA or its conjugate base.
When too much of a strong acid or base is added, a
buffer is said to have "collapsed" and no longer
resists changes in pH.
Ex.5) In terms of molarity, how could you make a high-
capacity buffer?
~ 38 ~
Explain why a buffer will eventually collapse if you
continue to add a strong acid or base:
Ex.6) The strong acid will eventually convert all of the ___
into its conjugate ______. Any further _____ added
will not react and will instead be in excess. As a
result the pH will ________ rapidly.
Ex.7) The strong base will eventually convert all the ____
into its conjugate ______. Any further _____ added
will not react and will instead be in excess. As a
result the pH will ________ rapidly.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
SC.1) Could a buffer be made from nitric acid and sodium
nitrate?
SC.2) Name the sodium salt which would form a buffer
with carbonic acid:
Answers: SC.1 / no, nitric acid is a SA and sodium nitrate is neutral. SC.2 / NaHCO3, sodium
hydrogen carbonate or sodium bicarbonate.
~ 39 ~
Buffers are typically prepared in one of three ways.
Options 2 and 3 are most common as it is rare to
have a weak acid and its conjugate both in stock:
1.) Dissolve a weak acid and is conjugate base (usually
a salt) in water.
2.) Start with a weak acid and add a smaller amount of
strong base such as NaOH, producing a mixture of
the surviving weak acid and its conjugate base.
3.) Start with a weak base (typically a salt), and add a
small amount of a strong acid such as HCl. This
produces a mixture of conjugate weak acid and the
surviving weak base.
Ex.1) What could you add to a solution of sodium cyanide
to produce a buffer?
Ex.2) Acetic acid solution and a small amount of _______
could be mixed to produce a buffer.
23.3.2 Basics of Buffers Preparation
~ 40 ~
Determine the products of each of the following
buffer preparation mixtures. Does a buffer result?
Ex.3) 1.40 mol NaAc →
0.80 mol HCl
Ex.4) 0.900 mol NaOH →
1.000 mol H3PO4
Ex.5) 2.40 mol HAc →
2.80 mol KOH
Ex.6) 0.100 mol NaF →
0.075 mol HCl
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
SC.1) 0.200 mol HCl + 0.240 mol NH3
SC.2) Name two ways you could make a buffer from 0.100
mol citric acid:
Answers: SC.1 / 0.200 mol NH4+ or NH4Cl and 0.040 mol NH3 SC.2 / Add a salt of citrate such
as sodium citrate OR add less than 0.100 mol of a strong base such as NaOH.
~ 41 ~
The easiest way of measuring the pH of a buffer
solution is to use the Henderson-Hasselbalch or "H-
H" equation shown below:
pH = pKa + log (A-/HA)
The value of A- and HA within this equation
represent the moles (or molarity) of the acid (HA)
and its conjugate base (A-) within the buffer.
Determine the pH of each of the following buffer
solutions when prepared as described:
Ex.1) 0.45 mol NaF is added to a solution containing 0.12
mol HF (Ka = 6.6e-4) in a total volume of 250 mL:
23.3.3 Measuring the pH of Buffer Sol'ns
~ 42 ~
Ex.2) 70.0 mL of 0.200 M HAc (Ka = 1.8e-5) is added to
100.0 mL of 0.100 M KAc:
Ex.3) 0.100 mol HAc + 0.040 mol NaOH
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
SC.1) What is the pH of a sol'n containing 0.500 mol NaAc
and 0.250 mol HCl?
SC.2) A solution with a pH of 5.61 contains 0.200 mol of
an unknown acid and 0.350 mol of its conjugate
base. What is the Ka for this unknown acid?
Answers: SC.1 / pH = 4.74 SC.2 / Ka = 4.3e-6
~ 43 ~
A buffer is said to be at the halfway point when
[HA] = [A-]. You can think of this as the point at
which the buffer is "centered" and is therefore most
effective at resisting large pH changes due to the
addition of strong acids or bases. When this occurs,
the log (A/HA) term in the H-H equation becomes
equal to zero (log 1 = 0). As a result,
pH = pKa (At the halfway point ONLY)
(1-3) Given a buffer made of HAc/Ac- (Ka = 1.8e-5):
Ex.1) What is the pH at the halfway point for this buffer?
By comparing the current pH to its center, we can
infer which species is in the majority:
Ex.2) At a pH of 5.000, [HAc] ___ [Ac-]
Ex.3) At a pH of 4.200, [HAc] ___ [Ac-]
23.3.4 The Halfway Point
~ 44 ~
Ex.4) 0.15 mol NaOH is added to a solution containing
0.50 mol HF (pKa = 3.14). What can be inferred
about the resulting solution?
[HF] ___ [F-] pH ___ pKa
A buffer's capacity or ability to resist changes in pH
is highest near its halfway point, as represented by
its pKa value. HAc, for example, is most effective at
pH = 4.74. For this reason, we tend to choose
buffers whose pKa values are very close to the pH
value we wish to use or create.
Ex.5) HAc buffers at pH = 6.0 are possible. How could this
be achieved?
Ex.6) The buffer in Ex.5 would have a substantially lower
capacity that one at a pH of 4.74. Why?
~ 45 ~
Ex.7) You wish to produce a buffer at pH = 9.0 for a
biochemical study. Which of the following
substances would be most appropriate?
HF (Ka = 7.2e-4)
HCN (Ka = 6.2e-10)
HSO4- (Ka = 1.2e-2)
HC2H3O2 (Ka = 1.8e-5)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
SC.1) Given that Ka = 6.2-10 for HCN, calculate the half-
way point for a HCN / CN- buffer:
SC.2) Which of the following would be true for a HCN
buffer at a pH of 9.00?
pH ___ pKa [HCN] ___ [CN-]
SC.3) What about a HCN buffer at pH = 10.5?
pH ___ pKa [HCN] ___ [CN-]
Answers: SC.1 / pKa = 9.21 SC.2 / pH < pKa [HCN] > [CN-] SC.3 / pH > pKa [HCN] < [CN
-]
~ 46 ~
Buffers are commonly made to have a highly
specific or "targeted" pH for a variety of uses. This
requires a system of equations and allows us to
solve for the exact amounts of HA and A- which
must be present in the solution:
Ex.1) You wish to prepare a HAc (Ka = 1.8e-5) buffer at a
pH of 5.25 with a total molarity of 1.00 M. What are
the [HAc] and [Ac-] necessary to do so?
Solve the H-H equation for "A/HA"
Create an equation for the total capacity of the
buffer, such that "HA + A = total molarity or moles"
23.3.5 Targeted Preparation of Buffers
~ 47 ~
Substitute the first equation into the second and
solve for either A or HA.
Solve for the remaining value, either A or HA
Ex.2) What mass of NaOH, in g, would have to be added
to 500.0 mL of 0.200 M HAc in order to produce a
buffer with pH = 4.10? (Assume no change in V)
~ 48 ~
Ex.3) Given a 1.00 M NH3 (Kb = 1.8e-5) solution and a 1.00
M HCl solution, how many mL of each will have to
be combined to produce a buffer with pH = 10.00
and a total volume of 2.00 L?
Ex.4) What masses of malic acid (HMal, Mr = 134.09 g
mol-1
, Ka = 3.5e-4) and potassium malate (Mr =
172.18 g mol-1
) would have to be added to 1.00 L of
water to produce a 0.500 M buffer at pH = 3.00?
~ 49 ~
SC.1) What mass of NaOH, in g, would have to be added
to 1.50 L of 1.00 M HAc to produce a buffer at pH =
5.00?
SC.2) What volumes of 1.00 M HF (Ka = 7.2e-4) and 1.00
M NaF would be combined to produce 250.0 mL of
a 1.00 M buffer with pH = 3.00?
Answers: SC.1 / ~38 g NaOH SC.2 / ~150 mL HF sol'n and 100 mL NaF sol'n