1. Chapter 10 Acids and Bases

2. Acids Arrhenius acids produce H+ ions in water H2O
HCl(g) H+(aq) + Cl(aq)
are electrolytes
have a sour taste
turn litmus red
neutralize bases

3. Bases Arrhenius bases produce OH− ions in water taste bitter or chalky
are electrolytes
feel soapy and slippery
neutralize acids

4. Brønsted–Lowry Acids and Bases
According to the Brønsted–Lowry theory,
acids donate a proton (H+)
bases accept a proton (H+)

5. NH3, A Brønsted–Lowry Base
In the reaction of ammonia and water,
NH3 is the base that accepts H+
H2O is the acid that donates H+

6. Conjugate Acid–Base Pairs
In the reaction of HF and H2O,
one conjugate acid–base pair is HF/F−
the other conjugate acid–base pair is H2O/H3O+
each pair is related by a loss and gain of H+

7. Conjugate Acid–Base Pairs (continued)
In the reaction of NH3 and H2O,
one conjugate acid–base pair is NH3/NH4+
the other conjugate acid–base is H2O/OH–
each pair is related by a loss and gain of H+

8. Chapter 10 Acids and Bases
10.2 Strengths of Acids and Bases

9. Strengths of Acids
Strong acids completely ionize (100%) in aqueous solutions.
HCl(g) + H2O(l) H3O+(aq) + Cl−(aq)
dissociates into ions
gives H3O+ and the anion (A–)

10. Weak Acids dissociate only slightly in water to form a solution of mostly molecules and a few ions.
remain mostly as the undissociated (molecular) form
have low concentrations of H3O+ and anion (A–)
HA(aq) + H2O(l) H3O+(aq) + A−(aq)
H2CO3(aq) + H2O(l) H3O+(aq) + HCO3−(aq)

11. Weak Acids Weak acids make up most of the acids
have strong conjugate bases

12. Strong Bases Strong bases
are formed from metals of Groups 1A(1) and 2A(2)
include LiOH, NaOH, KOH, and Ca(OH)2
dissociate completely in water
KOH(s) K+(aq) + OH−(aq)

13. Weak Bases Weak bases dissociate only slightly in water
form only a few ions in water
NH3(g) + H2O(l) NH4+(aq) + OH−(aq)

14. Acid Dissociation Constant, Ka
In a weak acid, the rate of the dissociation of the acid is equal to the rate of the association.
HA + H2O H3O+ + A–
The equilibrium expression is:
Keq = [H3O+][A–]
[HA][H2O]
The concentration of H2O is constant, the Ka expression for a weak acid is:
Ka = [H3O+][A–]
[HA]

15. Writing Ka for a Weak Acid
Write the Ka for H2S.
1. Write the equation for the dissociation of H2S:
H2S(aq) + H2O(l) H3O+(aq) + HS−(aq)
2. Set up the Ka expression:
Ka = [H3O+][HS−]
[H2S]

16. Chapter 10 Acids and Bases
10.3 Ionization of Water

17. Ionization of Water In the ionization of water,
H+ is transferred from one H2O molecule to another
one water molecule acts as an acid, while another acts as a base
H2O + H2O H3O+ + OH−
H:O: + H:O: H:O:H+ + :O:H−
H H H
water water hydronium hydroxide ion(+) ion(-)

18. Ion Product of Water, Kw The ion product constant, Kw, for water
is the product of the concentrations of the hydronium and hydroxide ions
Kw = [ H3O+][ OH−]
is obtained from the concentrations in pure water
Kw = [1.0 x 10−7 M][1.0 x 10−7 M]
= x 10−14

19. Guide to Calculating [H3O+]

20. Example of Calculating [H3O+]
What is the [H3O+] of a solution if [OH−] is 5.0 x 10−8 M?
STEP 1 Write the Kw for water:
Kw = [H3O+ ][OH− ] = 1.0 x 10−14
STEP 2 Rearrange the Kw expression for [H3O+]:
[H3O+] = 1.0 x 10−14
[OH−]
STEP 3 Substitute the known [OH−] and calculate:
[H3O+] = 1.0 x 10−14 = 2.0 x 10−7 M
[5.0 x 10−8]

21. Chapter 10 Acids and Bases
10.4 The pH Scale

22. Calculating pH
Mathematically, pH is the negative log of the hydronium ion concentration
pH = −log [H3O+]
For a solution with [H3O+] = 1 x 10−4,
pH = −log [1 x 10−4 ]
pH = [4.0]
pH = 4.0

23. Significant Figures in pH
When expressing log values, the number of decimal
places in the pH is equal to the number of significant
figures in the coefficient of [H3O+].
coefficient decimal places
[H3O+] = 1 x 10−4 pH = 4.0
[H3O+] = 8.0 x 10−6 pH = 5.10
[H3O+] = 2.4 x 10−8 pH = 7.62

24. Guide to Calculating pH

25. Example 1: Calculating [H3O+] from pH
Calculate the [H3O+] for a pH value of 8.0.
[H3O+] = 1 x 10−pH
For pH = 8.0, the [H3O+] = 1 x 10−8
STEP 1 Enter the pH value, change sign: –8.0
STEP 2 Convert pH to concentration:
Use 2nd function key and then10x key
or inverse key and then log key
1 −08
STEP 3 Adjust the significant figures in the coefficient (1 digit following decimal point = 1 digit in the coefficient): [H3O+] = 1 x 10−8 M

26. Example 2: Calculating [H3O+] from pH
Calculate the [H3O+] for a pH of 3.80.
STEP 1 Enter the pH value, change sign: –3.80
STEP 2 Convert pH to concentration:
Use 2nd function key and then10x key
or inverse key and then log key
−04
STEP 3 Adjust the significant figures in the coefficient (2 digit following decimal point = 2 digit in the coefficient):
[H3O+] = x 10−4 M

27. Chapter 10 Acids and Bases
Reactions of Acids and Bases

28. Acids and Metals Acids react with metals
such as K, Na, Ca, Mg, Al, Zn, Fe, and Sn
to produce hydrogen gas and the salt of the metal
Equations:
2K(s) + 2HCl(aq) 2KCl(aq) + H2(g)
Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g)

29. Acids and Carbonates
Acids react with carbonates and hydrogen carbonates to produce carbon dioxide gas, a salt, and water.
2HCl(aq) + CaCO3(s) CO2(g) + CaCl2(aq) + H2O(l)
HCl(aq) + NaHCO3(s) CO2(g) + NaCl (aq) + H2O(l)

30. Neutralization Reactions
Neutralization is the reaction of
an acid such as HCl and a base such as NaOH
HCl(aq) + H2O(l) H3O+ (aq) + Cl−(aq)
NaOH(aq) Na+ (aq) + OH−(aq)
the H3O+ from the acid and the OH− from the base to form water
H3O+(aq) + OH−(aq) H2O(l)

31. Neutralization Equations
In the equations for neutralization, an acid and a base produce a salt and water.
acid base salt water
HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)
2HCl(aq) + Ca(OH)2(aq) CaCl2(aq) + 2H2O(l)

32. Guide to Balancing an Equation for Neutralization

33. Balancing Neutralization Reactions
Write the balanced equation for the neutralization of
magnesium hydroxide and nitric acid.
STEP 1 Write the base and acid formulas:
Mg(OH)2(aq) + HNO3(aq)
STEP 2 Balance OH– and H+:
Mg(OH)2(aq) + 2HNO3(aq)
STEP 3 Balance with H2O:
Mg(OH)2(aq) + 2HNO3(aq) salt + 2H2O(l)
STEP 4 Write the salt from remaining ions:
Mg(OH)2(aq) + 2HNO3(aq) Mg(NO3)2(aq) + 2H2O(l)

34. Guide to Calculating Molarity

35. Calculating Molarity from a Titration with a Base
What is the molarity of an HCl solution if 18.5 mL of a
0.225 M NaOH are required to neutralize 10.0 mL HCl?
HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)
STEP 1 Given: 18.5 mL ( L) of M NaOH
10.0 mL of HCl
Need: M of HCl
STEP 2 Plan:
L of NaOH moles of NaOH moles of HCl
M of HCl

36. Calculating Molarity from a Titration with a Base (continued)
STEP 3 State equalities and conversion factors:
1 L of NaOH = mole of NaOH
1 L of NaOH and mole NaOH
0.225 mole NaOH L of NaOH
1 mole of NaOH = 1 mole of HCl
1 mole of NaOH and 1 mole HCl
1 mole HCl mole of NaOH
STEP 4 Set up the problem to calculate moles of HCl:
L NaOH x mole NaOH x 1 mole HCl
1 L NaOH mole NaOH
= mole of HCl

37. Calculating Molarity from a Titration with a Base (continued)
STEP 4 (continued)
Calculate the volume in liters of HCl:
10.0 mL HCl = L HCl
Calculate the molarity of HCl:
mole HCl = M HCl
L HCl

38. Chapter 10 Acids and Bases
10.7 Buffers

39. Buffers Buffers resist changes in pH from the addition of acid or base
in the body absorb H3O+ or OH from foods and cellular processes to maintain pH
are important in the proper functioning of cells and blood
in blood maintain a pH close to 7.4; a change in the pH of the blood affects the uptake of oxygen and cellular processes

40. Components of a Buffer The components of a buffer solution
are acid–base conjugate pairs
can be a weak acid and a salt of its conjugate base
typically have equal concentrations of the weak acid and its salt
can also be a weak base and a salt of its conjugate acid

41. Summary of Buffer Action
Buffer action occurs because
the weak acid in a buffer neutralizes base
the conjugate base in the buffer neutralizes acid
the pH of the solution is maintained

42. pH of a Buffer
The [H3O+] in the Ka expression is used to determine the pH of a buffer.
Weak acid + H2O H3O+ + conjugate base
Ka = [H3O+][conjugate base]
[weak acid]
[H3O+] = Ka x [weak acid]
[conjugate base]
pH = log [H3O+]

43. Guide to Calculating pH of a Buffer

44. Example of Calculating Buffer pH
The weak acid H2PO4 in a blood buffer H2PO4/HPO42
has a Ka = 6.2 x 108. What is the pH of the buffer if
[H2PO4] = 0.20 M and [HPO42] = 0.20 M?
STEP 1 Write the Ka expression for:
H2PO4(aq) + H2O(l) HPO42(aq) + H3O+(aq)
Ka = [HPO42][H3O+]
[H2PO4]
STEP 2 Rearrange the Ka for [H3O+]:
[H3O+] = Ka x [H2PO4]
[HPO42]
Dihydrogen phosphate ion and hydrogen phosphate ion

45. Example of Calculating Buffer pH (continued)
STEP 3 Substitute [HA] and [A]:
[H3O+] = 6.2 x 108 x [0.20 M] = 6.2 x 108
[0.20 M]
STEP 4 Use [H3O+] to calculate pH:
pH =  log [6.2 x 10 8] = 7.21