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Applications of Equilibrium and Bases Chapter 14 Acids and Bases 14.1 The Nature of Acids and Bases Acids -

Bases Acids and Bases 14.1 The Nature of Acids and Bases Acids - first recognized as class of substances that sour. Bases ( are soluble bases) - characterized by bitter taste + slippery feel (drain cleaners are highly basic mixtures) Arrhenius “Acids produce in aqueous solutions, while bases produce ions”

Limited application Bronsted-Lowry “An is a proton donor and a is a proton acceptor.”

⇋ + - HA(aq) + H2O(l) H3O (aq) + A (aq) Acid Base conjugate base N.B. The conjugate base is all that is left of the acid molecule after a proton is lost. N.B. The conjugate base is all that is left of the acid molecule after a proton is lost. The conjugate acid is formed when the proton is transferred to the base. N.B. The conjugate base is all that is left of the acid molecule after a proton is lost. The conjugate acid is formed when the proton is transferred to the base. A conjugate acid-base pair consists of two substances related to each other by the donating and accepting of a single proton. 2 conjugate acid-base pairs: HA and A-

H2O and H3O+ ( )* (*hydronium ion is actually more complicated)

The equilibrium system is a competition for the proton between the two bases: H2O and A- If H2O is a much stronger base than A- (if H2O has a greater affinity for H+ than A- does) the equilibrium position will be far to the right (favoring products); most of the acid dissolved will be in the ionized form. (For strong acids like , , , up to now we have assumed the acids are fully ionized in .)

⇋ + - HA(aq) + H2O(l) H3O (aq) + A (aq)

Equilibrium expression for system above:

Ka = [H3O+][A-] = [H+] [A-] [HA] [HA]

Ka is called the acid Ka has the same form as that for the simple dissociation into ions HA(aq) ⇋ H+ (aq) + A-(aq)

But, don’t forget the important role the plays in causing the acid to ionize in the first place. Write the simple dissociation (ionization) reactions (omitting water) for each of the following acids. a) Hydrochloric acid HCl(aq)

b) HC2H3O2(aq) Ethanoic acid + c) The ion NH4 + d) The anilinium ion C6H5NH3 (aq) 3+ e) The hydrated aluminum(III) ion [Al(H2O)6] Write the simple dissociation (ionization) reactions (omitting water) for each of the following acids. a) Hydrochloric acid HCl(aq) HCl(aq) ⇋ H+ (aq) + Cl-(aq) Write the simple dissociation (ionization) reactions (omitting water) for each of the following acids.

b) Acetic acid HC2H3O2(aq) Ethanoic acid ⇋ + - HC2H3O2(aq) H (aq) + C2H3O2 (aq) Write the simple dissociation (ionization) reactions (omitting water) for each of the following acids. + c) The ammonium ion NH4

+ ⇋ + NH4 (aq) H (aq) + NH3(aq) Write the simple dissociation (ionization) reactions (omitting water) for each of the following acids. + d) The anilinium ion C6H5NH3 (aq)

+ ⇋ + C6H5NH3 (aq) H (aq) + C6H5NH2(aq) Write the simple dissociation (ionization) reactions (omitting water) for each of the following acids. 3+ e) The hydrated aluminum(III) ion [Al(H2O)6]

3+ ⇋ + 2+ [Al(H2O)6] H (aq) + [Al(H2O)5OH] (aq) BrØnsted-Lowry model extends to reactions in phase. ⇋ HCl(g) + NH3(g) NH4Cl (s) 14.2 Defined by equilibrium position of its dissociation (ionization) reaction: ⇋ + - HA(aq) + H2O(l) H3O (aq) + A (aq)

Strong Acids: Strong Acids Ka large Equilibrium lies far to the right[H+] approx. equal to

[HA]0 Almost all the original HA is dissociated (ionized) at equilibrium A strong acid yields a weak conjugate base (which means the water molecules win the competition for the H+ ions) Weak Acids Ka small

Equilibrium lies far to the left [H+]<< [HA]0 Almost all of the original HA is still present at equilibrium A weak acid yields a conjugate base, A-, that is a stronger base than water. Weak Acids Strong Acids

Phosphoric acid H3PO4

Nitrous acid HNO2 Hypochlorous HOCl Organic acids - have a carboxyl group Acetic (ethanoic acid) Weak Acids Strong Acids

Phosphoric acid H3PO4 Hydrochloric acid

Nitrous acid HNO2 HCl(aq) Hypochlorous HOCl Nitric acid HNO3 (aq) Organic acids - have a carboxyl group HClO4(aq) Acetic (ethanoic acid) Benzoic acid Sulfuric Acid - diprotic acid (2 acidic protons)

+ - H2SO4(aq) → H (aq) + HSO4 (aq) Strong acid, virtually 100% ionized in water Sulfuric Acid - diprotic acid (2 acidic protons)

+ - H2SO4(aq) → H (aq) + HSO4 (aq) Strong acid, virtually 100% ionized in water

- However, the HSO4 ion is only a weak acid ⇋ 2- HSO4-(aq) H+(aq) + SO4 (aq)

Notice most acids are - Sulfuric Acid - diprotic acid (2 acidic protons)

+ - H2SO4(aq) → H (aq) + HSO4 (aq) Strong acid, virtually 100% ionized in water

- However, the HSO4 ion is only a weak acid ⇋ 2- HSO4-(aq) H+(aq) + SO4 (aq) Notice most acids are oxyacids - acidic proton is attached to an atom. Page 628 Table 14.2 Values of Ka for some common monoprotic acids Note that Ka values for strong acids are not listed - why? Problems finding Ka for strong acids

Ka = [H+] [A-] [HA] very small and hence very uncertain

So far to right, prevents accurate calculation of Ka. Relative Base Strength Using table 14.2 arrange the following species according to their strengths as BASES. (proton acceptors)

H2O, F-, Cl-, NO2- and CN- Relative Base Strength

H2O, F-, Cl-, NO2- and CN- Water is a stronger base than the conjugate base of a strong acid, but a weaker base than the conjugate base of a weak acid.

Cl- < H2O < conjugate bases of weak acids Weakest base strongest bases From table:

Ka HF > Ka for HNO2 > Ka for HCN Relative Base Strength

H2O, F-, Cl-, NO2- and CN-

Ka HF > Ka for HNO2 > Ka for HCN

So combined order of increasing base strength

Cl- < H2O < F- < NO2- < CN- Note HF is a weak acid - why might that be? Water - as an acid and a base Water - as an acid and a base Water is Amphoteric - can behave as acid and base Autoionization of water Kw ion product or dissociation constant for water

Kw = [H3O+] [OH-] = [H+][OH-]

For water at 25oC experiments show that [H+] = [OH-] = 1.0 x 10-7 M

At 25oC Kw = [H+][OH-] = (1.0 x 10-7)(1.0 x 10-7) = 1 x 10-14

Important to remember Kw is temperature dependent. When we find pH - will also be temperature dependent. Kw cont. In any aqueous solution at 25oC, the product of [H+] and [OH-] must always equal 1.0 x 10-14. Neutral solution, [H+] = [OH-] Acidic solution [H+] > [OH-] Basic solution [H+] < [OH-] But in each case, at 25oC, Kw = [H+][OH-] = 1 x 10-14 Calculating [H+] and [OH-]

Example 14.3 Calculate [H+] or [OH-] as required for each of the following solutions at 25oC, and state whether the solution is neutral, acidic, or basic. a) 1.0 x 10-5 M OH- -14 Kw = [H+][OH-] = 1 x 10 [H+] = Calculating [H+] and [OH-]

Example 14.3 Calculate [H+] or [OH-] as required for each of the following solutions at 25oC, and state whether the solution is neutral, acidic, or basic. a) 1.0 x 10-5 M OH- -14 Kw = [H+][OH-] = 1 x 10 [H+] = 1 x 10 -14 / 1.0 x 10-5 = 1.0 x 10-9M [OH-] > [H+] basic b) 1.0 x 10-7 M OH- -14 Kw = [H+][OH-] = 1 x 10 [H+] = 1 x 10 -14 / 1.0 x 10-7 = 1.0 x 10-7M

[OH-]= [H+] neutral c) 10.0 M H+

[OH-] = 1 x 10 -14 / 10 = 1.0 x 10-15 M

[OH-] < [H+] acidic Effect of Temperature on Kw Example 14.4 o -13 At 60 C the value of Kw is 1 x 10 . a) Using Le Chatelier’s principle, predict whether the reaction ⇋ + - 2H2O (l) H3O (aq) + OH (aq) Is exothermic or endothermic. Effect of Temperature on Kw ⇋ + - 2H2O (l) H3O (aq) + OH (aq)

Kw increases from 1 x 10-14 to 1 x 10-13 as temperature increases from 25 to 60oC. Effect of Temperature on Kw ⇋ + - 2H2O (l) H3O (aq) + OH (aq) Le Chatelier’s principle states that a system at equilibrium will respond to stress on the equilibrium in such a way as to oppose the stress imposed. \ Effect of Temperature on Kw ⇋ + - 2H2O (l) H3O (aq) + OH (aq) Here, we heated the system so the system would have been trying to overcome the increase in temperature by favoring the endothermic direction for the equilibrium system (since this would consume energy). The reaction as written must be endothermic from left to right. Effect of Temperature on Kw b) Calculate [H+] and [OH-] in a neutral solution at 60oC. For a neutral solution [H+] =[OH-] -13 and since Kw = 1 x 10 = [H+] [OH-]

[H+] = √1 x 10 -13 = 3 x 10-7 M 14.3 The pH Scale pH = -log[H+] For a solution where [H+] = 1.0 x 10-7 M pH = - (-7.00) = 7.00 14.3 The pH Scale pH = -log[H+] For a solution where [H+] = 1.0 x 10-7 M pH = - (-7.00) = 7.00 Significant figures for logarithms: The number of decimal places in the log is equal to the number of significant figures in the original number. 14.3 The pH Scale pH changes by 1 for every power of 10 change in [H+] hence pH decreases as [H+] increases Kw = [H+][OH-] log Kw = log [H+] + log [OH-]

-log Kw = -log[H+] - log [OH-]

Thus pKw = pH + pOH -14 and since Kw = 1.0 x 10 -14 pKw = -log (1.0 x 10 ) = 14.00

-14 pKw = -log (1.0 x 10 ) = 14.00

Thus for any aqueous solution at 25oC, pH and pOH add up to 14.00 pH + pOH = 14.00 Write out Sample Exercise 14.5 Calculating pH and pOH Write out Sample Exercise 14.6 Calculating pH

14.4 Calculating the pH of Strong Acid Solutions Page 634 - 635 - read and make own notes. Work through the sample exercise. 14.5 Calculating the pH of weak Acid Solutions

Look out for the “5% rule” Find the pH of a 1.00 M solution of HF

-4 (Ka = 7.2 x 10 ) Step 1. Write out the major species in the solution. (Sketches of particles a good idea). Major species

HF and H2O (Ka is small so little ionization will occur)

Step 2. Decide which of the major species can donate H+ Both Major species can do so: ⇌ + - -4 HF H (aq) + F (aq) (Ka = 7.2 x 10 ) ⇌ + - -14 H2O H (aq) + OH (aq) (Ka = 1.0 x 10 )

Compare Kas and we see that HF is a stronger acid than water. Thus we can assume that HF will be the dominant source of H+. (We ignore the contribution from water.) Equilibrium of H+ (and hence pH) will be determined by HF ⇌ H+ (aq) + F-(aq)

+ - -4 Ka = [ H ] [ F ] = 7.2 x 10 ) [HF] The set up ICE table Equilibrium concentration of H+ (and hence pH) will be determined by ⇌ + - -4 HF H (aq) + F (aq) (Ka = 7.2 x 10 ) I 1.00 M 10-7M 0 Approx. 0 Equilibrium concentration of H+ (and hence pH) will be determined by ⇌ + - -4 HF H (aq) + F (aq) (Ka = 7.2 x 10 ) I 1.00 M approx. 0 0 Initial value for [H+] is an approximation since we are ignoring the H+ ions from the autoionization of water Equilibrium concentration of H+ (and hence pH) will be determined by ⇌ + - -4 HF H (aq) + F (aq) (Ka = 7.2 x 10 ) I 1.00 M 0 0 C - x +x +x E 1.00 - x x x Substitute the equilibrium into the equilibrium expression

-4 + - Ka = 7.2 x 10 = [ H ] [ F ] = (x)(x) [HF] 1.00 - x This expression produces a quadratic equation, that can be solved using the quadratic formula.

HOWEVER, since Ka is so small, HF will dissociate only slightly, and so x is expected to be small which allows us to simplify the calculation. If x is very small compared to 1.00, the term in the denominator can be approximated 1.00 - x approx = 1.00 So then the equilibrium expression becomes:

7.2 x 10-4=approx. (x)(x) 1.00 Hence x2 approx = 7.2 x 10-4 x approx. = sq. rt (7.2 x 10-4) = 2.7 x 10-2- How valid is this approximation? Since Ka values for acids are known to an accuracy of only about +/- 5%, it is reasonable to apply this figure when determining the validity of the approximation [HA]0 - x approx.= [HA]0 Test: First calculate the value of x by making the approximation. Which we did above, x approx. = 2.7 x 10-2 Then compare the sizes of x and [HA]0 using x x 100% = 2.7 x 10-2x 100% = 2.7%

[HA]0 1.00 If this expression yields a value less than or equal to 5% (which is does in this case) then the value of x is indeed small enough that the approximation [HA]0 - x approx.= [HA]0 will be considered valid. So then x = [H+] = 2.7 x 10-2M pH = -log[H+] = -log(2.7 x 10-2) = 1.57 Solving Weak Acid Equilibrium Problems

Look back and write out a list of steps for solving this type of problem - then compare your list to the one given in the textbook - add in any you missed - but use your own words going forward when you practice problems - until the steps are embedded in your brain muscle memory. 14.5 cont. (skip to page 641) Percent dissociation

Percent dissociation = amount dissociated (mol/L) x 100% Initial concentration (mol/L)

Calculate the percent dissociation of acetic acid

-5 (Ka = 1.8 x 10 ) in 1.00 M HC2H3O2. Acetic acid is a weak acid, so the major species in this solution are HC2H3O2 and H2O

The acetic acid is however a stronger acid than water so the dominant equilibrium will be ⇋ + - HC2H3O2(aq) H (aq) + C2H3O2 (aq) And the equilibrium expression is

-5 - Ka = 1.8 x 10 = [H+][C2H3O2 ]

[ HC2H3O2 ] ⇋ + - HC2H3O2(aq) H (aq) + C2H3O2 (aq) Initial 1.00 0 0

Change -x x x

Equilibrium 1.00 - x x x

-5 - 2 Ka = 1.8 x 10 = [H+][C2H3O2 ] = (x)(x) x

[ HC2H3O2 ] 1.00 - x 1.00

(assuming the x is small compared with [HA]0 Thus x 4.2 x 10-3 Is the approximation 1.00 - x 1.00 valid by the 5% rule? Is the approximation 1.00 - x 1.00 valid by the 5% rule?

x x 100 = 4.2 x 10-3 x 100 = 0.42 % which is less than 5%.

[HA]0 1.0 Yes, the approximation is valid.

So [H+] = x =

The percent dissociation is

[H+] x 100 =

[HC2H3O2]0 Is the approximation 1.00 - x 1.00 valid by the 5% rule?

x x 100 = 4.2 x 10-3 x 100 = 0.42 % which is less than 5%.

[HA]0 1.0 Yes, the approximation is valid.

So [H+] = x = 4.2 x 10-3 M

The percent dissociation is

[H+] x 100 = 4.2 x 10-3 = 0.42 %

[HC2H3O2]0 1.0 Calculate the percent dissociation of acetic acid

-5 (Ka = 1.8 x 10 ) in 0.100 M HC2H3O2. ⇋ + - HC2H3O2(aq) H (aq) + C2H3O2 (aq) Initial 0.100 0 0

Change -x x x

Equilibrium 0.100 - x x x

-5 - 2 Ka = 1.8 x 10 = [H+][C2H3O2 ] = (x)(x) x

[ HC2H3O2 ] 0.100 - x 0.100 [H+] = x = 1.3 x 10-3 M

The percent dissociation is

[H+] x 100 =

[HC2H3O2]0 [H+] = x = 1.3 x 10-3 M

The percent dissociation is

[H+] x 100 = 1.3 x 10-3 = 1.3%

[HC2H3O2]0 0.100

Notice: The concentration of the H+ is smaller in the 0.10M acetic acid solution than in 1.0M solution BUT the percent dissociation is significantly greater in the 0.10M solution than the 1.0M solution. -3 [H+] = x = 1.3 x 10 M The more dilute the weak The percent dissociation is acid solution, [H+] x 100 = 1.3 x 10-3 = 1.3% the greater the percent [HC2H3O2]0 0.100 dissociation.

Notice: The concentration of the H+ is smaller in the 0.10M acetic acid solution than in 1.0M solution BUT the percent dissociation is significantly greater in the 0.10M solution than the 1.0M solution.

Try exercises 14.63 and 14.64 Explanation for increase in percent dissociation as initial [HA] decreases.

A weak acid, HA, with initial concentration [HA]0 is at equilibrium where

[HA] = [HA]0 - x [HA]0 [H+] = [A-] = x

Thus Ka = [H+][A-] (x)(x)

[HA] [HA]0 Now imagine enough water is added suddenly to dilute by a factor of 10 The new concentrations before any adjustment can occur will be [A-]new = [H+]new = x / 10

[HA]new = [HA]0 / 10

Thus Q = (x/10)(x/10) = 1 (x) (x) = 0.1 Ka

[HA]0 10 10 [HA]0 Since Q (reaction quotient) is less than Ka, the system must adjust to the right to reach the new equilibrium position. Thus the percent dissociation increases when the acid is diluted. Calculating Ka from percent dissociation

Lactic acid (HC3H5O3) 3.7% dissociated in a 0.100M aqueous solution.

Calculate Ka for this acid. Small % value for dissociation tells us this is a weak acid.

Major species in solution are HC3H5O3 and H2O Still is stronger acid than water ⇋ + - HC3H5O3(aq) H (aq) + C3H5O3 (aq) Initial 0.100 0 0

Change -x x x

Equilibrium 0.100 - x x x

- 2 Ka = = [H+][C3H5O3 ] = (x)(x) x

[ HC3H5O3 ] 0.100 - x 0.100 The change needed to reach equilibrium can be obtained from the percent dissociation: Percent dissociation = 3.7% = x x 100% =

[HC3H5O3]0 x = 3.7 (0.10) = 3.7 x 10-3 mol/L

100 Now find equilibrium concentrations:

[ HC3H5O3 ] = 0.10 - x = 0.10M - -3 [H+]= [C3H5O3 ] = 3.7 x 10 M

0.100 - x 0.100 Now calculate Ka

- -3 2 -4 Ka = = [H+][C3H5O3 ] = (3.7 x 10 ) = 1.4 x 10

[ HC3H5O3 ] 0.10

Try exercises 14.65 and 14.66 14.6 Bases Arrhenius model “Acids produce hydrogen ions in aqueous solutions, while bases produce hydroxide ions” Bronsted-Lowry model Base is a proton acceptor. Strong bases: (NaOH) and potassium hydroxide (KOH) satisfy both criteria. NaOH(s) → Na+ (aq) + OH- (aq)

Dissociate completely when dissolved in aqueous solution; behave as strong 1.0M NaOH solution really contains 1.0M Na+ and 1.0M OH-. Strong bases: All of group 1A elements are strong bases - Only NaOH and KOH common lab reagents (LiOH, RbOH, CsOH - expensive) Group 2A hydroxides

Ca(OH)2 and Ba(OH)2 and Sr(OH)2 also strong bases. 2 moles of OH- for every mole of cmpd dissolved. (Not very soluble, only used when solubility factor not important.) Many are suspensions of hydroxides such as aluminum hydroxide and hydroxide. The low solubility of these compounds prevents a large hydroxide ion concentration that would harm the tissues of the mouth, , and . Yet these suspensions provide plenty of hydroxide ions to react with the stomach acid, since the dissolve as the reaction happens. Calculating the pH of strong bases Calculate pH of a 5.0 x 10-2 M NaOH solution. Major species? Calculating the pH of strong bases Calculate pH of a 5.0 x 10-2 M NaOH solution. Major species?

Na+ OH- and H2O pH will be dominated by the OH- ions from the dissolved NaOH. In the solution [OH-] = 5.0 x 10-2 M [H+] can be calculated from Kw Kw = [H+][OH-] = 1.0 x 10-14 Calculating the pH of strong bases [H+] = Kw = 1.0 x 10-14 = [OH-] 5.0 x 10-2 Calculating the pH of strong bases [H+] = Kw = 1.0 x 10-14 = 2.0 x 10-13M [OH-] 5.0 x 10-2 pH = -log [H+] = pH = Calculating the pH of strong bases [H+] = Kw = 1.0 x 10-14 = 2.0 x 10-13M [OH-] 5.0 x 10-2 pH = -log [H+] = - log 2.0 x 10-13 pH = 12.70

Basic solution [OH-] > [H+] pH > 7 ⇋ + - H2O (l) H (aq) + OH (aq)

The added OH- from the has shifted the water autoionization equilibrium to the left, significantly lowering [H+] compared with that in pure water.

Exercises: 14.77 through 14.80 ⇋ + - NH3 (aq) + H2O(l) NH4 (aq) + OH (aq)

Note that concentrated solution will often be labelled as ammonium hydroxide.

The ammonia molecule accepts a proton and thus functions as a base, with water acting as the acid (proton donor).

Note that even though the ammonia contains no hydroxide ions, it increases the concentration of hydroxide ions to yield a basic solution.

Kb = [BH+][OH-] [B] pH calculations for solutions of weak bases - similar to those for weak acids.

Review and make notes from pages 647 -650

Exercises 14.83, 14.84, 14.85, 14.86

Read “Chemical Impact” and make notes about amines 14. 7 Polyprotic acids

H2SO4 can supply more than one proton

H3PO4

Stepwise dissociation constants for several common polyprotic acids Table 14.4 page 651 14. 7 Polyprotic acids

For a typical weak polyprotic acid

Ka1 > Ka2 > Ka3 In other words, the acid in each step of the dissociation is successively weaker. The loss of the second or third proton occurs less readily than the loss of the first proton; not surprising when consider the negative charge on the acid increases, so more difficult to remove the positively charged proton. Calculating the pH of a polyprotic acid

For a typical weak polyprotic acid

Phosphoric acid - follow the example in text pages 651 - 652 N.B. Sulfuric acid

Unique among the common acids in that it is a strong acid in its first dissociation step (and is a strong ) but a weak acid in its second step:

H2SO4(aq) H+ (aq) + HSO4- (aq) Ka very large

⇋ + - -2 HSO4- (aq) H (aq) + SO4 (aq) Ka = 1.2 x 10 N.B. Sulfuric acid

Work through the example (page 653) to see how to calculate the pH of sulfuric acid and how the second step does not make a significant contribution to the concentration of H+ for most common cases.

Also work through example (page 654) to see how for solutions more dilute than 1.0M we do have to consider the dissociation of the hydrogen ion (involves use of the quadratic formula). 14.8 Acid Base Properties of Salts

Salt is another name for .

When a salt dissolves in water we assume it breaks up into its ions, which then move about independently, at least in dilute solutions. Salts that produce neutral solutions

Remember that the conjugate base of a strong acid has virtually no affinity for protons in water. Strong acids completely dissociate in aqueous solution.

So when anions like Cl-, and NO3- (anions from strong acids) are placed in water, they do not combine with H+ and have no effect on the pH. Salts that produce neutral solutions

When cations like K+, and Na+ (cations from strong bases) are placed in water, they have no affinity for H+ nor can they produce H+ ions so they too have have no effect on the pH.

So aqueous solutions of salts such as KCl, NaCl, NaNO3, and

KNO3 are neutral (have a pH of 7). Salts that produce neutral solutionsThe salt of a strong acid When cations like K+, and Na+ (cationsand a strong from strong bases) base gives a are placed in water, they have noneutral affinity for H+ nor can they produce H+ ions so they too havesolution have no effect on the pH.

So aqueous solutions of salts such as KCl, NaCl, NaNO3, and

KNO3 are neutral (have a pH of 7). Salts that produce basic solutions

Consider a solution of (NaC2H3O2)

The major species are Na+ C2H3O2- and H2O

Na+ neither acid nor base properties

C2H3O2- conjugate base of a weak acid (acetic acid) - has a significant affinity for a proton and acts as a base.

Water - weakly amphoteric substance. Salts that produce basic solutions

Consider a solution of sodium acetate (NaC2H3O2)

The major species are Na+ C2H3O2- and H2O

Na+ neither acid nor base properties

C2H3O2- conjugate base of a weak acid (acetic acid) - has a significant affinity for a proton and acts as a base.

Water - weakly amphoteric substance. Salts that produce basic solutions

-5 Kb = [HC2H3O2 ][OH-] Ka = 1.8 x10 for acetic acid

[C2H3O2-]

Notice Ka x Kb = [H+][C2H3O2-] x [HC2H3O2 ][OH-]

[HC2H3O2 ] [C2H3O2-]

= [H+] [OH-] = Kw

VIP For a weak acid and its conjugate base Ka x Kb = Kw Salts that produce basic solutions

VIP For a weak acid and its conjugate base Ka x Kb = Kw

Taking -log of both sides gives us pKa + pKb = pKw

o At 25 C, Kw = 1.0 x 10-14, so pKw = 14.00. In solutions at 25oC we can use the relationship

pKa + pKb = 14.00 Thus when either Ka or Kb is known, the other can be calculated. So, for the acetate ion

Ka x Kb = Kw

-14 -10 Kb= Kw = 1.0 x 10 = 5.6 x 10 -5 Ka 1.8 x 10

Generally, for any salt whose cation has neutral properties (such as Na+ or K+) and whose anion is the conjugate base of a weak acid, the aqueous solution will be basic. Salts that produce acidic solutions

In general, salts in which the anion is not a base and the cation is the conjugate acid of a weak base produce acidic solutions. ⇋ + NH4+ H (aq) + NH3(aq)

Summary table 14.6 on page 660. 14.9 The Effect of Structure on Acid-Base properties Any molecule containing a hydrogen atom is potentially an acid. 2 main factors determine whether it will behave as a Bronsted-Lowry acid: ● Strength of the bond to H ● Polarity of the bond Consider bond polarity series for hydrogen halides H-F > H-Cl > H-Br > H-I Most polar least polar (electronegativity decreases down the group) Based on the high polarity of the H-F bond we might expect to be a very strong acid. In fact, HF is the only weak acid (Ka = 7.2 x 10-4) among HX molecules when dissolved in water. H-F > H-Cl > H-Br > H-I 565 427 363 295 Bond strength values in kJ/mol The HF bond is unusually strong, and thus is difficult to break, and contributes to the lack of dissociation of HF molecules in water. Increasing acid strength of oxyacids 14.10 Acid-Base properties of Read and make your own notes

14.11 The Lewis Acid-Base Model Read and make notes.