Models of Acids and Bases the Reaction of an Acid HA with Water to Form H3O+ and a Conjugate Base A-. Conjugate Acid/Base Pairs

Home , Conjugate acid, Degree of ionization

Models of Acids and Bases

•Arrhenius Concept: Acids produce H+ in solution, bases produce OH− ion. ACIDS •Brønsted-Lowry: Acids are H+ donors, bases are proton acceptors. and •Review Chapter 4 – Acids & Bases, Section 4.8 BASES − + HCl + H2O º Cl + H3O • acid base

The reaction of an acid HA with Conjugate Acid/Base Pairs

water to form H3O+ and a + − •HA(aq) + H2O(l) ↔ H3O (aq) + A (aq) conjugate base A-. • conj conj conj conj acid 1 base 2 acid 2 base 1

•conjugate base: everything that remains of the acid molecule after a proton is lost.

HA (aq) + H O (l) º H O+ (aq) + A- (aq) 2 3 •conjugate acid: formed when the proton is transferred to the base.

Acid Dissociation Constant (Ka) Representation of the reaction + H3O + NH3.

+ − •HA(aq) + H2O(l) ↔ H3O (aq) + A (aq)

+− +− HO3 A HA K == a HA HA Acid Base Conj. Base Conj. Acid

1 Acid Strength Concept Check 16.1

Strong Acid:

Its equilibrium position lies far to the right.

(HNO3) − - + Yields a weak conjugate base. (NO3 ) HCOOH + H2O º HCOO + H3O formic acid water formate ion hydronium ion + - HNO3 (aq) + H2O (l) º H3O (aq) + NO3 (aq)

Ka >>1 We do not usually consider the reactants in reactions because the are present in very small amounts!

Acid Strength (continued)

Weak Acid:

Its equilibrium lies far to the left.

(CH3COOH) Yields a much stronger (it is relatively strong) conjugate base than water. − (CH3COO ) + - HC2H3O2 (aq) + H2O (l) º H3O (aq) + C2H3O2 (aq)

Ka << 1 Equilibrium favors reactants – but the solution is still acidic – so acetic acid IS an acid!

Weak & Strong Acids: (a) A strong acid HA is completely ionized in water. (b) A weak acid HB exists mostly as undissociated HB molecules in water. Note that the water molecules are not shown in this figure.

+ - Ka = [H ][A ]/[HA]

2 Figure 14.4: Graphic representation of the behavior of acids of different strengths in aqueous solution. (a) A strong acid. (b) A weak acid.

Bases Bases (continued)

•“Strong” and “weak” are used in the same sense for bases as for acids.

•strong = complete dissociation (hydroxide •weak = very little dissociation (or reaction ion supplied to solution) with water)

•NaOH(s) → Na+(aq) + OH−(aq) + − •CH3NH2(aq) + H2O(l) º CH3NH3 (aq) + OH (aq) K >> 1 b Kb << 1

Figure 14.5: The relationship of acid strength and conjugate base strength for the reaction

K = [OH-][BH+]/[B] In general, the b strength of a weak acid is inversely related to the strength of its conjugate base

3 -5 Acetic Acid: Ka = 1.8 x 10 -4 Formic Acid: Ka = 1.8 x 10

-10 Acetate ion: Kb = 5 x 10 -11 Formate ion: Kb = 5 x 10

Acetate ion!

Figure 14.7: Two water molecules Self-ionization of Water + - react to form H3O and OH . • These ions are produced in equal numbers in pure water, so if we let x = [H+] = [OH-] 1.0×10−14 = (x)(x) at 25 oC x = 1.0×10−14 = 1.0×10−7

+ - 2 H2O (l) º H3O (aq) + OH (aq) • Thus, the concentrations of H+ and OH- in pure water are both 1.0 x 10-7 M. [H O+][OH-] = K = 1.01 x 10-14 3 w • If you add acid or base to water they are no

longer equal but the Kw expression still holds.

Solutions of Strong Acid or Solutions of Strong Acid or Base Base • In a solution of a strong acid you can • As an example, calculate the concentration of OH- normally ignore the self-ionization of water ion in 0.10 M HCl. + Because you started with 0.10 M HCl (a strong as a source of H (aq). acid) the reaction will produce 0.10 M H+(aq). • The H+(aq) concentration is usually determined + − by the strong acid concentration. HCl(aq) → H (aq) + Cl (aq) • However, the self-ionization still exists and is • Substituting [H+]=0.10 into the ion-product responsible for a small concentration of OH- expression, we get:

ion. + + H (aq) + H2O (l) º H3O (aq) Ka >> 1 −14 − + - -14 1.0×10 = (0.10)[OH ] 2 H2O (l) º H3O (aq) + OH (aq) Ka = 1.0 x 10

4 Solutions of Strong Acid or Solutions of Strong Acid or Base Base • As an example, calculate the concentration of OH- • Similarly, in a solution of a strong base you ion in 0.10 M HCl. can normally ignore the self-ionization of Because you started with 0.10 M HCl (a strong - acid) the reaction will produce 0.10 M H+(aq). water as a source of OH (aq). - + − • The OH (aq) concentration is usually HCl(aq) → H (aq) + Cl (aq) determined by the strong base concentration. • Substituting [H+]=0.10 into the ion-product • However, the self-ionization still exists and is expression, we get: responsible for a small concentration of H+ ion. 1.0×10-14 [OH − ] = = 1.0×10-13 M 0.10

Solutions of Strong Acid or Solutions of Strong Acid or Base Base • As an example, calculate the concentration of H+ • As an example, calculate the concentration of H+ ion in 0.010 M NaOH. ion in 0.010 M NaOH. Because you started with 0.010 M NaOH (a strong Because you started with 0.010 M NaOH (a strong base) the reaction will produce 0.010 M OH-(aq). base) the reaction will produce 0.010 M OH-(aq). H O + − H O + − NaOH(s) →2 Na (aq) + OH (aq) NaOH(s) →2 Na (aq) + OH (aq) • Substituting [OH-]=0.010 into the ion-product • Substituting [OH-]=0.010 into the ion-product expression, we get: expression, we get: 1.0×10-14 1.0×10−14 = [H+ ](0.010) [H + ] = = 1.0×10-12 M 0.010

Solutions of Strong Acid or The pH Scale Base • At 25 oC, you observe the following conditions. • In an acidic solution, [H+] > 1.0 x 10-7 M. · pH ≈−log[H+] • In a neutral solution, [H+] = 1.0 x 10-7 M. • pH in water ranges from 0 to 14. + -7 −14 + − • In a basic solution, [H ] < 1.0 x 10 M. • Kw = 1.00 × 10 = [H ] [OH ] •pKw = 14.00 = pH + pOH • As pH rises, pOH falls (sum = 14.00).

5 The pH of a Solution The pH of a Solution

• For a solution in which the hydrogen-ion •In a neutral solution, whose hydrogen-ion concentration is 1.0 x 10-3, the pH is: concentration is 1.0 x 10-7, the pH = 7.00. −3 • For acidic solutions, the hydrogen-ion pH = −log(1.0×10 ) = 3.00 concentration is greater than 1.0 x 10-7, so the pH is less than 7.00. • Note that the number of decimal places in the • Similarly, a basic solution has a pH greater pH equals the number of significant figures in than 7.00. the hydrogen-ion concentration. • Figure 16.6 shows a diagram of the pH scale and the pH values of some common solutions.

Figure 14.8: The pH scale A Problem to Consider and pH values of some • A sample of orange juice has a hydrogen-ion -4 common concentration of 2.9 x 10 M. What is the pH? + substances. pH = −log[H ] pH = −log(2.9×10−4 ) pH = 3.54

A Problem to Consider The pH of a Solution

• The pH of human arterial blood is 7.40. What is • A measurement of the hydroxide ion the hydrogen-ion concentration? concentration, similar to pH, is the pOH. [H+ ] = antilog(−pH) • The pOH of a solution is defined as the negative logarithm of the molar hydroxide-ion [H+ ] = antilog(−7.40) concentration. − [H+ ] = 10−7.40 = 4.0×10−8 M pOH = −log[OH ]

6 The pH of a Solution A Problem to Consider

• A measurement of the hydroxide ion • An ammonia solution has a hydroxide-ion concentration, similar to pH, is the pOH. concentration of 1.9 x 10-3 M. What is the pH of + - -14 the solution? • Then because Kw = [H ][OH ] = 1.0 x 10 at 25 oC, you can show that You first calculate the pOH: pOH = −log(1.9×10−3 ) = 2.72 pH + pOH = 14.00 Then the pH is: pH = 14.00 − 2.72 = 11.28

Figure 14.9: pH meters are used The pH of a Solution to measure acidity. • The pH of a solution can accurately be measured using a pH meter (see Figure 14.9). • Although less precise, acid-base indicators are often used to measure pH because they usually change color within a narrow pH range. • Figure 16.8 shows the color changes of various acid-base indicators.

Figure 16.8: Color changes of some acid- based indicators. Percent Dissociation (Degree of Ionization)

amount dissociated(M ) % dissociation =×100% initial concentration(M )

Weak acids and bases have low values for % dissociation - their degree of ionization is small.

7 A Problem To Consider A Problem To Consider

• Nicotinic acid is a weak monoprotic acid with the • Nicotinic acid is a weak monoprotic acid with the

formula HC6H4NO2. A 0.012 M solution of nicotinic formula HC6H4NO2. A 0.012 M solution of nicotinic acid has a pH of 3.39 at 25 oC. acid has a pH of 3.39 at 25 oC. Calculate the acid- o [H+] = 10-3.39 = 0.00041 ionization constant for this acid at 25 C. • To obtain the degree of dissociation: • It is important to realize that the solution was 0.00041 made 0.012 M in nicotinic acid, however, some Degree of dissociation = = 0.034 molecules ionize making the equilibrium 0.012 concentration of nicotinic acid less than 0.012 M. • The percent ionization is obtained by multiplying by • We will abbreviate the formula for nicotinic acid 100, which gives 3.4%. as HNic.

A Problem To Consider A Problem To Consider

• Nicotinic acid is a weak monoprotic acid with the • Nicotinic acid is a weak monoprotic acid with the

formula HC6H4NO2. A 0.012 M solution of nicotinic formula HC6H4NO2. A 0.012 M solution of nicotinic acid has a pH of 3.39 at 25 oC. Calculate the acid- acid has a pH of 3.39 at 25 oC. Calculate the acid- ionization constant for this acid at 25 oC. ionization constant for this acid at 25 oC. • Let x be the moles per liter of product formed. • We can obtain the value of x from the given pH. HNic(aq) + H O(l) H O+ (aq) + Nic− (aq) + 2 3 x = [H3O ] = antilog(−pH) Initial 0.012 00 x = antilog(−3.39) Change -x +x +x −4 Equilibrium 0.012-x x x x = 4.1×10 = 0.00041

A Problem To Consider A Problem To Consider

• Nicotinic acid is a weak monoprotic acid with the • Nicotinic acid is a weak monoprotic acid with the

formula HC6H4NO2. A 0.012 M solution of nicotinic formula HC6H4NO2. A 0.012 M solution of nicotinic acid has a pH of 3.39 at 25 oC. Calculate the acid- acid has a pH of 3.39 at 25 oC. Calculate the acid- ionization constant for this acid at 25 oC. ionization constant for this acid at 25 oC. • The equilibrium-constant expression is: • Substituting the expressions for the equilibrium concentrations, we get + − 2 [H3O ][Nic ] x Ka = K = [HNic] a (0.012 − x)

8 A Problem To Consider A Problem To Consider

• Nicotinic acid is a weak monoprotic acid with the • Nicotinic acid is a weak monoprotic acid with the

formula HC6H4NO2. A 0.012 M solution of nicotinic formula HC6H4NO2. A 0.012 M solution of nicotinic acid has a pH of 3.39 at 25 oC. Calculate the acid- acid has a pH of 3.39 at 25 oC. Calculate the acid- ionization constant for this acid at 25 oC. ionization constant for this acid at 25 oC. • Substitute this value of x in our equilibrium • Substitute this value of x in our equilibrium expression. expression. • Note first, however, that 2 2 x (0.00041) −5 (0.012 − x) = (0.012 − 0.00041) = 0.01159 ≅ 0.012 K = ≅ = 1.4×10 a (0.012 − x) (0.012) the concentration of unionized acid remains virtually unchanged.

Solving Weak Acid Equilibrium Solving Weak Acid Equilibrium Problems Problems (continued)

List major species in solution. Define change at equilibrium (as “x”). Choose species that can produce H+ and write Write equilibrium concentrations in terms of x. reactions. Substitute equilibrium concentrations into Based on K values, decide on dominant equilibrium expression. equilibrium. Solve for x the “easy way.” Write equilibrium expression for dominant equilibrium. Verify assumptions using 5% rule.

List initial concentrations in dominant equilibrium. Calculate [H+] and pH.

Calculations With Ka Calculations With Ka

• If you know the value of Ka, you can • Note that in our previous example, the degree of calculate the equilibrium concentrations dissociation was so small that “x” was negligible - + compared to the concentration of nicotinic acid. of species HA, A , and H3O for solutions of different molarities. • It is the small value of the degree of ionization that allowed us to ignore the subtracted x in the denominator of our equilibrium expression. • The degree of ionization of a weak acid depends

on both the Ka and the concentration of the acid solution.

9 Calculations With Ka Calculations With Ka

• How do you know when you can use this • How do you know when you can use this simplifying assumption? simplifying assumption? • It can be shown that if the acid concentration, • If the simplifying assumption is not valid, you

Ca, divided by the Ka exceeds 100, that is, can solve the equilibrium equation exactly by using the quadratic equation. Ca if > 100 • The next example illustrates this with a Ka solution of aspirin (acetylsalicylic acid), then this simplifying assumption of ignoring HC H O , a common headache remedy. the subtracted x gives an acceptable error 9 7 4 of less than 5%.

A Problem To Consider A Problem To Consider

• What is the pH at 25 oC of a solution obtained by • What is the pH at 25 oC of a solution obtained by dissolving 0.325 g of acetylsalicylic acid (aspirin), dissolving 0.325 g of acetylsalicylic acid (aspirin),

HC9H7O4, in 0.500 L of water? The acid is HC9H7O4, in 0.500 L of water? The acid is -4 o -4 o monoprotic and Ka=3.3 x 10 at 25 C. monoprotic and Ka=3.3 x 10 at 25 C. • The molar mass of HC9H7O4 is 180.2 g. • The molar mass of HC9H7O4 is 180.2 g. From this we find that the sample contained Hence, the concentration of the acetylsalicylic 0.00180 mol of the acid. acid is 0.00180 mol/0.500 L = 0.0036 M (Retain two significant figures, the same

number of significant figures in Ka).

A Problem To Consider A Problem To Consider

HC9H7O4, in 0.500 L of water? The acid is HC9H7O4, in 0.500 L of water? The acid is -4 o -4 o monoprotic and Ka=3.3 x 10 at 25 C. monoprotic and Ka=3.3 x 10 at 25 C. • Note that • We will abbreviate the formula for C acetylsalicylic acid as HAcs and let x be the a 0.0036 + = −4 = 11 amount of H3O formed per liter. Ka 3.3×10 • The amount of acetylsalicylate ion is also x mol; which is less than 100, so we must solve the the amount of nonionized acetylsalicylic acid is equilibrium equation exactly. (0.0036-x) mol.

10 A Problem To Consider A Problem To Consider

HC9H7O4, in 0.500 L of water? The acid is HC9H7O4, in 0.500 L of water? The acid is -4 o -4 o monoprotic and Ka=3.3 x 10 at 25 C. monoprotic and Ka=3.3 x 10 at 25 C. • These data are summarized below. • The equilibrium constant expression is + − HAcs(aq) + H2O(l) H3O (aq) + Acs (aq) [H O+ ][Acs− ] Initial 3 0.0036 00 = Ka Change -x +x +x [HAcs] Equilibrium 0.0036-x x x

A Problem To Consider A Problem To Consider

HC9H7O4, in 0.500 L of water? The acid is HC9H7O4, in 0.500 L of water? The acid is -4 o -4 o monoprotic and Ka=3.3 x 10 at 25 C. monoprotic and Ka=3.3 x 10 at 25 C. • If we substitute the equilibrium concentrations • You can solve this equation exactly by using

and the Ka into the equilibrium constant the quadratic formula. expression, we get • Rearranging the preceding equation to put it in x2 the form ax2 + bx + c = 0, we get = 3.3×10−4 (0.0036 − x) x2 + (3.3×10−4 )x − (1.2×10−6 ) = 0

A Problem To Consider A Problem To Consider

HC9H7O4, in 0.500 L of water? The acid is HC9H7O4, in 0.500 L of water? The acid is -4 o -4 o monoprotic and Ka=3.3 x 10 at 25 C. monoprotic and Ka=3.3 x 10 at 25 C. • Now substitute into the quadratic formula. • Now substitute into the quadratic formula.

2 − b ± b − 4ac − (3.3×10−4 ) ± (3.3×10−4 )2 − 4(1.2×10−6 ) x = x = 2a 2 • The lower sign in ± gives a negative root which we can ignore

11 A Problem To Consider A Problem To Consider

• What is the pH at 25 oC of a solution obtained by • What is the pH of a 0.20 M solution of pyridine,

dissolving 0.325 g of acetylsalicylic acid (aspirin), C5H5N, in aqueous solution? The Kb for pyridine -9 HC9H7O4, in 0.500 L of water? The acid is is 1.4 x 10 . -4 o • As before, we will follow the three steps in monoprotic and Ka=3.3 x 10 at 25 C. • Taking the upper sign, we get solving an equilibrium. x = [H O+ ] = 9.4×10−4 1. Write the equation and make a table of 3 concentrations. • Now we can calculate the pH. 2. Set up the equilibrium constant expression. pH = −log(9.4×10−4 ) = 3.03 3. Solve for x = [OH-].

A Problem To Consider A Problem To Consider (Weak Base) • What is the pH of a 0.20 M solution of pyridine, • What is the pH of a 0.20 M solution of pyridine,

C5H5N, in aqueous solution? The Kb for pyridine C5H5N, in aqueous solution? The Kb for pyridine is 1.4 x 10-9. is 1.4 x 10-9. • Pyridine ionizes by picking up a proton from • Note that water (as ammonia does). + − C 8 C H N(aq) H O(l) C H NH (aq) OH (aq) a 0.20 5 5 + 2 5 5 + = −9 = 1.4×10 Ka 1.4×10 Initial 0.20 00 which is much greater than 100, so we Change -x +x +x may use the simplifying assumption Equilibrium 0.20-x x x that (0.20-x) ≅ (0.20).

A Problem To Consider A Problem To Consider

• What is the pH of a 0.20 M solution of pyridine, • What is the pH of a 0.20 M solution of pyridine,

C5H5N, in aqueous solution? The Kb for pyridine C5H5N, in aqueous solution? The Kb for pyridine is 1.4 x 10-9. is 1.4 x 10-9. • The equilibrium expression is • If we substitute the equilibrium concentrations

and the Kb into the equilibrium constant expression, we get [C H NH+ ][OH− ] 5 5 = K 2 b x −9 [C5H5N] = 1.4×10 (0.20 − x)

12 A Problem To Consider A Problem To Consider

• What is the pH of a 0.20 M solution of pyridine, • What is the pH of a 0.20 M solution of pyridine,

C5H5N, in aqueous solution? The Kb for pyridine C5H5N, in aqueous solution? The Kb for pyridine is 1.4 x 10-9. is 1.4 x 10-9. • Using our simplifying assumption that the x in • Solving for x we get the denominator is negligible, we get x2 ≅ (0.20)×(1.4×10−9 )

− −9 −5 x2 x = [OH ] ≅ (0.20)×(1.4×10 ) = 1.7×10 ≅ 1.4×10−9 (0.20)

Polyprotic Acids A Problem To Consider

• What is the pH of a 0.20 M solution of pyridine, + C5H5N, in aqueous solution? The Kb for pyridine • . . . can furnish more than one proton (H ) is 1.4 x 10-9. to the solution. • Solving for pOH −5 HCO↔+ H+ HCO− (K ) pOH = −log[OH−] = −log(1.7×10 ) = 4.8 23 3 a1 • Since pH + pOH = 14.00 HCO− ↔+ H+ CO2− (K ) 33a2 pH = 14.00 − pOH = 14.00 − 4.8 = 9.2

Polyprotic Acids Polyprotic Acids • Some acids have two or more protons • Some acids have two or more protons (hydrogen ions) to donate in aqueous solution. (hydrogen ions) to donate in aqueous solution. These are referred to as polyprotic acids. These are referred to as polyprotic acids. • Sulfuric acid, for example, can lose two protons • For a weak diprotic acid like carbonic acid,

in aqueous solution. H2CO3, two simultaneous equilibria must be + − considered. H2SO4 (aq) + H2O(l) → H3O (aq) + HSO4 (aq) − + 2− + − HSO4 (aq) + H2O(l) H3O (aq) + SO4 (aq) H2CO3 (aq) + H2O(l) H3O (aq) + HCO3 (aq)

• The first proton is lost completely followed by a − + 2− - HCO3 (aq) + H2O(l) H3O (aq) + CO3 (aq) weak ionization of the hydrogen sulfate ion, HSO4 .

13 Polyprotic Acids Polyprotic Acids • Some acids have two or more protons • Some acids have two or more protons (hydrogen ions) to donate in aqueous solution. (hydrogen ions) to donate in aqueous solution. These are referred to as polyprotic acids. These are referred to as polyprotic acids. • Each equilibrium has an associated acid- • Each equilibrium has an associated acid- ionization constant. ionization constant. For the loss of the first proton For the loss of the second proton

+ − + 2− [H3O ][HCO3 ] −7 [H3O ][CO3 ] −11 Ka1 = = 4.3×10 Ka2 = − = 4.8×10 [H2CO3 ] [HCO3 ]

Polyprotic Acids • Some acids have two or more protons (hydrogen ions) to donate in aqueous solution. These are referred to as polyprotic acids.

• The second ionization constant, Ka2, for a polyprotic acid is always smaller than the first

ionization constant, Ka1.

• In the case of a triprotic acid, such as H3PO4, the third ionization constant, Ka3, is smaller than the second one, Ka2.

Brønsted-Lowry Concept of Brønsted-Lowry Concept of Acids and Bases Acids and Bases • Some species can act as an acid or a base. • Some species can act as an acid or a base. •An amphoteric species is a species that can act •An amphoteric species is a species that can act either as an acid or a base (it can gain or lose a either as an acid or a base (it can gain or lose a proton). proton). - Alternatively, HCO can act as a proton acceptor (a • For example, HCO3 acts as a proton donor (an 3 acid) in the presence of OH- base) in the presence of HF. − − −2 − − HCO3 (aq) + OH (aq) → CO3 (aq) + H2O(l) HCO3 (aq) + HF(aq) → H2CO3 (aq) + F (aq)

H+ H+

14 Brønsted-Lowry Concept of Brønsted-Lowry Concept of Acids and Bases Acids and Bases • The amphoteric characteristic of water is • The amphoteric characteristic of water is important in the acid-base properties of important in the acid-base properties of aqueous solutions. aqueous solutions. • Water can also react as a base with the acid HF. • Water reacts as an acid with the base NH3.

− + + − HF(aq) + H2O(l) → F (aq) + H3O (aq) NH3 (aq) + H2O(l) → NH4 (aq) + OH (aq)

H+ H+

Acid-Base Properties of a Salt Acid-Base Properties of a Salt Solution Solution • One of the successes of the Brønsted-Lowry • One of the successes of the Brønsted-Lowry concept of acids and bases was in pointing out concept of acids and bases was in pointing out that some ions can act as acids or bases. that some ions can act as acids or bases. • Consider a solution of sodium cyanide, NaCN. • Sodium ion, Na+, is unreactive with water, but the cyanide ion, CN-, reacts to produce HCN H2O + − NaCN(s) → Na (aq) + CN (aq) and OH-. − − • A 0.1 M solution has a pH of 11.1 and is CN (aq) + H2O(l) HCN(aq) + OH (aq) therefore fairly basic.

Acid-Base Properties of a Salt Predicting Whether a Salt is Solution Acidic, Basic, or Neutral • The hydrolysis of an ion is the reaction of an • These rules apply to normal salts (those in ion with water to produce the conjugate acid which the anion has no acidic hydrogen) and hydroxide ion or the conjugate base and 1. A salt of a strong base and a strong acid. hydronium ion. The salt has no hydrolyzable ions and so + • The NH4 ion hydrolyzes to the conjugate base gives a neutral aqueous solution. (NH3) and hydronium ion. + + NH4 (aq) + H2O(l) NH3 (aq) + H3O (aq) An example is NaCl.

15 Predicting Whether a Salt is Predicting Whether a Salt is Acidic, Basic, or Neutral Acidic, Basic, or Neutral • These rules apply to normal salts (those in • These rules apply to normal salts (those in which the anion has no acidic hydrogen) which the anion has no acidic hydrogen) 2. A salt of a strong base and a weak acid. 3. A salt of a weak base and a strong acid. The anion of the salt is the conjugate of the The cation of the salt is the conjugate of the weak acid. It hydrolyzes to give a basic weak base. It hydrolyzes to give an acidic solution. solution.

An example is NaCN. An example is NH4Cl.

Predicting Whether a Salt is Acidic, Basic, or Neutral • These rules apply to normal salts (those in which the anion has no acidic hydrogen) 4. 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.

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. (see Figure 17.8) • 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.

16 The pH of a Salt Solution The pH of a Salt Solution

• To see the relationship between Ka and Kb for • To see the relationship between Ka and Kb for conjugate acid-base pairs, consider the acid conjugate acid-base pairs, consider the acid ionization of HCN and the base ionization of CN-. ionization of HCN and the base ionization of CN-. + − + − HCN(aq) + H2O(l) H3O (aq) + CN (aq) Ka HCN(aq) + H2O(l) H3O (aq) + CN (aq) Ka − − − − CN (aq) + H2O(l) HCN(aq) + OH (aq) Kb CN (aq) + H2O(l) HCN(aq) + OH (aq) Kb

+ − + − 2H2O(l) H3O (aq) + OH (aq) Kw 2H2O(l) H3O (aq) + OH (aq) Kw • When these two reactions are added you get the • When two reactions are added, their ionization of water. equilibrium constants are multiplied.

The pH of a Salt Solution The pH of a Salt Solution

• To see the relationship between Ka and Kb for • For a solution of a salt in which only one ion conjugate acid-base pairs, consider the acid hydrolyzes, the calculation of equilibrium ionization of HCN and the base ionization of CN-. composition follows that of weak acids and bases. + − • The only difference is first obtaining the Ka or HCN(aq) + H2O(l) H3O (aq) + CN (aq) Ka Kb for the ion that hydrolyzes. CN− (aq) + H O(l) HCN(aq) + OH− (aq) K 2 b • The next example illustrates the reasoning and + − calculations involved. 2H2O(l) H3O (aq) + OH (aq) Kw • Therefore, Ka × Kb = K w

A Problem To Consider A Problem To Consider

• What is the pH of a 0.10 M NaCN solution at 25 oC? • What is the pH of a 0.10 M NaCN solution at 25 oC? -10 -10 The Ka for HCN is 4.9 x 10 . The Ka for HCN is 4.9 x 10 . • Sodium cyanide gives Na+ ions and CN- ions in • The CN- ion is acting as a base, so first, we - solution. must calculate the Kb for CN . • Only the CN- ion hydrolyzes. −14 K 1.0×10 −5 − − w CN (aq) + H2O(l) HCN(aq) + OH (aq) Kb = = −10 = 2.0×10 Ka 4.9×10 • Now we can proceed with the equilibrium calculation.

17 A Problem To Consider A Problem To Consider

• What is the pH of a 0.10 M NaCN solution at 25 oC? • What is the pH of a 0.10 M NaCN solution at 25 oC? -10 -10 The Ka for HCN is 4.9 x 10 . The Ka for HCN is 4.9 x 10 . • Let x = [OH-] = [HCN], then substitute into the • This gives equilibrium expression. 2 [HCN][OH− ] x −5 =K = 2.0×10 [CN− ] b (0.10 − x)

Structure and Acid-Base A Problem To Consider Properties

• What is the pH of a 0.10 M NaCN solution at 25 oC? -10 The Ka for HCN is 4.9 x 10 . • Two factors for acidity in binary • Solving the equation, you find that compounds: x = [OH− ] = 1.4×10−3 Bond Polarity (high is good) • Hence, pH = 14.00 − pOH = 14.00 + log(1.4×10−3 ) = 11.2 Bond Strength (low is good) • As expected, the solution has a pH greater than 7.0.

Molecular Structure and Acid Molecular Structure and Acid Strength Strength • Two factors are important in determining • Two factors are important in determining the relative acid strengths. the relative acid strengths. • One is the polarity of the bond to which the • The second factor is the strength of the bond. hydrogen atom is attached. Or, in other words, how tightly the proton is held. • The H atom should have a partial positive charge: • This depends on the size of atom X. δ+ δ- δ+ δ- H − X H − X • The more polarized the bond, the more easily the • The larger atom X, the weaker the bond and the proton is removed and the greater the acid strength. greater the acid strength.

18 Molecular Structure and Acid Strength • Consider a series of binary acids from a given column of elements. • As you go down the column of elements, the radius increases markedly and the H-X bond strength decreases. • You can predict the following order of acidic strength. HF < HCl < HBr < HI

Molecular Structure and Acid Molecular Structure and Acid Strength Strength • As you go across a row of elements, the • Consider the oxoacids. An oxoacid has the polarity of the H-X bond becomes the structure: dominant factor. H − O − Y − • As electronegativity increases going to the • The acidic H atom is always attached to an O right, the polarity of the H-X bond increases atom, which in turn is attached to another atom Y. and the acid strength increases. • Bond polarity is the dominant factor in the relative • You can predict the following order of acidic strength of oxoacids. strength. • This, in turn, depends on the electronegativity of H3N < H2O < HF the atom Y.

Molecular Structure and Acid Strength • Consider the oxoacids. An oxoacid has the structure: H − O − Y • If the electronegativity of Y is large, then the O-H bond is relatively polar and the acid strength is greater. · We can predict the acid strength for compounds with Y in the same group: HOCl > HOBr > HOI

19 Molecular Structure and Acid Molecular Structure and Acid Strength Strength • Consider the oxoacids. An oxoacid has the • Consider the oxoacids. An oxoacid has the structure: H − O − Y − X structure: H − O − Y − • Other groups, such as O atoms or O-H groups, • As a result, the H atom becomes more acidic. may be attached to Y. • The acid strengths of the oxoacids of chlorine • With each additional O atom, Y becomes effectively increase in the following order. more electronegative. HClO < HClO2 < HClO3 < HClO4

Models of HOCl HClO,

HClO2, HOClO HClO3, and

HClO4.

HOClO2

HOClO3

117

Figure 14.11: Molecular Structure and Acid The effect of Strength the number of • Consider polyprotic acids and their attached corresponding anions. oxygens on the • Each successive H atom becomes more difficult to O—H bond in a remove. series of • Therefore the acid strength of a polyprotic acid chlorine and its anions decreases with increasing negative oxyacids. charge. 2− − HPO4 < H2PO4 < H3PO4

20 Oxides Lewis Acids and Bases

• Acidic Oxides (Acid Anhydrides): • Lewis Acid: electron pair acceptor O−X bond is strong and covalent. • Lewis Base: electron pair donor

•SO2, NO2, CrO3 • Basic Oxides (Basic Anhydrides): 3+ O−X bond is ionic. H H Al3+ + 6 O Al O •KO, CaO 2 H H 6

Lewis Concept of Acids and Bases • The Lewis concept defines an acid as an electron pair acceptor and a base as an electron pair donor. • This concept broadened the scope of acid-base theory to include reactions that did not involve H+. • The Lewis concept embraces many reactions that we might not think of as acid-base reactions. 3+ 3+ Al + 6 H2O ö Al(OH2)6

A model of 3+ A(OH2)6 . A lone pair from the oxygen atom in each water molecule is donated to the Al3+ to coordinate covalent bonds between the water molecules and aluminum ion