Acids and Bases
Calculations for a Titration Plot
Some basics from General Chem
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HCl ↔ H+ + ClHOAc ↔ H+ + OAcH2O ↔ Η+ + ΟΗ−
Kw = [H+][OH-]
Ka and Kw are the
molar equilibrium
constants
Ionization of Water
• [H+][OH-] = 1.0 x 10-14
= Kwat 25 C
• In pure water
[H+]=[OH-] = 1.0 x 10-7M
Calculate [OH-] of acid in water
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1- x 10-3 M solution of HCl. [OH-]?
HCl completely ionized
H+ is 1.0 x 10-3
1.0 x 10-3[OH-] = 1.0 x 10-14
[OH-] = 1.0 x 10-11 M
Calculate pH and pOH of
1.0 x 10-7 M HCl in water
• solve quadratically
• x = 6.2 x 10-8
• total H+
= 1.00 x 10-7 + 6.2 x 10-8
= 1.62 x 10-7
• pH = 6.79
• pOH = 14.00 - 6.79
= 7.21
• but CO2 acidic in water
Weak Acids and Weak Bases
• HOAc ↔
H+ + OAc-
(C - x)
x
x
• Calculate pH and POH
1.00 x 10-3 M acetic acid
• from Ka expression:
• (x)(x)/ (1.00 x 10-3 - x)
= 1.75 x 10-5 = (Ka)
• pH = - lg 1.32 x 10-4 = 3.88
• pOH = 14.00 - 3.88 = 10.12
Kb example, Calc pOH, pH of
1.0 x 10-3 M soln of NH3
• Kb = 1.75 x 10-5 = [NH4+][OH-]/[NH3]
• NH3
+ H2O ← ΝΗ4+ + ΟΗ−
-3
(1.00 x 10 - x)
x
x
• (x)(x)/1.00 x 10-3 = 1.75 x 10-5
(x neglected because small relative to 10-3 )
• x = 1.32 x 10-4 = [OH-]
• pOH = - lg 1.32 x 10-4 = 3.88
• pH = 14.00 - 3.88 = 10.12
pH of salts of weak acids/bases
• some salts when dissolved react with
water
• salts of strong acids and strong bases are
completely ionized; don’t react to give H+
or OH- e.g. NaCl
• but salt of weak acid reacts with water:
• OAc- + H20 ↔ HOAc + OHsalt of weak acid
undissociated
weak acid
Salts of Weak Acids/Bases
• sodium acetate is a salt of the weak acetic
acid, HOAc:
• HOAc ↔
H+ + OAc-
• i.e. sodium acetate is a weak base
• OAc- + H20 ↔ HOAc + OH• ionization constant is Kb
• reaction is called hydrolysis of salt ion
Calc. of Kb of salt of weak acid
• Kb can be calculated from Ka of acetic
acid and Kw
• multiply top and bottom by [H+]
KaKb = Kw
• inside the dashed line is Kw, the rest is 1/Ka:
• small Kb indicates acetate is a weak base
• product of Ka of a weak acid and Kb of its
conjugate base is always = Kw
• note “conjugate” terminology (weak base formed
from weak acid)
Calc. of pH of salt of weak acid
• calc. same as any weak base
• approximation: x neglected if C A- is > 100 K b
• best to do quadratic - trivial for calculator
• and
• or
compare with NH 3 (weak base) example - identical
Salts of Weak Bases
• Similar equations can be written for salts
of weak bases which also ionize in water
• BH+ + H2O ↔ B ++ H3O+
Homework
• Study these expressions in Harris,
e.g. 187
• Work on the examples in Harris
Summary - Salts Weak Acids/Bases
• Salts of weak acids or weak bases change pH of
water. Salts of strong acids and bases don’t
• e.g. Na+Cl- + H2O ↔(H+ + OH-) + Cl- + Na+
• HCl completely dissociated so no hydrolysis solution pH determined by [HCl]
• Na+OAc- + H2O ↔ HOAc + OH- + Na+
• OAc- - salt of weak acid, Na+ - salt of strong
base
• HOAc not completely dissociated, some OHwill form - solution basic
Summary (cont.)
• NH4+(Cl-) + H2O ↔ NH4OH + H+
• NH4+ is salt of a strong acid (HCl) and weak
base (NH3)
• solution is acidic
• salts of weak acids and weak bases, pH
depends of relative strengths of acids and
bases
Buffers
• solution that resists change in pH if small
amount of acid or base is added
• consists of mixture of weak acid and its
conjugate base, or weak base and its
conjugate acid
• acid equilibrium: HOAc ↔ H+ + OAc• if add acetate OAc- anions to system:
• H+ no longer = OAc-
Henderson-Hasselbalch eqn.
• take -ve log of each side to obtain pH:
i.e.
• inversion of last log term gives:
Ways to remember HH expn
or:
pH = pKa + log [salt]/[acid]
What do buffers do to resist pH?
• H+ is added - taken up by OH• H+ + OAc- ↔ HOAc
• OH- added - taken up by HOAc
• OH- + HOAc ↔ OAc-
• If buffer solution is 0.2 M in acetic acid
and in sodium acetate, calculate the
change in pH on adding 2.0 ml of 0.1 M
HCl to 10 ml of this solution
volume
• 10 mls of 0.2 M is 10 mls of 200 mM
1/100 liter
• = 200/100 or 2 mmoles of acetic acid and
sodium acetate at start
Buffer Example (cont.)
pH = pKa + log [salt]/[acid]
Initially pH = pKa + log 0.2M/0.2M
• i.e. pH = pKa because [salt] = [acid]
• pKa acetic acid = 4.76, so initially pH = 4.76
• If 2.0 ml of 0.1 M HCl is added
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0.2 mmoles in 2 ml added
0.2 mmoles OAc- forms 0.2 mmole of HOAc
leaves 2 mmoles - 0.2 mmole = 1.8 mmole OAcand forms 2 + 0.2 mmole = 2.2 mmole HOAcpH = 4.76 + log 1.8/2.2 = 4.67
Buffer Example (cont.)
• If add 0.2 mmoles strong base (2 ml 0.1M
NaOH)
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0.2 mmoles HOAc reacts to form 0.2 mmole OActhis gives 2 mmole + 0.2 mmole = 2.2 mmole OAcand leaves 2 - 0.2 mmole = 1.8 mmole HOAc
i.e. pH = 4.76 + lg 2.2/1.8 = 4.84
• If we had added HCl to neutral (pH 7 water)
• 2.0 ml of 0.1 M diluted to 12 mls
• MHCl = 2 x 0.1/12 = 1.66 x 10-2
• pH = 1.8
Buffer Capacity
• The amount of acid or base that can be added
without causing a large change in pH
• Determined by concentrations of HA and A• the higher [weak acid] & [conjugate base] the
more acid or base can be taken up
• If HA = A-
i.e. log [salt]/[acid] = log 1
• buffering capacity maxium for acid or base
• [salt] and [acid] must be high or adequate
• Effective range is tenfold i.e. pH = pKa ± 1
• ratio [salt]/[acid] = < 10
After [salt] or [acid] used up?
• previous example: if form more than 2
mmoles HOAc, OAc- will be used up, just
pure acid determines pH
Example - ionic strength specified
• Prepare 250 ml of 0.0500 M acetate
buffer (total acetate) pH 5.0, from glacial
acetic acid - 17 M and solid sodium
acetate Ka = 1.75 x 10-5, pKa = 4.76
[salt]
first find ratio of [acid]
pH = 4.76 + lg
[OAc − ]
[HOAc]
ionic strength (cont.)
[OAc − ]
5.0 = 4.76 + lg [HOAc]
0.24 = lg
[OAc − ]
[HOAc]
[OAc − ]
1.7 = [HOAc]
[OAc-] = 1.7 [HOAc]
[OAc-] + [HOAc] = total acetate 0 0.0500M (given)
ionic strength (cont.)
• 1.7 [HOAc] + [HOAc] = 0.0500 M
• [HOAc] = 0.0500/2.7 = 0.0185 M
• [OAc-] = 1.7 x 0.0185 = 0.0320 M
• to obtain mls and grams:
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0.0185 M acetic acid, i.e. (0.25L x 0.0185 M)/17M
= 0.00027L = 0.27 ml
0.0320 M sodium acetate is:
0.0320M x 0.25L x 82.08 g = 0.656g NaOAc
Acid-base Titrations
• titrant
always
strong acid
or strong
base
• analyte
strong base
or acid
weak base
or acid
strong acid vs. strong base
Indicators
• Equivalence point - where reaction is
theoretically complete
• End point - where reaction is observed to
be complete
• difference between them is the end point
error (p 283 Harris)
Endpoint determination
• measure pH with pH meter as function of
titrant added, or
• use indicator to visually detect endpoint
• indicator is colored weak acid or weak base
• color of base and acid forms - different
• one form may be colorless
• if indicator is a weak acid
HIn
(red)
↔
H+
+
In(blue)
• two forms of phenolphthalein:
2OH-
↔
2H+
• Henderson Hasselbalch
−
[In ]
[HIn]
pH = pKa + lg
• indicator changes color over a range
• eyes see color if one form of indicator present
at 10 times concentration of other form
if [HIn] = 10 and [In-] = 1
−
[In ]
pH = pKa + lg[HIn]
[1]
= pKa + lg[10]
[10]
and pH = pKa + lg [1]
then:
and
pH = pKa -1
pH = pKa + 1
• pH = pKa ± 1 (transition range of indicator)
Choice of indicator
• transition range should overlap steepest
part of titration curve
• pKa of indicator close to equivalence
point
• pH = pKa at middle of transition range
• similarly for a weak base indicator
pKb = pOH (at end point)
pH = 14 - pKb
Transition
ranges of
indicators
Caveats
• indicator is weak acid or base so:
• amount added should be low
• prevent contribution to pH
• color change sharper if [In] is low
• less acid/base required to change forms
• few tenths percent wt/vol of indicator used
End point and concentration
1- 100 mL 1M
HCl vs 0.1 M
NaOH
2 - 100 mL 0.01M
HCl vs 0.01M
NaOH
3 - 100 mL
0.001M HCl vs
0.001M NaOH
100 mL 0.1M NaOH vs 0.1M HCl
color change in
opposite direction
Weak acid vs strong base
same as titration of strong acid
pH at end point alkaline
(hydrolysis of conjugate
base)
buffer region
midpoint pH = pKa
[salt]
pH = pKa + log
[acid]
[salt] = [acid]
Calculations - four types
• before base added:
• solution is HA in water
• weak acid - pH determined by equilibrium
constant
HA
↔ H+ + AKa
• immediately base added, and before
equivalence, mixture of H+ and A• use Henderson Hasselbalch
[salt]
• pH = pKa + log [acid]
• at equivalence point, all HA converted to A• solution could have been made by dissolving Ain water
• weak base, pH determined by hydrolysis of AA- + H20
↔ HA + OHKb
• beyond equivalence point, excess OH• pH determined by pH of strong base
• hyrolysis of A- suppressed can be neglected
weak acid/strong base calculation
• Calculate pH of 0, 10.0, 25.0, 50.0, 60.0
mL titrant in titration of 50 mL of
0.100M acetic acid with 0.100M NaOH
• at 0mL, solution is only 0.100M HOAc
HOAc ↔
H+ + OAc(C - x)
x
x
(x)(x)
= 1.75x10 −5
0.100 − x
[H + ] = x = 1.32 x 10−3 M
pH = 2.88
equilibrium constant
expression
• at 10 mL, started with
0.100 M x 50 mL = 5 mmol HOAc
part reacted with OH- and converted to OAc• mmol HOAc at start = 5.00 mmol HOAc
• mmol OH- added
= 0.100M x 10.00 mL = 1.00 mmol OH= mmol OAC- formed in 60 mL
mmol HOAc left = 4.00 mmol HOAC in 60 mL
• This is a buffer. Volumes cancel, use mmoles
[OAc − ]
[HOAc]
1.00
pH = 4.76 + log
= 4.16
4.00
pH = pK a + lg
• At 25 mL, one-half the HOAc has been
converted to OAc-, so pH = pKa
• mmol HOAc at start = 5.00 mmol HOAc
• mmol OH- = 0.100M x 25.0 mL
= 2.50 mmol OAc- formed
• mmol HOAc left = 2.50 mmole HOAC
pH = 4.76 + lg
2.50
= 4.76
2.50
• at stoichiometric end point, 50 mL
all HOAc converted to NaOAc
• HOAc + OH- ↔ OAc- + H20
• pH given by salt of weak acid, determined by
hyrdolysis of that salt
• OAc- + H20 ↔ HOAc + OH(c-x)
x
x
where [OH-] calculated from Kb = Kw/Ka
Kb =
[OH − ][HOAc]
[OAc − ]
Kw OH
=
Ka
C
2
(x is small)
[OH − ] =
1.0 x 10 −1 4
x 0.0500
1.75 x 10 −5
= 5.35 x 10−6 M
pOH = 5.27; pH = 8.73
• At 60 mL, solution of NaOAc and excess
NaOH. Hydrolysis of acetate is negligible
in presence of added OH-. pH
determined by concentration of OH• mmol OH- = 0.100M x 10.0 mL = 1.00 mmol
in 110 mL
• [OH-] = 0.00909 M
• pOH = 2.04; pH = 11.96
titration curve and [HOac]
(1) 100 mL 0.1 M HOAc
vs 0.1 M NaOH
(2) 100 mL 0.01 M HOAc
vs 0.01M NaOH
(3) 100 mL 0.001 M HOAc
vs 0.001 M NaOH
• above, equivalence point decreases as
system becomes more dilute - does not
happen in strong acid/base systems
• phenolphthalein not useful in dilute
solutions
• equivalence point for titration of weak
acid is alkaline. For weaker acid, Kb of
salt is larger, equivalence point more
alkaline
Effect of Ka on equivalence point
100 mL 0.1 M weak acids
Polyprotic Acids
• more than one ionizable proton or OH• ionize stepwise
• equilibrium constant for each step
H3PO4 ↔ H+ + H2PO4H2PO4 ↔ H+ + HPO42HPO42- ↔ H+ + PO43Overall ionization:
H3PO4 ↔ H+ + PO43-
Ka1 = 1.1 x 10-2
Ka2 = 7.5 x 10-8
Ka3 = 4.8 x 10-13
Ka = Ka1 Ka2 Ka3
= 4.0 x 10-22
Titrations of Polyprotic Acids
• neutralized by stepwise reaction with H+
• e.g. maleic acid (H2C4H2O4)
• 2 eqivalence points
H2C4H2O4 + NaOH ↔ NaHC4H2O4 + H2O
NaHC4H2O + NaOH ↔ Na2C4H2O4 + H2O
• to see equivalence point - must be
sufficiently sharp break between the two
protons
• sufficient if pK values differ by 4 or more
• Maleic pK1 = 2.0 and pK2 = 6.26 OK
• Carbonic acid pK1 = 6.37 and pK2 = 10.25
• only one seen pK2 is too weak to see
• Phosphoric acid - only two equivalence
points seen (Ka , Ka , Ka too weak) pK1 =
2.12, pK2 = 7.21, pK3 = 12.32
2
3
3
Mixtures of acids
• same principles as polyprotic acids
• mixtures of strong and weak acids give
two equivalence points
• mixtures of two strong or two weak acids
in general only give one equivalence point
Polyprotic titration calculations
calcs work well only
for some wellbehaved diprotic
acids
Hwk, due date Mon. March 22
• for acetic acid, oxalic acid, and phosphoric acid
do the calculations for their titration with
NaOH
• plot the fractional composition diagram and
the titration curve
• use a spreadsheet to do the calculations
• assume 0.1M NaOH and titrations with 20 mL
of 0.1 M of the acids
• assume a 5 ml increment
• near the end point assume smaller increments
sufficient to define the curve
• useful web site:
http://chemistry.beloit.edu/Rain/pages/titr.html