Simplified understandable Acid and bases For nutrition and dietetics students at North Coast Medical Training.

1. Walter Waswa
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2. ACID
Arrhenius : an acid is a material that can release a proton or
hydrogen ion (H +)
Brønsted : an acid is a material that donates a proton.
Lewis definitions are: Acids are electron pair acceptors
CONUGATE BASE:Each acid has a proton available (an
ionizable hydrogen) and another part.

3. When the acid ionizes, the hydrogen ion is the acid and the
rest of the original acid is the conjugate base.
Nitric acid, HNO 3, dissociates (splits) into a hydrogen ion
and a nitrate ion.
The hydrogen almost immediately joins to a water molecule
to make a hydronium ion.
 The nitrate ion is the conjugate base of the hydrogen ion.
 In the second part of the reaction, water is a base (because it
can accept a proton) and the hydronium ion is its conjugate
acid.

4. HNO3+ H2O (NO3)- + (H3O)+
ACID BASE CONUGATE BASES CONJUGATE ACID

5. PROPERTIES OF ACIDS
1. Acids release a hydrogen ion into water (aqueous)
solution
2. Acids neutralize bases in a neutralization
reaction.
3. Acids corrode active metals.
4. Acids turn blue litmus to red.
5. Acids taste sour..

6. Common Acids:
Strong Acids The Formula
Sulphuric acid
Hydrochloric acid
Hybrobromic acid
Hydroiodic acid
Nitric acid
Perchloric acid
H2SO4
HCl
HBr
HI
HNO3
HClO4
All others considered Weak (examples)
Weak Acid The Formula
Acetic acid (vinegar) HC2H3O2
Carbonic acid HCO3
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7. bases
Arrhenius: is a material that can donate a hydroxide ion (OH-
Brønsted: base is a material that can accept a proton
Lewis: bases are electron pair donors.

8. Properties of bases
1. Bases release a hydroxide ion into water solution
2. Bases neutralize acids in a neutralization reaction.
3. Bases denature protein.
4. Bases turn red litmus to blue.
5. Bases taste bitter

9. Common Bases:
Strong Bases The Formulae
Lithium hydroxide
Sodium hydroxide
Potassium hydroxide
Rubidium hydroxide
Caesium hydroxide
Barium hydroxide
Calcium hydroxide
Strontium hydroxide
LiOH
NaOH
KOH
RbOH
CsOH
Ba(OH)2
Ca(OH)2
Sr(OH)2
(Hydroxides of Group 1 and Group 2 Metals
are STRONG)
All others are WEAK
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10. STRENGTH OF ACIDS AND BASES
A strong acid is one that fully ionizes eg
HNO3 - nitric acid
HCl1 - hydrochloric acid
H2SO4 - sulfuric acid
HClO4 - perchloric acid
HBr1 - hydrobromic acid
HI1 - hydroiodic acid

11. Incompletely ionized acids are called weak acids.
because there is a smaller concentration of ionized hydrogens
available in the solution
differences in concentration of the entire acid will be termed
dilute or concentrated

12. ones that completely ionize into hydroxide ions and a
conjugate acid=Strong base
All of the bases of Group I and Group II metals except for
beryllium are strong bases
The bases of Group I metals are all monobasic. The bases of
Group II metals are all dibasic. Aluminum hydroxide,
Al(OH)3 is tribasic.
 Any material with two or more ionizable hydroxyl groups
would be called polybasic

13. Most of the alkaline organic compounds have an amino group
-(NH2) rather than an ionizable hydroxyl group.
The amino group attracts a proton (hydrogen ion) to become
-(NH3 )+.
 (The dash before the (NH3)+ or (NH2) indicates a single
bonding electron, so this is attached to something else by a
covalent bond.)
By the Lowry- Brønsted definition, an amino group
definitely acts as a base, and the effect of removing hydrogen
ions from water molecules is the same as adding hydroxide
ions to the solution.

15. Ph
The pH of a solution is the negative log of the hydrogen ion
concentration.
The hydrogen ion concentration is inversely proportional to
the hydroxide ion concentration, and the two of them
multiplied together give the number 1 E-14.

16. WEAK ACID AND WEAK BASES
We can write the chemical equation for the dissociation of a
weak acid, using 'A-' to represent the conjugate base, as;
HA A- + H+
we can write the chemical equation for the dissociation of a
weak base, using 'X+' to represent the conjugate acid, as;
XOH (OH)- + X+

17. The equilibrium expression for the dissociation of a weak acid is
The dissociation constant of an acid is equal to the concentration
of hydrogen ions times the concentration of the conjugate base of
the acid divided by the concentration of un-ionized acid."

18. Equilibrium expression for a weak
base reads: "The dissociation
constant of a base equals
concentration of hydroxide ions
times concentration of conjugate
acid divided by the concentration
of un-ionized base."

19. The kA of an acid or the kB of a base are properties of that
acid or base at the given temperature. The temperature at
which these dissociation constants are listed is usually near
room temperature.

20. INDICATORS An indica tor is a compound tha t will change
color in the presence of an a cid or ba se
litmus phenophthalein
 Acid-red
 Base-blue
 Neutral-purple
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 Acid-colourles
 Base-pink
 Neutral-colourles
Methyl orange
Acid-pink
Base-yellow
Neutral-orange
Universal indicator
P 59 sec.chem f1 5th ed.by klb 2003

21. Acid Strength and pKa:
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Acids

22. Figure 2.5 Summary of the factors that determine acidity
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Acids and Bases
Factors that Determine Acid Strength

23. 2major Factors that affect strength of
an acid
1) Bond strength :The
strength of the bond
between the acidic
proton and the rest of the
molecule will have an effect on
acidity.
The weaker the bond, the
more acidic the acid will be
generally.
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 2) Bond polarity :The polarity of a
bond is the distribution of the
electrons between the two bonded atoms.
 If the electrons are fairly equally
distributed, the bond is not very polar. As
the electron distribution gets weighted
towards one atom, the bond becomes
more polar .
 A highly polar bond between an acidic
hydrogen and another atom tends to make
it more easy for the proton to leave the
molecule than would happen for a non-polar
bond.

24. Brønsted-Lowry Theory of Acids & Bases
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Conjugate Acid-Base Pairs
General Equation

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28. Brønsted-Lowry Theory of Acids & Bases
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29. Brønsted-Lowry Theory of Acids & Bases
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30. Brønsted-Lowry Theory of Acids & Bases
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Notice that water is both an
acid & a base = amphoteric
Reversible reaction

31. EELLEECCTTRROOLLYYTTEESS
Electrolytes are species which conducts electricity when
dissolved in water. Acids, Bases, and Salts are all electrolytes.
Salts and strong Acids or Bases form Strong Electrolytes. Salt and
strong acids (and bases) are fully dissociated therefore all of
the ions present are available to conduct electricity.
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HCl(s) + H2O ® H3O+ + Cl-
Weak Acids and Weak Bases for Weak Electrolytes. Weaks
electrolytes are partially dissociated therefore not all species
in solution are ions, some of the molecular form is present.
Weak electrolytes have less ions avalible to conduct
electricity.
NH3 + H2O ® NH4
+ + OH-

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33. AAcciiddss && BBaasseess
SSTTRROONNGG vvss WWEEAAKK
__ ccoommpplleetteellyy iioonniizzeedd __ ppaarrttiiaallllyy iioonniizzeedd
__ ssttrroonngg eelleeccttrroollyyttee __ wweeaakk eelleeccttrroollyyttee
__ iioonniicc//vveerryy ppoollaarr bboonnddss __ ssoommee ccoovvaalleenntt bboonnddss
SSttrroonngg AAcciiddss:: SSttrroonngg BBaasseess::
HHCCllOO44 LLiiOOHH
HH22SSOO44 NNaaOOHH
HHII KKOOHH
HHBBrr CCaa((OOHH))22
HHCCll SSrr((OOHH))22
HHNNOO33 BBaa((OOHH))22
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34. AAcciiddss && BBaasseess
 One ionizable proton:
HCl → H+ + Cl-
 Two ionizable protons:
H2SO4 → H+ + HSO4
- → H+ + HPO4
2- → H+ + PO4
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-
HSO4
- → H+ + SO4
2-
 Three ionizable protons:
H3PO4 → H+ + H2PO4
–
H2PO4
2-
HPO4
-3
Combined:
H2SO4 → 2H+ + SO4
2-
Combined:
H3PO4 → 3H+ + PO4
3-

35. AAcciiddss && BBaasseess
For the following identify the acid and the base as strong or weak .
a. Al(OH)3 + HCl ®
Weak base Strong acid
b. Ba(OH)2 + HC2H3O2 ®
c. KOH + H2SO4 ®
d. NH3 + H2O ®
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Weak acid
Strong acid
Strong base
Strong base
Weak base Weak acid

36. AAcciiddss && BBaasseess
For the following predict the product. To check your answer left click on the
mouse. Draw a mechanism detailing the proton movement.
a. Al(OH)3 + HCl ®
3
b. Ba(OH)2 + HC2H3O2 ®
c. KOH + H2SO4 ®
d. NH3 + H2O ®
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AlCl3 + 3 H2O
Ba(C2H3O2)2 + 2 H2O
K2SO4 + 2 H2O
NH4
+ + OH-
2
2

37. Reactions between
acids and bases
When and acid and a base
react with each other, the
characteristic properties of
both are destroyed. This is
called neutralization.
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38. Reactions between
acids and bases
General formula for acid base reaction:
Acid + Base → HO + Salt
2“Salt” means any ionic
compound formed from
NOT JUST
NaCl !!
an acid/base reaction
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Neutralization Reaction Animation

39. Neutralization
HCl + NaOH → H2O + NaCl
acid base water salt
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40. Neutralization
Another Example
HNO3 + KOH → H2O + KNO3
HNO KOH 3
acid base water salt
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41. Concentration
Strength of an acid or base is determined by the
amount of ionization. Concentration is determined
by the amount of water added to the substance.
Molarity (M)
The number of moles of solute dissolved in
each liter of solution.
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Molarity = moles of solute
liters of solution

42. Example Problem #1
If 1.00 liter of sugar water contains
exactly 1.00 mole of sugar, what is its
molarity?
Molarity = 1.00 mol
1.00 L
Molarity = 1.00 M
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43. Example Problem #2
If 1.00 liter of sugar water contains
exactly 2.00 mole of sugar, what is its
molarity?
Molarity = 2.00 mol
1.00 L
Molarity = 2.00 M or 2.00 mol/L
(Twice as concentrated…)
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44. Example Problem #3
What is the molarity when 0.75 mol is
dissolved in 2.50 L of solution?
Molarity = 0.75 mol = 0.30 mol/L or 0.30M
2.50 L
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45. In Lab, grams are typically
used in place of moles.
If you wanted to make 2.00L of a 6M
HCl solution, how much HCl would you
need?
First, calculate the molar mass of the
acid.
H 1 x 1.00795 = 1.00795
Cl 1 x 35.453 = 35.453
36.46095 =
36.461
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46. If you wanted to make 2.00L of a 6M HCl
solution, how much HCl would you need?
First, calculate the molar mass of the acid.
HCl contains 36.461 g/mol
It would take 36.461 g of HCl to make 1 liter of a 1M
HCl solution. How many grams would it take to make
2L of a 1M solution?
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2 x 36.461g = 72.922g

47. If you wanted to make 2.00L of a 6M HCl
solution, how much HCl would you need?
It takes 72.922g of HCl to make 2 liters of a 1M solution.
How much would it take to make 2 liters of a 6M solution?
6 x 72.922g = 437.532 g
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48. Try One More
Suppose you wanted to make 2 liters of a 0.5 M solution
of HCl. How much HCl would you need?
Each mole of HCl is equal to 36.461g
For a 0.5 M solution, you would need half that much.
36.461 x 0.5 = 18.2305g.
However, you want to make 2 liters, so double that
amount. 18.2305 x 2 = 36.461g.
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49. TTIITTRRAATTIIOONN
Titration of a strong acid with a strong base
EENNDDPPOOIINNTT == PPOOIINNTT OOFF NNEEUUTTRRAALLIIZZAATTIIOONN ==
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EEQQUUIIVVAALLEENNCCEE PPOOIINNTT
At the end point for the titration of a strong acid with a strong
base, the moles of acid (H+) equals the moles of base (OH-) to
produce the neutral species water (H2O). If the mole ratio in
the balanced chemical equation is 1:1 then the following
equation can be used.
MOLES OF ACID = MOLES OF BASE
nnaacciidd == nnbbaassee
Since M=n/V
MMAAVVAA == MMBBVVBB

50. TTIITTRRAATTIIOONN
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MMAAVVAA == MMBBVVBB
1. Suppose 75.00 mL of hydrochloric acid was required to
neutralize 22.50 mLof 0.52 M NaOH. What is the molarity of
the acid?
HHCCll ++ NNaaOOHH ® HH22OO ++ NNaaCCll
MMaa VVaa == MMbb VVbb rreeaarrrraannggeess ttoo MMaa == MMbb VVbb // VVaa
ssoo MMaa == ((00..5522 MM)) ((2222..5500 mmLL)) // ((7755..0000 mmLL))
== 00..1166 MM
Now you try:
2. If 37.12 mL of 0.843 M HNO3 neutralized 40.50 mL of KOH,
what is the molarity of thMe base? b = 0.773 mol/L

51. Molarity and Titration
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52. TTIITTRRAATTIIOONN
Titration of a strong acid with a strong base
EENNDDPPOOIINNTT == PPOOIINNTT OOFF NNEEUUTTRRAALLIIZZAATTIIOONN ==
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EEQQUUIIVVAALLEENNCCEE PPOOIINNTT
At the end point for the titration of a strong acid with a strong
base, the moles of acid (H+) equals the moles of base (OH-) to
produce the neutral species water (H2O). If the mole ratio
in the balanced chemical equation is NOT 1:1 then you must
rely on the mole relationship and handle the problem like
any other stoichiometry problem.
MMOOLLEESS OOFF AACCIIDD == MMOOLLEESS OOFF BBAASSEE
nnaacciidd == nnbbaassee

53. TTIITTRRAATTIIOONN
1. If 37.12 mL of 0.543 M LiOH neutralized 40.50 mL of
H2SO4, what is the molarity of the acid?
22 LLiiOOHH ++ HH22SSOO44 ® LLii22SSOO44 ++ 22 HH22OO
First calculate the moles of base:
00..0033771122 LL LLiiOOHH ((00..554433 mmooll//11 LL)) == 00..00220022 mmooll LLiiOOHH
Next calculate the moles of acid:
00..00220022 mmooll LLiiOOHH ((11 mmooll HH22SSOO44 // 22 mmooll LLiiOOHH))== 00..00110011 mmooll HH22SSOO44
Last calculate the Molarity::
MMaa == nn//VV == 00..001100 mmooll HH22SSOO44 // 00..44005500 LL == 00..224488 MM
2. If 20.42 mL of Ba(OH)2 solution was used to
titrate29.26 mL of 0.430 M HCl, what is the molarity of
the barium hydroxide solution?
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Mb = 0.308 mol/L

54. ppHH
AA mmeeaassuurree ooff tthhee hhyyddrroonniiuumm iioonn
 The scale for measuring the hydronium ion concentration [H3O+]
in any solution must be able to cover a large range. A
logarithmic scale covers factors of 10. The “p” in pH stands for
log.
 A solution with a pH of 1 has [H3O+] of 0.1 mol/L or 10-1
 A solution with a pH of 3 has [H3O+] of 0.001 mol/L or 10-3
 A solution with a pH of 7 has [H3O+] of 0.0000001 mol/L or 10-7
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ppHH == -- lloogg [[HH33OO++]]

55. TThhee ppHH ssccaallee
TThhee ppHH ssccaallee rraannggeess ffrroomm 11 ttoo 1100--1144 mmooll//LL oorr ffrroomm 11 ttoo
1144..
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pH = - log [H3O+]
1 2 3 4 5 6 7 8 9 10 11 12 13 14
aacciidd nneeuuttrraall bbaassee

56. MMaanniippuullaattiinngg ppHH
AAllggeebbrraaiicc mmaanniippuullaattiioonn ooff::
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pH = - log [H3O+]
aalllloowwss ffoorr::
[H3O+] = 10-pH
IIff ppHH iiss aa mmeeaassuurree ooff tthhee hhyyddrroonniiuumm iioonn
ccoonncceennttrraattiioonn tthheenn tthhee ssaammee eeqquuaattiioonnss ccoouulldd bbee
uusseedd ttoo ddeessccrriibbee tthhee hhyyddrrooxxiiddee ((bbaassee))
ccoonncceennttrraattiioonn..
[OH-] = 10-pOH pOH = - log [OH-]
tthhuuss::
pH + pOH = 14 ; the entire pH range!

57. PRACTICE PROBLEM #25
1. How many milliliters of 1.25 M LiOH must be added to neutralize 34.7 mL
of 0.389 M HNO3?
2. What mass of Sr(OH)2 will be required to neutralize 19.54 mL of 0.00850 M
HBr solution?
3. How many mL of 0.998 M H2SO4 must be added to neutralize 47.9 mL of
1.233 M KOH?
4. What is the molar concentration of hydronium ion in a solution of pH
8.25?
5. What is the pH of a solution that has a molar concentration of hydronium
ion of 9.15 x 10-5?
6. What is the pOH of a solution that has a molar concentration of
hydronium ion of 8.55 x 10-10?
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10.8 mL
0.0101 g
29.6 mL
5.623 x 10-9 M
pH = 4.0
pOH = 4.9

58. Molarity and Titration
A student finds that 23.54 mL of a 0.122 M NaOH solution
is required to titrate a 30.00-mL sample of hydr acid
solution. What is the molarity of the acid?
A student finds that 37.80 mL of a 0.4052 M NaHCO3
solution is required to titrate a 20.00-mL sample of sulfuric
acid solution. What is the molarity of the acid?
The reaction equation is:
H2SO4 + 2 NaHCO3 → Na2SO4 + 2 H2O + 2 CO2
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Water Equilibrium

60. Water Equilibrium
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Kw = [H+] [OH-] = 1.0 x 10-14
Equilibrium constant for water
Water or water solutions in which [H+] = [OH-] = 10-7 M
are neutral solutions.
A solution in which [H+] > [OH-] is acidic
A solution in which [H+] < [OH-] is basic

61. GROUP STUDY PROBLEM #25
______1. How many milliliters of 0.75 M KOH must be added to neutralize 50.0 mL
of 2.50 M HCl?
______2. What mass of Ca(OH)2 will be required to neutralize 100 mL of 0.170 M
HCl solution?
______3. How many mL of 0.554 M H2SO4 must be added to neutralize 25.0 mL of
0.9855 M NaOH?
______ 4. What is the molar concentration of hydronium ion in a solution of pH
2.45?
______ 5. What is the pH of a solution that has a molar concentration of
hydronium ion of 3.75 x 10-9?
______ 6. What is the pOH of a solution that has a molar concentration of
61 hydronium WALTER ion WASWA
of 4.99 x 10-4?

62. KWISHA!
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63. Kwisha!
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ANY QUESTIONS?
PROFESA WALTER WAKHUNGU

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