Chapter 14 acids and bases
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Chapter 14 Acids and Bases. Some Definitions. Arrhenius Acid:Substance that, when dissolved in water, increases the concentration of hydrogen ions. Base:Substance that, when dissolved in water, increases the concentration of hydroxide ions. Some Definitions. Brønsted – Lowry

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Chapter 14 Acids and Bases

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Chapter 14Acids and Bases


Some Definitions

  • Arrhenius

    • Acid:Substance that, when dissolved in water, increases the concentration of hydrogen ions.

    • Base:Substance that, when dissolved in water, increases the concentration of hydroxide ions.


Some Definitions

  • Brønsted–Lowry

    • Acid:Proton donor

    • Base:Proton acceptor


A Brønsted–Lowry acid…

…must have a removable (acidic) proton.

A Brønsted–Lowry base…

…must have a pair of nonbonding electrons.


If it can be either…

...it is amphiprotic.

HCO3−

HSO4−

H2O


What Happens When an Acid Dissolves in Water?

  • Water acts as a Brønsted–Lowry base and abstracts a proton (H+) from the acid.

  • As a result, the conjugate base of the acid and a hydronium ion are formed.


Acid Dissociation (Ionization) Reactions

  • Write the simple dissociation (ionization) reaction (omitting water) for each of the following acids:

    • a. Hydrochloric acid

    • b. Acetic acid

    • c. Ammonium ion

    • d. Anilinium ion (C6H5NH3)

    • e. Hydrated aluminum (III) ion [Al(H2O)6]3+


Conjugate Acids and Bases:

  • From the Latin word conjugare, meaning “to join together.”

  • Reactions between acids and bases always yield their conjugate bases and acids.


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.


Acid and Base Strength

  • Substances with negligible acidity do not dissociate in water.

    • Their conjugate bases are exceedingly strong.


Acid and Base Strength

In any acid-base reaction, the equilibrium will favor the reaction that moves the proton to the stronger base.

HCl(aq) + H2O(l) H3O+(aq) + Cl−(aq)

H2O is a much stronger base than Cl−, so the equilibrium lies so far to the right K is not measured (K>>1).


C2H3O2(aq) + H2O(l)

H3O+(aq) + C2H3O2−(aq)

Acid and Base Strength

Acetate is a stronger base than H2O, so the equilibrium favors the left side (K<1).


Relative Base Strength

Using the following Ka values, arrange the following species according to their strength as bases:


H2O(l) + H2O(l)

H3O+(aq) + OH−(aq)

Autoionization of Water

  • As we have seen, water is amphoteric.

  • In pure water, a few molecules act as bases and a few act as acids.

  • This is referred to as autoionization.


Ion-Product Constant

  • The equilibrium expression for this process is

    Kc = [H3O+] [OH−]

  • This special equilibrium constant is referred to as the ion-product constant for water, Kw.

  • At 25°C, Kw = 1.0  10−14


Calculating [H+] & [OH-]

  • Calculate [H+] & [OH-] as required for each of the following solutions at 250C, & state whether the solution is neutral, acidic, or basic.

    • a. = 1.0 x 10-5 M OH-

    • b. a. = 1.0 x 10-7 M OH-

    • c. 10.0 M H+


pH

pH is defined as the negative base-10 logarithm of the hydronium ion concentration.

pH = −log [H3O+]


pH

  • In pure water,

    Kw = [H3O+] [OH−] = 1.0  10−14

  • Because in pure water [H3O+] = [OH−],

    [H3O+] = (1.0  10−14)1/2 = 1.0  10−7


pH

  • Therefore, in pure water,

    pH = −log (1.0  10−7) = 7.00

  • An acid has a higher [H3O+] than pure water, so its pH is <7

  • A base has a lower [H3O+] than pure water, so its pH is >7.


pH

These are the pH values for several common substances.


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


Watch This!

Because

[H3O+] [OH−] = Kw = 1.0  10−14,

we know that

−log [H3O+] + −log [OH−] = −log Kw = 14.00

or, in other words,

pH + pOH = pKw = 14.00


Calculating pH, pOH

pH = -log10(H3O+)

pOH = -log10(OH-)

Relationship between pH and pOH

pH + pOH = 14

Finding [H3O+], [OH-] from pH, pOH

[H3O+] = 10-pH

[OH-] = 10-pOH


Calculating pH & pOH

  • Calculate pH & pOH for each of the following solutions at 250C.

    • a. = 1.0 x 10-3 M OH-

    • b. a. = 1.0 M H+


Calculating pH

  • The pH a sample of human blood was measured to be 7.41 at 250C. Calculate pOH, [H+], & [OH-] for the sample.


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 pH?

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


Strong Acids

  • You will recall that the seven strong acids are HCl, HBr, HI, HNO3, H2SO4, HClO3, and HClO4.

  • These are, by definition, strong electrolytes and exist totally as ions in aqueous solution.

  • For the monoprotic strong acids,

    [H3O+] = [acid].


pH of Strong Acids

  • Calculate pH of 0.10 M HNO3.

  • Calculate pH of 1.0 x 10-10 M HCl.


Kc =

[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 Constants

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


[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 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 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 Ka from pH

Now we can set up a table…


[4.2  10−3] [4.2  10−3]

[0.10]

Ka =

Calculating Ka from pH

= 1.8  10−4


Solving Weak Acid Equilibrium Problems

  • 1. List the major species in the solution.

  • 2. Choose the species that can produce H+ and write balanced equations for the reactions producing H+ .

  • 3. Using the values of the equilibrium constants for the reactions you have written, decide which equilibrium will dominate in producing H+ .

  • 4. Write the equilibrium expression for the dominant equilibrium

  • 5. ICE the problem

  • 6. Substitute the equilibrium [ ] into the equilibrium expression

  • 7. Solve for x the “easy” way; that is, by assuming [HA]0-x  [HA]0

  • 8. Use the 5% rule to verify whether the approximation is valid

  • 9. Calculate [H+] and pH


The pH of Weak Acids

  • The hypochlorite ion (OCl-) is a strong oxidizing agent often found in household bleaches & disinfectants. It is also the active ingredient that forms when swimming pool water is treated with chlorine. In addition to its oxidizing abilities, the hypochlorite ion has a relatively high affinity for protons (it is a much stronger base than Cl-for example) & forms the weakly acidic hypochlorous acid (HOCl, Ka = 3.5 x 10-8 ). Calculate pH of 0.100 M aqueous solution of hypochlorous acid.


The pH of Weak Acids continued

  • HOCl, Ka = 3.5 x 10-8 Calculate pH of 0.100 M aqueous solution of hypochlorous acid.


The pH of Weak Acid Mixtures

  • Calculate the pH of a solution that contains 1.00 M HCN (Ka = 6.2 x 10-10) and 5.00 M HNO2 (Ka = 4.0 x 10-4) . Also calculate the concentration of cyanide ion in this solution at equilibrium.


Percent Dissociation

  • In general, the more dilute the weak acid solution, the greater the percent dissociation of the weak acid.


Calculate Percent Dissociation

  • Calculate the percent dissociation of acetic acid (Ka = 1.8x 10-5) in each of the following solutions.

  • a. 1.00 M HC2H3O2

  • b. 0.100 M HC2H3O2


Calculation Ka from Percent Dissociation

  • Lactic acid (HC3H5O3) is a waste product that accumulates in muscle tissue during exertion, leading to pain & feeling of fatigue. In a 0.100 M aqueous solution, lactic acid is 3.7% dissociated. Calculate the value of Ka for this acid.


Strong Bases

  • Strong bases are the soluble hydroxides, which are the alkali metal and heavier alkaline earth metal hydroxides (Ca2+, Sr2+, and Ba2+).

  • Again, these substances dissociate completely in aqueous solution.


The pH of strong bases

  • Calculate the pH of a 5.0x 10-2 M NaOH solution.


Weak Bases

  • Many types of proton acceptors (bases) do not contain hydroxide ions. When dissolved in water, they increase the concentration of hydroxide ions because of their reaction with water.

  • Ex.

  • Bases such as ammonia typically have at least one unshared pair of electrons that is capable of forming a bond with a proton.


Weak Bases

  • We will solve weak base problems in the same manner we solved weak acid problems (look back over the steps you were given

  • We will use Kb instead of Ka and will find [OH-] instead of [H+]

    • Remember the process for “switching” from pOH to pH &/or from [OH-] [H+]


pH of a Weak Base (I)

  • Calculate the pH for a 15.0 M NH3 solution (Kb = 1.8x 10-5).


pH of a Weak Base (II)

  • Calculate the pH for a 1.0 M methylamine solution (Kb = 4.38x 10-4).


Polyprotic Acids

  • Some acids furnish more than one acidic proton such as H2SO4, H3PO4.

    • Ex. H2CO3


pH of a polyprotic acid

  • Calculate the pH of a 5.0 M H3PO4 solution and the equilibrium concentrations of the species H3PO4, H2PO4-, HPO42-, & PO43- .


pH of sulfuric acid

  • Calculate the pH of a 1.0 M H2SO4 solution.

  • Calculate the pH of a .0100 M H2SO4 solution.


Acid-Base Properties of Salts


Acid-Base Properties of Salts

These salts simply dissociate in water:

KCl(s)  K+(aq) + Cl-(aq)


Acid-Base Properties of Salts

The basic anion can accept a proton from water:

C2H3O2- + H2O  HC2H3O2 + OH- base acid acid base


Acid-Base Properties of Salts

The acidic cation can act as a proton donor:

NH4+(aq)  NH3(aq) + H+(aq)

Acid Conjugate Proton

base


Acid-Base Properties of Salts

  • IF Ka for the acidic ion is greater than Kb for the basic ion, the solution is acidic

  • IF Kb for the basic ion is greater than Ka for the acidic ion, the solution is basic

  • IF Kb for the basic ion is equal to Ka for the acidic ion, the solution is neutral


Acid-Base Properties of Salts

Step #1:

AlCl3(s) + 6H2O  Al(H2O)63+(aq) + Cl-(aq)

Salt water Complex ion anion

Step #2:

Al(H2O)63+(aq)  Al(OH)(H2O)52+(aq) + H+(aq)

Acid Conjugate base Proton


Salts as Weak Bases

Calculate the pH of a 0.30 M NaF solution. The Ka value for HF is 7.2 x 10-4


Salts as Weak Acids I

Calculate the pH of a 0.10 M NH4Cl solution. The Kb value for NH3 is 1.8 x 10-5


Salts as Weak Acids II

Calculate the pH of a 0.010 M AlCl3solution. The Kavalue for Al(H2O)63+ is 1.4 x 10-5


The Acid/Base Properties of Salts

Predict whether an aqueous solution of each of the following salts will be acidic, basic, or neutral.

NH4C2H3O2

NH4CN

Al2(SO4)3


Effect of Structure on Acid-Base Properties

Hypochlorous

acid

Chlorous

acid

Chloric

acid

Perchloric

acid

Increasing Acidity


The Lewis Acid-Base Model

  • Lewis Acid: An electron pair acceptor

    • Has an empty orbital to accept a pair of electrons

  • Lewis Base: An electron pair donor

    • Has a lone pair of electrons


Lewis Acids & Bases

  • For each reaction, identify the Lewis acid & base.

  • Ni2+(aq) + 6NH3(aq) Ni(NH3)62+(aq)

  • H+(aq) + H2O(l) H3O+(aq)


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