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AP Chemistry. Ch. 14 -Acids, Bases, and Acid-Base Equilibrium. Properties of Acids. pH below 7 Conduct electricity in solution React with to form Products Metals to form H 2 Carbonates to form CO 2 , H 2 0, and a salt Bicarbonates to form CO 2 , and H 2 0

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ap chemistry

AP Chemistry

Ch. 14 -Acids, Bases, and Acid-Base Equilibrium

properties of acids
Properties of Acids
  • pH below 7
  • Conduct electricity in solution
  • React with to form Products
    • Metals to form H2
    • Carbonates to form CO2, H20, and a salt
    • Bicarbonates to form CO2, and H20
    • A Base to form H20 and a salt
properties of bases
Properties of Bases
  • pH above 7
  • Turns red litmus paper to blue
  • “Basic Blue”
  • Conduct electricity in solution
  • React with to form Products
    • Metal ions to form Precipitates
    • Bicarbonates to form CO3, and H20
    • An acid to form H20 and a salt
definitions of acids bases
Definitions of Acids & Bases
  • Arrhenius Definition of Acids and Bases:
  • Most elementary of ideas about acids
  • a) Acids produce H+ ions in solutions
  • b) Bases produce (OH)- ions in solutions
strong acids
HI – hydroiodic 7 you need

HBr-hydrobromic to remember.

HCl - hydrochloric

HClO4 - perchloric

H2SO4 - sulfuric

HClO3 - chloric

HNO3 - nitric

Strong Acids
acid and base strength
Acid and Base Strength
  • Strong acids are completely dissociated in water.
    • Their conjugate bases are quite weak.
  • Weak acids only dissociate partially in water.
    • Their conjugate bases are weak bases.
strong bases
LiOH – lithium hydroxide

NaOH Sodium Hydroxide

KOH Potassium Hydroxide

Ca(OH)2 Calcium Hydroxide

Sr(OH)2 Strontium Hydroxide

Ba(OH)2 Barium Hydroxide

Strong Bases

6 you need to remember

problems with arrhenius
Problems with Arrhenius
  • The Arrhenius definitions of acids and bases did not properly explain why other substances are acids or bases
  • Example) NH3
bronsted lowry definition
Bronsted-Lowry Definition
  • These two scientists devise better definitions for acids and bases which would encompass more substances
  • Bronsted-Lowry Acid: A proton donor
  • Bronsted-Lowry Base: A proton acceptor
more about bronsted lowry
More About Bronsted-Lowry
  • A proton is simply a H+ ion
  • The hydrogen in substances is described as being ionizable
  • This ionizable hydrogen is attracted to a center of negative charge (lone pairs of electrons)
more about bronsted lowry1

H

H

O

O+

H

H

Cl

Cl-

H

H

More About Bronsted-Lowry
  • HCl + H2O  H3O+ + Cl-
more about bronsted lowry2
More About Bronsted-Lowry
  • 1) Show the reaction between HCl and NH3 and clearly identify the acid and the base
  • 2) Identify the acid and the base when NH3 is placed in water
conjugate acids and bases
Conjugate Acids and Bases
  • When an acid-base reactions occurs, a conjugate acid and base is formed
  • Conjugate acids are the original base plus a hydrogen ion (ie it is the acid on the product side of the equation)
conjugate acids and bases1
Conjugate Acids and Bases
  • Conjugate bases are the original acid minus a hydrogen ion (ie it is the base on the product side of the equation)
conjugate acids and bases2
Conjugate Acids and Bases:
  • Reactions between acids and bases always yield their conjugate bases and acids.
identify the bronsted lowry acids and bases their conjugates
Identify the Bronsted-Lowry Acids and Bases & their conjugates
  • 1) H2S + NH3 NH4+ + HS-

Acid base con. Base con. base

  • 2) OH- + H2PO4-  H2O + HPO42

base acid con. acid con. base

significance of conjugate acid bases
Significance of Conjugate Acid-Bases
  • The stronger an acid, the weaker is its conjugate base
  • The stronger a base, the weaker is its conjugate acid
  • Acid-Base reactions favor the direction of the stronger member to the weaker member of each pair
acids base definitions
Acids & Base Definitions

Definition #3 – Lewis

Lewis acid - a substance that accepts an electron pair

Lewis base - a substance that donates an electron pair

lewis acids bases
Lewis Acids & Bases

Formation ofhydronium ion is also an excellent example.

  • Electron pair of the new O-H bond originates on the Lewis base.
ap practice
AP Practice
  • In the lab H2(g) can be produced by adding which of the following to 1.0M HCl(aq)?
  • I. 1 M NH3 (aq)
  • II. Zn (s)
  • III. NaHCO3 (s)

A. I only C. III only E. I, II, and III

B. II only D. I and II only

ap practice1
AP Practice
  • In liquid ammonia, the reaction represented below occurs.

2NH3 NH4+ + NH2-

In the reaction, NH4+ acts as:

  • A. a catalyst
  • B. Both an acid and a base
  • C. the conjugate acid of NH3
  • D. The reducing agent
  • E. the oxidizing agent
ap practice 4
AP Practice #4
  • Write the balance net ionic equation for :

A 0.1M nitrous acid solution is added to the same volume of a 0.1M sodium hydroxide solution.

HNO2 + OH- NO2- + H2O

ap practice 41
AP Practice #4
  • Write the net ionic equation for:

Hydrogen iodide gas is bubbled into a solution of lithium carbonate.

2 HI + CO3-2 2 I- + H20 + CO2

ap practice 42
AP Practice #4
  • Write the balanced net ionic equation for: Concentrated hydrochloric acid is added to a solution of sodium sulfide.

2 H+ + S-2 H2S

ap practice 43
AP Practice #4
  • Write the balanced net ionic equation for: A solution of ethanoic (acetic) acid is added to a solution of barium hydroxide:

HC2H3O2 + OH- H2O + C2H3O2-

slide28

Strong and Weak Acids/Bases

The strength of an acid (or base) is determined by the amount of IONIZATION.

slide29

Strong and Weak Acids/Bases

  • Weak acids are much less than 100% ionized in water.

*One of the best known is acetic acid = CH3CO2H

slide30

Strong Acid

Weak Acid

15.4

slide31

CaO

Strong and Weak Acids/Bases

  • Strong Base:100% dissociated in water. Have very weak conjugate acids.

NaOH (aq) ---> Na+ (aq) + OH- (aq)

Other common strong bases include KOH andCa(OH)2.

CaO (lime) + H2O -->

Ca(OH)2 (slaked lime)

slide32

Strong and Weak Acids/Bases

  • Weak base:less than 100% ionized in water

One of the best known weak bases is ammonia

NH3 (aq) + H2O (l)  NH4+ (aq) + OH- (aq)

significance of acid base strength
Significance of Acid/Base Strength
  • When equilibrium is established, the side in which there are stronger acids/bases will shift toward the weaker sides (LeChatlier Principle)
  • Thus the concentration of substances will favor the weaker members
factors affecting acid strength
Factors Affecting Acid Strength
  • A) Binary Acids: The lower the bond dissociation energy, the easier the bond is broken.
  • More likely that acid will donate a H ion.
  • (Low BE = Stronger Acid (Weaker Conjugate Base)
  • ***Low BE = Stronger Acid***
slide36

H+ + X-

H X

The stronger the bond

Molecular Structure and Acid Strength

  • Bond strength
  • Polarity

The weaker the acid

HF << HCl < HBr < HI

factors affecting acid strength1
Factors Affecting Acid Strength
  • The larger the anion, the stronger that acid is (the easier the bond is broken)
  • Acid strength increases going across the table right to left while increasing going down the table
  • {the greater the distribution in charge (polarity) the more likely the substance will lose H ion}
factors affecting acid strength2
Factors Affecting Acid Strength
  • Nonpolar covalent acids are weaker than polar covalent acids which are weaker than ionic acids
  • Essentially, the greater the dipole in the acid, the more likely the acid is strong!
slide39

d-

d+

O-

Z

+ H+

O

Z

H

Molecular Structure and Acid Strength

  • The O-H bond will be more polar and easier to break if:
  • Z is very electronegative or
  • Z is in a high oxidation state
ap chemistry quiz
AP Chemistry Quiz
  • 1. Explain how to determine between a Bronsted-Lowry Acid and Base.
  • 2. Why is HBr is a stronger acid than H2S?
oxoacids
Oxoacids:
  • B) Oxoacids: Contain hydrogen, oxygen, and some other element (nonmetal). At least one H bonded to an O.
  • The other element’s tendency to attract other electrons assists in determining the strength of the acid
oxoacids1
Oxoacids:
  • If the other element attracts electrons very strongly, electrons are withdrawn from Oxygen – Hydrogen bond. This weakens the O-H bond and results in stronger acids.
slide43

O

H O Cl O

O

H O Br O

Molecular Structure and Acid Strength

1. Oxoacids having different central atoms (Z) that are from the same group and that have the same oxidation number.

Acid strength increases with increasing electronegativity of Z

Cl is more electronegative than Br

HClO3 > HBrO3

slide44
Adding more oxygen atoms that are added (H2SO4 and H2SO3) is the same as adding more electronegative elements.
  • Since oxygen has a high electronegativity, this withdraws electrons from the O-H bond and results in stronger acids
slide45

Molecular Structure and Acid Strength

2. Oxoacids having the same central atom (Z) but different numbers of attached groups.

Acid strength increases as the oxidation number of Z increases.

HClO4 > HClO3 > HClO2 > HClO

ph scale
pH Scale
  • The pH Scale runs 0-14
  • pH values < 7 are acidic
  • pH values > 7 are basic (alkaline)
  • pH values = 7, Neutral
how do we measure ph
How Do We Measure pH?
  • For less accurate measurements, one can use
    • Litmus paper
      • “Red” paper turns blue above ~pH = 8
      • “Blue” paper turns red below ~pH = 5
    • An indicator
how do we measure ph1
How Do We Measure pH?

For more accurate measurements, one uses a pH meter, which measures the voltage in the solution.

mathematical determination of ph
Mathematical Determination of pH
  • pH = - log [H+] OR
  • - log [H3O+]
mathematical determination of poh other relationships
Mathematical Determination of pOH & Other Relationships
  • pOH = - log [OH-]
  • pH + pOH = 14
  • [H+] * [OH-] = 1 x 10-14 = Kw
pk a and pk b values
pKa and pKb Values
  • pKa = - log Ka
  • pKb = - log Kb
  • Low values for pKa & pKb correspond to large values for Ka and Kb
other p scales
Other “p” Scales
  • The “p” in pH tells us to take the negative log of the quantity (in this case, hydrogen ions).
  • Some similar examples are
    • pOH = −log [OH−]
    • pKw = −log Kw
homework 38
Homework #38

Kw= [H+] [OH-]

a. [H+] = = 6.7 × 10-15M

pH=-log [H+] =14.1 Basic

b. [H+] = = 2.8 M; acidic

38 continued
38 Continued
  • [H+] = = 1.0 × 10-7M; neutral

d. [H+] = = 1.4 × 10-11M; basic

number 44
Number 44

pH pOH

  • 9.63 14.00 – 9.63 = 4.37
  • [H]
  • 10-pH =[H+] = = 2.3 × 10-10M
  • [OH]
  • 10-pOH =[ OH] = = 4.3 × 10-5M;
  • acid/base/neutral
  • BASIC
calculating the ph of strong acids or bases
Calculating the pH of strong acids or bases
  • Since strong acids and bases completely ionize, we assume that no reactant remains together.
  • Focus on the major species present in solution.
  • To help understand lets calculate the pH of a 1M HCl solution.
stong acid ph
Stong acid pH
  • Major species in solution:
  • H+, Cl- H20 (does not contribute enough H)
  • 1M [H+] and 1M of [Cl-] are in solution
  • pH= -log [H+]
  • So pH= -log[1]
  • pH=0
problem 48
Problem 48
  • Strong acids are assumed to completely dissociate in water: HCl(aq) → H+(aq) + Cl (aq)

a. 0.10 M HCl solution gives 0.10 M H+ and 0.10 M Cl since HCl completely dissociates. pH = -log [H+] = -log (0.10) = 1.00

b. 5.0 M H+ is produced when 5.0 M HCl completely dissociates. pH = -log (5.0) = -0.70 (Negative pH values just indicate very concentrated acid solutions).

48 cont
48 Cont.
  • C. 1.0 × M H+ is produced when 1.0 × M HCl completely dissociates. This gives pH = 11.00. This is impossible!
  • We dissolved an acid in water and got a basic pH. What we must consider in this problem is that water by itself donates 1.0 × 10-7 M H+.
  • We can normally ignore the small amount of H+ from H2O except when we have a very dilute solution of an acid (as is the case here). Therefore, the pH is that of neutral water (pH = 7.00) since the amount of HCl present is insignificant.
weak acids
Weak Acids
  • Weak acids do not ionize 100% but instead establish an equilibrium or a Ka value.
  • Still use products over reactants to calculate Ka.
  • Must now focus on major species in a reaction and use ICE charts.
dissociation constants

Ka =

[H3O+] [A−]

[HA]

HA(aq) + H2O(l)

A−(aq) + H3O+(aq)

Dissociation Constants
  • For a generalized acid dissociation,

the equilibrium expression would be

  • This equilibrium constant is called the acid-dissociation constant, Ka.
dissociation constants1
Dissociation Constants

The greater the value of Ka, the stronger the acid.

equilibria involving a weak acid
Equilibria Involving A Weak Acid

You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H3O+, OAc-, and the pH.

Step 1.Define equilibrium conc. in ICE table.

[HOAc] [H3O+] [OAc-]

initial

change

equilib

1.00 0 0

-x +x +x

1.00-xx x

equilibria involving a weak acid1
Equilibria Involving A Weak Acid

You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H3O+, OAc-, and the pH.

Step 2.Write Ka expression and obtain Ka value from a table.

This is a quadratic. Solve using quadratic formula.

or you can make an approximation if x is very small! (Rule of thumb: 10-5 or smaller is ok this will usually satisfy the 5% rule)

equilibria involving a weak acid2
Equilibria Involving A Weak Acid

You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H3O+, OAc-, and the pH.

Step 3.Solve Ka expression

First assume x is very small because Ka is so small.

Now we can more easily solve this approximate expression.

equilibria involving a weak acid3
Equilibria Involving A Weak Acid

You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H3O+, OAc-, and the pH.

Step 3.Solve Kaapproximateexpression

x =[H3O+] = [OAc-] = 4.2 x 10-3 M

pH = - log [H3O+] = -log (4.2 x 10-3) =2.37

equilibria involving a weak acid4
Equilibria Involving A Weak Acid

Calculate the pH of a 0.0010 M solution of formic acid, HCO2H.

HCO2H + H2O  HCO2- + H3O+

Ka = 1.8 x 10-4

Approximate solution

[H3O+] = 4.2 x 10-4 M,pH = 3.37

Exact Solution

[H3O+] = [HCO2-] = 3.4 x 10-4 M

[HCO2H] = 0.0010 - 3.4 x 10-4 = 0.0007 M

pH = 3.47

problem 54
Problem #54
  • a. HOC6H5 (Ka = 1.6 × 10-10 ) and H2O (Ka = Kw = 1.0 × 10-14 ) are the major species. The major equilibrium is the dissociation of HOC6H5. Solving the weak acid problem:
  • OC6H5 ⇌ H+ + OC6H5

Initial 0.250 M ~0 0

Change -x → +x +x

Equil. 0.250 - x xx

problem 54 continued
Problem #54 continued

Ka = 1.6 × 10-10 = = =

(assuming x << 0.250)

  • x = [H+] = 6.3 × 10-6M
  • Checking assumption: x is 2.5 × 10-3 % of 0.250, so assumption is valid by the 5% rule.
  • pH = -log(6.3 × 10-6) = 5.20
slide71
54 (b)
  • x = [H+] = 1.2 × 10-5M;
  • Checking assumption: x is 4.8 × 10-3 % of 0.250
  • Assumptions good.
  • pH = -log (1.2 × 10-5 ) = 4.92
slide72
58
  • Major species: HIO3, H2O; Major source of H+: HIO3 (a weak acid, Ka = 0.17)
  • HIO3 ⇌ H+ + IO3
  • Initial 0.20 M ~0 0
  • Change -x → +x +x
  • Equil. 0.20 –xxx
58 cont
58 Cont.
  • Ka = 0.17 = = x = 0.18; Check assumption.

Assumption is horrible (x is 90% of 0.20).

  • 0.17 =
  • x2 = 0.17(0.20 – x)
  • x2 + 0.17 x - 0.034 = 0
  • x =
  • x = 0.12 or -0.29 Only x = 0.12 makes sense.
58 answer
58 Answer
  • x = 0.12 M = [H+];
  • pH = -log (0.12) = 0.92
equilibria involving a weak base
Equilibria Involving A Weak Base

You have 0.010 M NH3. Calc. the pH.

NH3 + H2O  NH4+ + OH-

Kb = 1.8 x 10-5

Step 1.Define equilibrium concs. in ICE table

[NH3] [NH4+] [OH-]

initial

change

equilib

0.010 0 0

-x +x +x

0.010 - x x x

equilibria involving a weak base1
Equilibria Involving A Weak Base

You have 0.010 M NH3. Calc. the pH.

NH3 + H2O  NH4+ + OH-

Kb = 1.8 x 10-5

Step 2.Solve the equilibrium expression

Assume x is small, so

x = [OH-] = [NH4+] = 4.2 x 10-4 M

and [NH3] = 0.010 - 4.2 x 10-4 ≈ 0.010 M

The approximation is valid !

equilibria involving a weak base2
Equilibria Involving A Weak Base

Step 3.Calculate pH

[OH-] = 4.2 x 10-4 M

so pOH = - log [OH-] = 3.37

Because pH + pOH = 14,

pH = 10.63

problem 88
Problem 88
  • Kb = 1.3 × 10-3
  • (C2H5)2NH + H2O ⇌ (C2H5)2NH2+ + OH
  • Initial 0.050 M 0 ~0
  • Change -x → +x +x
  • Equil. 0.050 - xxx
88 cont
88 cont.
  • Kb = 1.3 × 10-3 = = ≈
  • x = 8.1 × 10-3;
  • Assumption is bad (x is 16% of 0.20).
  • Using Quadratic x= -b +/- (b2-4ac)1/2 / 2a
  • 0=x2 + .0013x - 6.5 E-5
  • X=.0067
  • [OH] = x = .0067M;
  • [H+] = Kw/[OH] = 1.49 × 10-12M;
  • pH = 11.83
calculating k a from the ph

[H3O+] [COO−]

[HCOOH]

Ka =

Calculating Ka from the pH
  • The pH of a 0.10 M solution of formic acid, HCOOH, at 25°C is 2.38. Calculate Ka for formic acid at this temperature.
  • We know that
calculating k a from the ph1
Calculating Ka from the pH
  • The pH of a 0.10 M solution of formic acid, HCOOH, at 25°C is 2.38. Calculate Ka for formic acid at this temperature.
  • To calculate Ka, we need the equilibrium concentrations of all three things.
  • We can find [H3O+], which is the same as [HCOO−], from the pH.
calculating k a from the ph2
Calculating Ka from the pH

pH = −log [H3O+]

2.38 = −log [H3O+]

−2.38 = log [H3O+]

10−2.38 = 10log [H3O+] = [H3O+]

4.2  10−3 = [H3O+] = [HCOO−]

calculating k a from ph
Calculating Ka from pH

Now we can set up a table…

homework 92
Homework #92
  • Remember this is an equation for OH but they give you a pH. Think about how to get to OH.
calculating percent ionization

[H3O+]eq

[HA]initial

Calculating Percent Ionization
  • Percent Ionization =  100
  • [H3O+]eq is obtained from your ICE chart
  • In this example

[H3O+]eq = 4.2  10−3 M

[HCOOH]initial = 0.10 M

calculating percent ionization1

4.2  10−3

0.10

Calculating Percent Ionization

Percent Ionization =  100

= 4.2%

polyprotic acids
Polyprotic Acids
  • Polyprotic acids are acids that have the ability to ionize more than once
  • Ex) H3PO4, H2SO4, H2CO3
  • The Ka for the second and third ionization is much smaller than the first so we generally ignore their contributions to the H+ conc.
polyprotic acids1
Polyprotic Acids
  • H3PO4 H3O+ + (H2PO4)-
  • (H2PO4)-  H3O+ + (HPO4)2-
  • H(PO4)2-  H3O+ + (PO4)3-
  • Ka1 = 7.1 x 10-3, Ka2 = 6.3 x 10-8, Ka3 = 4.3 x 10-13
problem 96
Problem #96
  • 96. The reactions are:
  • H3AsO4 ⇌ H+ + H2AsO4 = 5 × 10-3
  • H2AsO4 ⇌ H+ + HAsO42 = 8 × 10-8
  • HAsO42 ⇌ H+ + AsO43 = 6 × 10-10
96 cont
96 cont.
  • We will deal with the reactions in order of importance, beginning with the largest Ka, .
  • H3AsO4 ⇌ H+ + H2AsO4
  • Initial 0.20 M ~0 0
  • Equil. 0.20 – xxx
  •  By using quadratic formula. X= 3 x 10-2
  •  [H+] = [H2AsO4] = 3 × 10-2M;
  • [H3AsO4] = 0.20 - 0.03 = 0.17 M
96 cont1
96 cont.
  • Because Ka = = 8 × 10-8 is much smaller than the value, very little of H2AsO4 (and HAsO42) dissociates compared to H3AsO4. Therefore, [H+] and [H2AsO4 ] will not change significantly by the reaction. Using the previously calculated concentrations of H+ and H2AsO4 we can calculate the concentration of HAsO42:
slide94
96.
  • 8 × 10-8M =
  • [HAsO42] = 8 × 10-8M
  • Assumption that the Ka2 reaction does not change [H+] and [HAsO4] is good because it is very small compared to Ka1
  • We repeat the process using Ka3 to get [AsO43].
  • = =
  • [AsO43] = 1.6 × 10-15≈ 2 × 10-15 M
slide95
96.
  • So in 0.20 M analytical concentration of H3AsO4:
  •  [H3AsO4] = 0.17 M;
  • [H+] = [H2AsO4] = 3 × 10-2M
  •  [HAsO42] = 8 × 10-8 M;
  • [AsO43] = 2 × 10-15 M
  •  [OH] = Kw/[H+] = 3 × 10-13M
polyprotic acids2
Polyprotic Acids
  • Sulfuric acid (H2SO4): first ionization is strong
  • Ka2 = 1.1 x 10-2
  • In concentrated solutions, the first ionization produces all the H3O+
  • In dilute solutions (less than 0.0010 M), the second dissociation goes to completion
acid base properties of a salt solution

Consider a solution of sodium cyanide, NaCN.

Acid-Base Properties of a Salt Solution
  • One of the successes of the Brønsted-Lowry concept of acids and bases was in pointing out that some ions can act as acids or bases.
  • A 0.1 M solution has a pH of 11.1 and is therefore fairly basic.
acid base properties of a salt solution1
Acid-Base Properties of a Salt Solution
  • Sodium ion, Na+, is unreactive with water, but the cyanide ion, CN-, reacts to produce HCN and OH-.
  • You can also see that OH- ion is a product, so you would expect the solution to have a basic pH. This explains why NaCN solutions are basic.
  • The reaction of the CN- ion with water is referred to as the hydrolysis of CN-.
acid base properties of a salt solution2

The CN- ion hydrolyzes to give the conjugate acid and hydroxide.

  • The hydrolysis reaction for CN- has the form of a base ionization so you write the Kb expression for it.
Acid-Base Properties of a Salt Solution
  • The hydrolysis of an ion is the reaction of an ion with water to produce the conjugate acid and hydroxide ion or the conjugate base and hydronium ion.
predicting whether a salt is acidic basic or neutral
Predicting Whether a Salt is Acidic, Basic, or Neutral
  • How can you predict whether a particular salt will be acidic, basic, or neutral?
  • The Brønsted-Lowry concept illustrates the inverse relationship in the strengths of conjugate acid-base pairs.
  • Consequently, the anions of weak acids (poor proton donors) are good proton acceptors.
  • Anions of weak acids therefore, are basic.
acid base properties of a salt solution3
Acid-Base Properties of a Salt Solution
  • The NH4+ ion hydrolyzes to the conjugate base (NH3) and hydronium ion.
    • This equation has the form of an acid ionization so you write the Ka expression for it.
predicting whether a salt is acidic basic or neutral1
Predicting Whether a Salt is Acidic, Basic, or Neutral
  • These rules apply to normal salts (those in which the anion has no acidic hydrogen)
  • A salt of a strong base and a strong acid.
  • The salt has no hydrolyzable ions and so gives a neutral aqueous solution.
  • An example is NaCl.
predicting whether a salt is acidic basic or neutral2
Predicting Whether a Salt is Acidic, Basic, or Neutral
  • A salt of a strong base and a weak acid.
  • The anion of the salt is the conjugate of the weak acid. It hydrolyzes to give a basic solution.
  • An example is NaCN.
predicting whether a salt is acidic basic or neutral3
Predicting Whether a Salt is Acidic, Basic, or Neutral
  • A salt of a weak base and a strong acid.
  • The cation of the salt is the conjugate of the weak base. It hydrolyzes to give an acidic solution.
  • An example is NH4Cl.
predicting whether a salt is acidic basic or neutral4
Predicting Whether a Salt is Acidic, Basic, or Neutral
  • A salt of a weak base and a weak acid.
  • Both ions hydrolyze. You must compare the Ka of the cation with the Kb of the anion.
  • If the Ka of the cation is larger the solution is acidic.
  • If the Kb of the anion is larger, the solution is basic.
predicting whether a salt is acidic basic or neutral5

The potassium ion is the cation of a strong base (KOH) and does not hydrolyze.

Predicting Whether a Salt is Acidic, Basic, or Neutral
  • To predict the acidity or basicity of a salt, you must examine the acidity or basicity of the ions composing the salt.
  • Consider potassium acetate, KC2H3O2.
predicting whether a salt is acidic basic or neutral6
Predicting Whether a Salt is Acidic, Basic, or Neutral
  • Consider potassium acetate, KC2H3O2. The acetate ion, however, is the anion of a weak acid (HC2H3O2) and is basic.
  • A solution of potassium acetate is predicted to be basic
  • .
the ph of a salt solution
The pH of a Salt Solution
  • To calculate the pH of a salt solution would require the Ka of the acidic cation or the Kb of the basic anion.
  • The ionization constants of ions are not listed directly in tables because the values are easily related to their conjugate species.
  • Thus the Kb for CN- is related to the Ka for HCN.
the ph of a salt solution1

Kw

  • When these two reactions are added you get the ionization of water.
The pH of a Salt Solution
  • To see the relationship between Ka and Kb for conjugate acid-base pairs, consider the acid ionization of HCN and the base ionization of CN-.

Ka

Kb

the ph of a salt solution2

Kw

  • When two reactions are added, their equilibrium constants are multiplied.
  • Therefore,
The pH of a Salt Solution

Ka

Kb

the ph of a salt solution3
The pH of a Salt Solution
  • For a solution of a salt in which only one ion hydrolyzes, the calculation of equilibrium composition follows that of weak acids and bases.
  • The only difference is first obtaining the Ka or Kb for the ion that hydrolyzes.
  • The next example illustrates the reasoning and calculations involved.
a problem to consider

Only the CN- ion hydrolyzes.

A Problem To Consider
  • What is the pH of a 0.10 M NaCN solution at 25 oC? The Ka for HCN is 4.9 x 10-10.
  • Sodium cyanide gives Na+ ions and CN- ions in solution.
a problem to consider1
A Problem To Consider
  • The CN- ion is acting as a base, so first, we must calculate the Kb for CN-.
  • Now we can proceed with the equilibrium calculation.
a problem to consider2
A Problem To Consider
  • What is the pH of a 0.10 M NaCN solution at 25 oC? The Ka for HCN is 4.9 x 10-10.
  • Let x = [OH-] = [HCN], then substitute into the equilibrium expression.
a problem to consider3

Hence,

A Problem To Consider
  • What is the pH of a 0.10 M NaCN solution at 25 oC? The Ka for HCN is 4.9 x 10-10.
  • Solving the equation, you find that
  • As expected, the solution has a pH greater than 7.0.
homework 106
Homework 106
  • a. KNO2 → K+ + NO2: NO2 is a weak base. Ignore K+.
  •  NO2 + H2O ⇌ HNO2 + OH Initial 0.12 M 0 ~0
  • Equil. 0.12 - xxx
  • Kb = = =2.5 × 10-11
  • Kb = 2.5 × 10-11 = ≈
  • x = [OH] = 1.7 × 10-6M;
  • pOH = 5.77; pH = 8.23
slide117

2.9 ×

106
  • NaOCl → Na+ + OCl: OCl is a weak base. Ignore Na+.
  • OCl + H2O ⇌ HOCl + OH Initial 0.45 M 0 ~0
  • Equil. 0.45 – xxx
  • Kb = = = 2.9 x 10-7
  • Kb = 2.9 x 10-7 = ≈
  • x = [OH] = 3.6 × 10-4M;
  • pOH = 3.44;
  • pH = 10.56
slide118
106
  • NH4ClO4 → NH4+ + ClO4: NH4+ is a weak acid. ClO4 is the conjugate base of a strong acid. ClO4 has no basic (or acidic) properties.
  • NH4+ ⇌ NH3 + H+
  • Initial 0.40 M 0 ~0
  • Equil. 0.40 – xxx
  • Ka = = = 5.6 × 10-10
  • Ka = ≈
  • x = [H+] = 1.5 × 10-5M; pH = 4.82;