Chemistry 102 Chapter 15 ACID-BASE CONCEPTS

Transcription

1 General Properties: ACID-BASE CONCEPTS ACIDS BASES Taste sour Bitter Change color of indicators Blue Litmus turns red no change Red Litmus no change turns blue Phenolphtalein Colorless turns pink Neutralization React with bases to form salt and water React with acids to form salt and water Arrhenius Concept of Acids and Bases: Arrhenius definition of acids and bases in based on their behavior in water (aqueous solutions) ACIDS Substances that dissolved in water to increase the concentration of H 3 O (hydronium ions) Accepted simplification: H H is always attached to a water molecule. BASES Substances that dissolved in water to increase the concentration of OH (hydroxide ions) O H H O H H 1

2 STRONG ACIDS AND BASES Strong acids and bases completely ionize in aqueous solutions and are therefore strong electrolytes. STRONG ACIDS STRONG BASES Completely ionize in aqueous solution to form H 3 O and an anion Completely ionize in aqueous solution to form OH and a cation HA (aq) H 2 O (l) H 3 O A HO 2 (aq) BOH (s) B (aq) OH (aq) Examples: Examples: HCl (aq) H 2 O (l) H 3 O (aq) Cl (aq) HO 2 NaOH (s) Na (aq) OH (aq) HNO 3 (aq) H 2 O (l) H 3 O (aq) NO 3 (aq) HO 2 KOH (s) K (aq) OH (aq) H 2 SO 4 (aq) H 2 O (l) H 3 O (aq) HSO 4 (aq) HO 2 Ca(OH) 2 (s) Ca 2 (aq) 2 OH (aq) HClO 4 (aq) H 2 O (l) H 3 O (aq) ClO 4 (aq) HO 2 Ba(OH) 2 (s) Ba 2 (aq) 2 OH (aq) NEUTRALIZATION REACTIONS BETWEEN STRONG ACIDS AND STRONG BASES HA (aq) H 2 O (l) BOH (aq) BA (aq) 2 H 2 O (l) strong acid strong base soluble salt water H 3 O (aq) A (aq) B (aq) OH (aq) B (aq) A (aq) 2 H 2 O (aq) H 3 O (aq) A (aq) B (aq) OH (aq) B (aq) A (aq) 2 H 2 O (aq) Net Ionic Equation H 3 O (aq) OH (aq) 2 H 2 O (aq) Accepted simplification for N.I.E. H (aq) OH (aq) H 2 O(l) NOTE: It has been observed that all neutralization reactions between strong acids and bases: are exothermic have the same enthalpy of reaction per mol of H = kj CONCLUSION: The neutralization reaction between any strong acid and any strong base is: H (aq) OH (aq) H 2 O(l) H 0 = 55.90/mol H ions 2

3 WEAK ACIDS AND BASES Weak acids and bases partially ionize in aqueous solutions and are therefore weak electrolytes. Weak acids and bases exist in reversible reactions (equilibrium) with the corresponding ions. WEAK ACIDS: HC 2 H 3 O 2 (aq) H 2 O (l) H 3 O (aq) C 2 O 3 O 2 (aq) HF (aq) H 2 O (l) H 3 O (aq) F (aq) WEAK BASES: NH 3 (g) H 2 O (l) NH 4 (aq) OH (aq) Limitations of the Arrhenius Concept: 1. Considers acid-base reactions only in aqueous solutions 2. Singles out the OH ion as the source of base character (other species can play a similar role) 3

4 BRONSTED-LOWRY CONCEPT OF ACIDS AND BASES Bronsted and Lowry defined acids and bases in terms of H (proton) transfer. ACIDS BASES H (proton) donors H (proton) acceptors NEUTRALIZATION: A reaction in which a H (proton) is transferred HCl (aq) NH 3 (aq) NH 4 Cl (aq) strong acid weak base soluble salt H (aq) Cl (aq) NH 3 (aq) NH 4 (aq) Cl (aq) N.I.E. H H (aq) NH 3 (aq) NH 4 (aq) acid base Bronsted-Lowry neutralization can also occur in a non-aqueous system: H HCl (benzene) NH 3 (benzene) NH 4 Cl (s) acid base Examples: Identify the Bronsted-Lowry acid and base in each of the following equations: 1. C 6 H 5 OH C 2 H 5 O C 6 H 5 O C 2 H 5 OH 2. H 2 O Cl HCl OH 4

5 BRONSTED-LOWRY CONCEPT OF ACIDS AND BASES In a reversible acid-base reaction, both forward and reverse reactions involve H transfer H H NH 3 (aq) H 2 O (l) NH 4 (aq) OH (aq) base acid acid base NH 4 and NH 3 differ by a H are a conjugate acid-base pair NH 4 is the conjugate acid of NH 3 NH 3 is the conjugate base of NH 4 H 2 O and OH differ by a H are a conjugate acid-base pair H 2 O is the conjugate acid of OH OH is the conjugate base of H 2 O H H HC 2 H 3 O 2 (aq) H 2 O (l) C 2 H 3 O 2 (aq) H 3 O (aq) acid base base acid HC 2 H 3 O 2 and C 2 H 3 O 2 differ by a H are a conjugate acid-base pair HC 2 H 3 O 2 is the conjugate acid of C 2 H 3 O 2 C 2 H 3 O 2 is the conjugate base of HC 2 H 3 O 2 H 2 O and H 3 O differ by a H are a conjugate acid-base pair H 2 O is the conjugate base of H 3 O H 3 O is the conjugate acid of H 2 O In General: A conjugate acid-base pair consists of two species in an acid-base reaction that differ by the loss or gain of a H (proton) 5

6 Consider: Species that can act both as an acid and a base are called amphiprotic. H NH 3 (aq) H 2 O (l) NH 4 (aq) OH (aq) base acid acid base H HC 2 H 3 O 2 (aq) H 2 O (l) C 2 H 3 O 2 (aq) H 3 O (aq) acid base base acid H 2 O can OR lose a H and act as an acid (with the weak base NH 3 ) gain a H and act as a base (with the weak acid HC 2 H 3 O 2 ) H 2 O is an amphiprotic species Bicarbonate ion(hco 3 ) is another example of amphiprotic species, as illustrated by the following reactions: H HCO 3 (aq) OH (aq) CO 3 2 (aq) H 2 O (l) acid H base HCO 3 (aq) H (aq) CO 2 (g) H 2 O (l) base acid The Bronsted-Lowry Concept of Acids & Bases is more general than the Arrhenius concept. Bronstead-Lowry Concept introduces additional points of view: 1. A base is a species that accepts H ions (OH is only one example of a base) 2. Acids and Bases can be ions as well as molecular substances 3. Acid-Base reactions are not restricted to aqueous solutions 4. Amphiprotic species can act as either acids or bases depending on what the other reactant is. 6

7 LEWIS CONCEPT OF ACIDS AND BASES Gilbert Lewis defined acids and bases in terms of electron-pair transfer. HCl (aq) NH 3 (aq) NH 4 Cl (aq) strong acid weak base H (aq) Cl (aq) NH 3 (aq) NH 4 (aq) Cl (aq) N.I.E.: H (aq) NH 3 (aq) NH 4 (aq) H H H : N H H : N H electron pair acceptor LEWIS ACID H electron pair donor LEWIS BASE H Example 1: BF 3 NH 3 BF 3 NH 3 F H F H F B : N H F B N H F H F H electron deficient molecule electron pair acceptor LEWIS ACID electron pair donor LEWIS BASE 7

8 Example 2: Na 2 O(s) SO 3 (g) Na 2 SO 4 (s) : O : : O : : O : S = O : : O S O : : O : : O :.... electron pair donor LEWIS BASE Example 3: Formation of complex ions electron pair acceptor LEWIS ACID In General: Al 3 (aq) 6 H 2 O (l) 3 Al(H 2 O) 6 electron pair electron pair hydrated ion acceptor donor ACID BASE LEWIS ACIDS electron pair acceptors Usually positive ions or electron deficient molecules LEWIS BASES electron pair donors Usually negative ions or molecules with lone pairs NEUTRALIZATION: A reaction in which an electron pair is transferred SUMMARY OF ACID BASE CONCEPTS ACID BASE NEUTRALIZATION Arrhenius Concept (least general) Bronsted- Lowry Concept Lewis Concept Substance that dissolved in water increases concentration of H 3 O Substance that dissolved in water increases concentration of OH 8 H (aq) OH (aq) H 2 O (l) A H donor A H acceptor A H transfer An electron pair acceptor An electron pair donor An electron pair transfer

9 ACID AND BASE STRENGTHS Relative Strengths of Acids A strong acid completely ionizes in solution, whereas a weak acid only partially ionizes. Therefore, the strength of the acid depends on the equilibrium: HA (aq) H 2 O (l) strong acid if equibrium lies to the right weak acid if equilibrium lies to the left H 3 O (aq) A (aq) An acid is stronger than another acid if it lose H more readily. To estimate relative acidic strength, comparisons between pairs of acids are made. HCl (aq) and H 3 O (aq) Recall that HCl(aq) is a strong acid and as such it ionizes completely in water: HCl (aq) H 2 O (aq) H 3 O (aq) Cl (aq) The reaction goes completely to products. ASSUME that a reverse reaction were possible to a very small extent: H H HCl (aq) H 2 O(aq) H 3 O (aq) Cl (aq) loses H more readily HCl(aq) is a stronger acid than H 3 O (aq) loses H less readily HCl(aq) H 3 O (aq) Similar comparisons between other pairs of acids show the following results: H 3 O (aq) is a stronger acid than HC 2 H 3 O 2 (aq) H 3 O (aq) is a stronger acid than HF(aq) HF(aq) is a stronger acid than HC 2 H 3 O 2 (aq) HI is a stronger acid than HCl H 3 O (aq) HC 2 H 3 O 2 (aq) H 3 O (aq) HF(aq) HF(aq) HC 2 H 3 O 2 (aq) HI HCl 9

10 ACID AND BASE STRENGTHS SUMMING UP: 1. HCl(aq) is a stronger acid than H 3 O (aq) HCl(aq) H 3 O (aq) 2. H 3 O (aq) is a stronger acid than HC 2 H 3 O 2 (aq) H 3 O (aq) HC 2 H 3 O 2 (aq) 3. H 3 O (aq) is a stronger acid than HF(aq) H 3 O (aq) HF(aq) 4. HF(aq) is a stronger acid than HC 2 H 3 O 2 (aq) HF(aq) HC 2 H 3 O 2 (aq) 5. HI is a stronger acid than HCl HI HCl A Table of Relative Strengths of acids can now be constructed: Strongest Acid HI HCl H 3 O Strength of acids increases HF Weakest Acid HC 2 H 3 O 2 An acid-base reaction always goes in the direction of the weaker acid: HCl(aq) H 2 O(aq) stronger acid H 3 O (aq) Cl (aq) weaker acid HC 2 H 3 O 2 (aq) H 2 O(l) weaker acid HF (aq) H 2 O (l) weaker acid C 2 H 3 O 2 (aq) H 3 O (aq) stronger acid F (aq) H 3 O (aq) stronger acid 10

11 Relative Strengths of Bases ACID AND BASE STRENGTHS Similar to acids, a strong base completely ionizes in solution. NaOH (aq) Na (aq) OH (aq) Unlike strong bases that contain OH, weak bases produce hydroxide by reacting with water. B (aq) H 2 O (aq) BH (aq) OH (aq) When diprotic acids ionize, they do so in 2 different steps. H 2 SO 4 (aq) H 2 O (l) H 3 O (aq) HSO 4 2 HSO 4 H 2 O (l) H 3 O (aq) SO 4 When diprotic bases ionize, they do so in 1 step. Ca(OH) (aq) Ca 2 (aq) 2 OH (aq) A base is stronger than another base if it gains H more readily To estimate relative base strength, comparisons between pairs of bases are made. H 2 O (l) and Cl (aq) Recall that HCl(aq) is a strong acid and as such it ionizes completely in water: HCl(aq) H 2 O(aq) H 3 O (aq) Cl (aq) The reaction goes completely to products. ASSUME that a reverse reaction were possible to a very small extent: H H HCl (aq) H 2 O(l) H 3 O (aq) Cl (aq) gains H more readily gains H less readily H 2 O(l) is a stronger base than Cl(aq) H 2 O(l) Cl(aq) 11

12 ACID AND BASE STRENGTHS Similar comparisons between other pairs of bass show the following results: C 2 H 3 O 2 (aq) is a stronger base than H 2 O(l) F (aq) is a stronger base than H 2 O(aq) C 2 H 3 O 2 (aq) is a stronger base than F (aq) Cl is a stronger base than I C 2 H 3 O 2 (aq) H 2 O(l) F (aq) H 2 O (aq) C 2 H 3 O 2 (aq) F (aq) Cl I SUMMING UP: 1. H 2 O(l) is a stronger base than Cl (aq) H 2 O(l) Cl (aq) 2. C 2 H 3 O 2 (aq) is a stronger base than H 2 O(l) C 2 H 3 O 2 (aq) H 2 O(l) 3. F (aq) is a stronger base than H 2 O(aq) F (aq) H 2 O (aq) 4. C 2 H 3 O 2 (aq) is a stronger base than F (aq) C 2 H 3 O 2 (aq) F (aq) 5. Cl is a stronger base than I Cl I A Table of Relative Strengths of bases can now be constructed: Weakest Base I Cl H 2 O Strength of bases increases F Strongest Base C 2 H 3 O 2 The Tables of Relative Strengths of Acids and Relative Strengths of Bases can now be combined: HI I Strength of acids increases HCl H 3 O Cl H 2 O Strength of bases increases HF F HC 2 H 3 O 2 C 2 H 3 O 2 12

13 CONCLUSIONS: 1. When an acid loses its H readily, its conjugate base does not hold the H very tightly. Ex: HCl and Cl strong weak acid base The strongest acids have the weakest conjugate bases 2. When a base holds the H very tightly, its conjugate acid does not lose the H readily Ex: C 2 H 3 O 2 strong base and HC 2 H 3 O 2 weak acid The strongest bases have the weakest conjugate acids 3. The direction of an acid-base reaction always favors the formation of the weaker acid and the weaker base: H H HC l(aq) H 2 O (l) H 3 O (aq) Cl (aq) stronger stronger weaker weaker acid base acid base H H HC 2 H 3 O 2 (aq) H 2 O(l) C 2 H 3 O 2 (aq) H 3 O (aq) weaker weaker stronger stronger acid base base acid H H HF(aq) H 2 O(l) F (aq) H 3 O (aq) weaker weaker stronger stronger acid base base acid 13

14 Examples: For each reaction below, determine the conjugate acid-base pairs and their relative strengths: 1. HSO 4 CH 3 CO 2 CH 3 CO 2 H SO HCO 3 F HF CO NH 4 CN HCN NH 3 4. The hydrogen oxalate ion (HC 2 O 4 ) is amphiprotic. Write a balanced equation showing how it reacts as an acid towards water and another equation showing how it reacts as a base towards water. 5. For each acid shown below, write the formula of its conjugate base: HPO 4 2 CH 3 NH 3 NH 3 6. For each base shown, write the formula of its conjugate acid: HAsO 4 2 IO O 2 14

15 MOLECULAR STRUCTURE AND ACID STRENGTH The strength of an acid depends on how easily a H is removed from the H Y bond. Ease of H loss is determined by 2 factors: Polarity of the H Y bond Strength of the H Y bond To be acidic, the HY bond must be polar with H atom as the positive pole. H Y As a result, compare the following bonds and their corresponding dipole moments: H Li H C H F Not acidic Not acidic Acidic The Polarity of the H Y bond The polarity of the H Y bond depends on the EN (electronegativity difference) between the atoms involved in the bond The higher the E.N. of Y The higher the E.N. The more polar the bond The easier to lose H The stronger the acid The strength of the H X bond The strength of the H Y bond depends on the size of Y (smaller atoms form stronger bonds) The larger Y is The weaker the H Y bond The less tightly the H is held The easier to lose H The stronger the acid 15

16 MOLECULAR STRUCTURE AND ACID STRENGTH Trends among Binary Acids (HY) 1. Across a period going from left to right The polarity of the H Y bond determines this trend. Electronegativity of Y increases Polarity of the H Y increases Consequently: H 2 O is a weaker acid than HF H 2 S is a weaker acid than HCl H 2 O HF H 2 S HCl 2. Going down a group The strength of the H Y bond determines this trend F H F Atomic Strength Acidic Cl Size H Cl of Strength Increases Bond Increases Br (more H Br Decreases electron I shells) H I The diagram below illustrates the combined effect of bond polarity and strength on the acids of group VI and VII. 16

17 MOLECULAR STRUCTURE AND ACID STRENGTH Trends among Oxoacids Oxoacids are O containing acids H O Y acidic H is always attached to an O atom The polarity of the H O bond determines the relative acidic strength The polarity of the H O bond depends on 2 factors: The Electronegativity of the Y atom The more EN Y is The more polar the H O bond The easier to lose H The stronger the acid H O Cl H O Br H O I EN (Cl) = 3.16 EN (Br) = 2.96 EN (I) = 2.66 (strongest) (weakest) ACIDIC STRENGTH INCREASES The number of O atoms bonded to Y (excluding O in OH groups) The more O atoms bonded to Y The more electronegative Y becomes The more polar the H O Y bond becomes The stronger the acid HClO HClO 2 HClO 3 HClO 4 (weakest) (strongest) ACIDIC STRENGTH INCREASES O O H O Cl H O Cl O H O Cl O H O Cl O least polar most polar O 17

18 SUMMARY BINARY ACIDS MOLECULAR STRUCTURE AND ACID STRENGTH Acidic Strength Increases (H Y polarity increases) Acidic strength increases (H Y bonds weakens as size of Y increases) OXOACIDS (H attached to Y through O) Acidic strength increases (polarity of H O increases) The more O atoms attached to Y, the stronger the acid POLYPROTIC ACIDS contain more than one acidic H are stronger than their corresponding anions H 2 SO 4 HSO 4 strong weak acid acid H 3 PO 4 H 2 PO 4 HPO 4 2 Acidic Strength Increases Reason: The acid strength of a polyprotic acid and its anion(s) decreases with increasing negative charge (it becomes increasingly difficult to lose a H ) 18

19 MOLECULAR STRUCTURE AND ACID STRENGTH Examples: 1. Which member of each of the following pair is the stronger acid? a) NH 3 and PH 3 b) HI and H 2 Te c) H 2 SO 3 and H 2 SeO 3 d) HBrO 4 and HBrO 3 e) HBr and H 2 Se 2. List the following compounds in order of increasing acid strength: a) HBr, H 2 Se, H 2 S b) HBrO 3, HClO 2, HBrO 19

20 SELF-IONIZATION OF WATER Pure Water is generally considered a non-electrolyte. It is actually a very, very weak electrolyte: H H 2 O (l) H 2 O (l) H 3 O (aq) OH (aq) [H3O ][OH K c = 2 [H2O] ] K c is very small The equilibrium is strongly shifted to the left [H 2 O] 2 = is essentially constant [H 2 O] for 1L of water can be calculated as follows: moles 1000 g 1 mole? = 1 L x x 55 M L 1 L g [H 2 O] 2 x K c = [H 3 O ] [OH ] = K w = 1.0 x K w = is called THE ION-PRODUCT OF WATER = 1.0 x at 25 0 C (is temperature dependent) CONCLUSION: K w = [H 3 O ] [OH ] = 1.0 x (25 0 C) In Pure Water: [H 3 O ] = [OH ] = x K w = x 2 x = K w = 1.0x10 14 [H 3 O ] = [OH ] = 1.0 x 10 7 M 20

21 Solution of a Strong Acid: Example: A 0.1 M solution of HCl complete HCl (aq) H 2 O (l) ionization H 3 O (aq) Cl (aq) M 0.1 M The TOTAL CONCENTRATION OF THE [H 3 O ] can be thought of as the sum of the: concentration of [H 3 O ] provided by the acid; [H 3 O ] = 1.0 x 10 1 M concentration of [H 3 O ] provided by the self-ionization of water; [H 3 O ] = 1.0 x 10 7 M Total Conc. of [H 3 O ] = [H 3 O ] [H 3 O ] from acid from self-ionization of water = 1.0 x 10 1 M less than 1.0 x 10 7 M (negligible) = 1.0 x 10 1 M Concentration of the [OH ] can now easily be calculated: [H 3 O ] [OH ] = 1.0 x x 10 [OH] = [H O ] x 10 = 1.0 x 10 = 1.0 x M 1 In a solution of a strong acid the concentration of the [H 3 O ] produced by the self-ionization of water can be ignored. In a solution of a Strong Base: Example: A M solution of NaOH complete NaOH (aq) Na (aq) OH (aq) ionization M M The TOTAL CONCENTRATION OF THE [OH ] can be thought of as the sum of the : concentration of [OH ] provided by the base; [OH ] = 1.0 x 10 2 M concentration of [OH ] provided by the self-ionization of water; [OH ] = 1.0 x 10 7 M Total Conc. of [OH ] = [OH ] [OH ] from base from self-ionization of water = 1.0 x 10 2 M less than 1.0 x 10 7 M (negligible) = 1.0 x 10 2 M Concentration of the [H 3 O ] can now easily be calculated: [H 3 O ] [OH ] = 1.0 x x x 10 [H 12 3O ] = = = 1.0 x 10 2 [OH ] 1.0 x 10 In a solution of a strong base the concentration of the [OH ] produced by the self-ionization of water can be ignored. 21

22 CONCLUSIONS: 1. In a neutral solution: [H 3 O ] = [OH ] = 1.0 x 10 7 M 2. In an acidic solution: [H 3 O ] [OH ] [H 3 O ] 1.0 x 10 7 M [OH ] 1.0 x 10 7 M 2. In a basic solution: [H 3 O ] [OH ] [H 3 O ] 1.0 x 10 7 M [OH ] 1.0 x 10 7 M 22

23 Examples: 1. What is the concentration of the [H 3 O ] and [OH ] in a 0.25 M solution of Ba(OH) 2? Ba(OH) 2 (aq) Ba 2 (aq) 2OH (aq) [OH ] = 0.50 M M 2 x 0.25 M [H 3 O ] = 2. What is the concentration of the [H 3 O ] and [OH ] in a M solution of HNO 3? 3. Identify each of the following solutions are acidic, basic or neutral: a) [OH ] = 1.0 x10 5 M b) [H 3 O ] = 3.8 x10 9 M c) [H 3 O ] = 6.2 x10 3 M d) [OH ] = 4.5 x10 10 M 23

24 ph SCALE ph is an important chemical quantity introduced by Sorensen, to simplify handling of negative exponents, when expressing the acidity or basicity of aqueous solutions ph can be loosely referred to as power of hydrogen) By Sorensen s definition, widely accepted today: ph = log [H 3 O ] or simply: ph = log [H ] For a neutral solution: [H 3 O ] = 1.0 x 10 7 M ph = log (1.0 x 10 7 ) = 7.00 For an acidic solution: [H 3 O ] 1.0 x 10 7 M ph 7.00 For a basic solution: [H 3 O ] 1.0 x 10 7 M ph 7 ph calculations can be greatly simplified by a rearrangement of the expression: [H 3 O ] [OH ] = 1.0 x Taking logs of both sides: log [H 3 O ] log [OH ] = log (1.0 x ) ph poh = ph poh = NOTE: poh = (log [OH ]) A phorseshoe [H ] [OH ] Up arrows indicate changing the sign and then calculating inverse log Down arrows indicate calculating log and then changing the sign ph ph poh = 14 poh 24

25 Examples: 1. Obtain the ph corresponding to a hydroxide-ion concentration of 2.7 x10 10 M. poh = log (2.7 x ) = ph = = A wine was tested for acidity, and it ph was found to be 3.85 at 25 C. What is the hydronium ion concentration? log [H 3 O ] = ph = 3.85 [H 3 O ] = antilog ( 3.85) = = 1.4 x 10 4 M 3. A 1.00 L aqueous solution contains 6.78 g of Ba(OH) 2. What is the ph of the solution? First: Calculate the molarity of the Ba(OH) 2 solution: moles 6.78 g Ba(OH) 2 1mol Ba(OH) 2? = x = M Ba(OH) 2 L 1.00 L solution g Ba(OH) 2 2 moles OH [OH ] = M Ba(OH) 2 x = M 1 mole Ba(OH) x [H 3 O ] = = x M M ph = - log (1.264 x M) = What is concentration of a solution of KOH with a ph of 11.89? 25

26 MEASUREMENT OF ph Acid-Base Indicators Acid-Base indicators change color within a small range. Reason: Indicators establish an equilibrium between their acid form and their base form, respectively: HIn In = acid form of indicator = base form of indicator These two forms of the Indicator have different colors (not necessarily red and blue) HIn (aq) H 2 O (aq) H 3 O (aq) In (aq) Example: Phenolphthalein: HIn (aq) is colorless In (aq) is pink When a base is added to the acidic solution of phenolphthalein: HIn (aq) H 2 O (l) H 3 O (aq) In (aq) OH reacts with H 3 O (aq) to produce H 2 O Equilibrium shifts In (aq) is produced. Solution turns pink Phenolphthalein begins to turn pink at about ph = 8.0 Universal ph paper This paper is impregnated with several indicators It gives different colors to different ph ranges (to the nearest integer value) ph-meter Consists of two electrodes (or one combination electrode) dipped into the solution whose ph is to be measured. Measures the voltage that is generated between the electrodes. This voltage depends on ph and is read on a meter calibrated directly in ph units. 26

27 ACID IONIZATION CONSTANT Solutions of a Weak Acid or Base Weak acids ionize (dissociate) partially in water HA H 2 O H 3 O (aq) A (aq) K = c [H3O ][A ] [HA][H O] 2????? [H 2 O] can be considered constant for dilute solutions can be included in K c [H3O ][A [HA] ] = K c x [H 2 O] = K a = ACID IONIZATION CONSTANT K a expresses the extent of ionization of a weak acid is a measure of the relative strength of weak acids Weak Acids Acid-Ionization Constant (at 25 0 C) HF HC 2 H 3 O 2 HClO K a = 6.8 x 10 4 K a = 1.7 x 10 5 K a = 3.5 x 10 8 strongest weak acid Strength Of Weak Acids Increases HCN K a = 4.9 x weakest weak acid 27

28 ACID IONIZATION CONSTANT The smaller the acid ionization constant (K a ), the less the acid ionizes and the weaker the acid. The acid ionization constant can also be related to the ph of an acid solution. For a given concentration, the smaller the K a, the lower the [H 3 O ] and therefore the higher the ph. Examples: 1. Rank the following 0.1 M solutions of acids in order of increasing ionization (lowest to highest): H 2 S K a = 1.0 x 10 7 HCN K a = 6.2 x HNO 2 K a = 7.2 x 10 4 HOCl K a = 2.9 x Rank the following 0.1 M solutions of acids in order of decreasing ph (highest to lowest): H 2 S K a = 1.0 x 10 7 HCN K a = 6.2 x HNO 2 K a = 7.2 x 10 4 HOCl K a = 2.9 x

29 CALCULATIONS WITH K a Examples: 1. Lactic acid, HC 3 H 5 O 3, is a weak acid found in sour milk. A M solution of lactic acid has a ph of What is the ionization constant, K a for this acid? What is the degree of ionization? HC 3 H 5 O 3 (l) H 2 O (l) H 3 O (aq) C 3 H 5 O 3 (aq) Initial M x ---- x x Equilibrium x x x K = a 2 [H3O ][C2H3O2 ] x = [HC H O ] x [H 3 O ] = x = antilog (-ph) = antilog (- 2.75) = 1.78 x 10 3 [C 3 H 5 O 3 ] = x = 1.78 x 10 3 [HC 3 H 5 O 3 ] = (25 x 10 3 ) (1.78 x 10 3 ) = 23.2 x 10 3 K = a 3 2 (1.78 x 10 ) = 1.4 x x 10 3 Molar concentration of Ionized Acid Degree of Ionization = Total Molar concentration of Acid 1.78 x 10 3 Degree of Ionization = = or 7.1%

30 Examples: 2. What are the concentrations of hydrogen ion and acetate ion in a solution of 0.10 M acetic acid? What is the ph of the solution? What is the degree of ionization? (K a of acetic acid = 1.7 x 10 5 ) HC 2 H 3 O 2 (l) H 2 O (l) H 3 O (aq) C 2 H 3 O 2 (aq) Initial x ---- x x Equilibrium 0.10 x x x K = a 2 [H3O ][C2H3O2 ] x = [HC H O ] = 1.7 x 10 x 5 This yields a quadratic equation: x 2 = 1.7 x x 10 5 x x x 10 5 X 1.7 x 10 6 = 0 The equation can be greatly simplified, by realizing that: x 0.10 Therefore: 0.10 x 0.10 can be neglected x 2 x 2 It follows: = 1.7 x 10 5 x 2 = 1.7 x X 0.10 x = [H 3 O ] = [C 2 H 3 O 2 ] = 1.30 x 10 3 M Now we can check if the assumption: 0.10 X 0.10 is valid or not = = 0.10 YES, the assumption is valid! ph = - log (H 3 O ) = - log (1.30 x 10 3 ) = x 10 3 Degree of Ionization = = or 1.3% 1.0 x 10 1 This method is commonly referred to as the Approximation Method 30

31 Examples: 3. Calculate the ph and % ionization of a M solution of HF. (K a = 3.5x10 4 ) Initial Equilibrium 31

32 POLYPROTIC ACIDS Polyprotic acids are acids that can release more than one H per molecules of acid Examples: 1. H 2 SO 4 (strong acid) and HSO 4 (weak acid) complete dissociation H 2 SO 4 (aq) H 2 O (l) partial dissociation HSO 4 (aq) H 2 O (l) H 3 O (aq) H 3 O (aq) HSO 4 (aq) SO 4 2 (aq) 2. H 2 CO 3 (weak acid) and HCO 3 (weak acid) H 2 CO 3 (aq) H 2 O (l) H 3 O (aq) HCO 3 (aq) K a = 4.3 x 10 7 HCO 3 (aq) H 2 O (l) H 3 O (aq) CO 3 2 (aq) K a = 4.8 x H 3 PO 4 (weak acid), H 2 PO 4 (weak acid), and HPO 2 4 (weak acid) H 3 PO 4 (aq) H 2 O (l) H 3 O (aq) H 2 PO 4 (aq) K a = 6.9 x 10 3 H 2 PO 4 (aq) H 2 O (l) H 3 O (aq) HPO 2 4 (aq) K a = 6.2 x HPO 4 (aq) H 2 O (l) H 3 O 3 (aq) PO 4 (aq) K a = 4.8 x NOTE: K a1 K a2 K a3 Meaning: A diprotic acid - loses the first H easier than the second one. A triprotic acid: - loses the first H easier than the second one - loses the second H easier than the third one. Reason: The 1 st H separates from an ion of a single negative charge The 2 nd H separates from an ion of an ion of a double negative charge The 3 rd H separates from an ion of a triple negative charge 32

33 CALCULATING THE CONCENTRATION OF VARIOUS SPECIES IN A SOLUTION OF A POLYPROTIC ACID Ascorbic acid (H 2 Asc) is a diprotic acid with K a1 = 7.9 x 10 5 and K a2 = 1.6 x What is the ph of a 0.10 M solution of ascorbic acid? What is the concentration of the ascorbate ion (Asc 2 ), in the solution? First Part: Calculate the ph of the solution: Note that K a2 (1.6 x ) K a1 (7.9 x 10 5 ) As such: - the 2 nd ionization can be neglected - any H produced by the 2 nd ionization can be ignored. H 2 Asc (aq) H 2 O (l) H 3 O (aq) HAsc (aq) Initial 0.10 M x ---- x x Equilibrium 0.10 x x x K = a1 [H3O ][HAsc ] x = [H Asc] = 7.9x10 x 5 x 2 = (0.10)(7.9x10 5 ) = 7.9x10 6 x = [H 3 O ] = 2.8x10 3 ph = - log (2.8x10 3 ) = 2.55 Second Part: Calculate the concentration of the sulfite ion (SO 2 3 ) : H 2 Asc(aq) H 2 O(l) HAsc (aq) H 3 O (aq) K a1 = 7.9 x10 5 HAsc (aq) H 2 O(l) Asc 2 (aq) H 3 O (aq) K a2 = 1.6 x10 12 HAsc (aq) H 2 O (l) H 3 O (aq) Asc 2 (aq) Initial y ---- y y Equilibrium y y y K = a2 2 [H3O ][Asc ] ( y)(y) = = 1.6x10 [HAsc ] y 12 33

34 Assume that: y <<<<<< Therefore: y and y K = a2 ( y)(y) (0.0028)(y) = = 1.6x y y = [Asc 2 ] = 1.6 x M General Rule: The concentration of the ion A 2 equals the second ionization constant, Ka 2 Examples: 1. Carbonic acid (H 2 CO 3 ) is a weak acid with K a1 = 4.3x10 7 and K a2 = 4.8x Calculate the ph and the carbonate ion concentration of a 0.050M solution of carbonic acid. 34

35 BASE IONIZATION EQUILIBRIA Weak bases ionize (dissociate) partially in water. NH 3 (aq) H 2 O (l) NH 4 (aq) OH (aq) K = c [NH 4 ][OH ] [NH ][H O] 3 2 [H 2 O] can be considered constant for dilute solutions can be included in K c [NH 4 ][OH [NH ] 3 ] = K x [H O] c 2 = K b = BASE IONIZATION CONSTANT K b expresses the extent of ionization of a weak base is a measure of the relative strength of weak bases Weak Bases Formula Base-Ionization Constant (at 25 C) Dimethylamine Ammonia Aniline (CH 3 ) 2 NH NH 3 C 6 H 5 NH 2 K b = 5.1 x 10 4 K b = 1.8 x 10 5 K b = 4.2 x strongest weak base Strength Of Weak Bases Increases Urea NH 2 CO NH 2 K b = 1.5 x weakest weak base 35

36 CALCULATIONS WITH K b Examples: 1. Quinine is an alkaloid, or naturally occurring base, used to treat malaria. A M solution of quinine has a ph of What is K b of quinine? Step 1: Convert the ph to poh, and then to [OH ] poh = ph = = 4.16 [OH ] = antilog ( 4.16) = 6.92 x 10 5 M Step 2: Using the symbol Qu for quinine, assemble reaction table Qu (aq) H 2 O (l) HQu (aq) OH (aq) Initial M x ---- x x Equilibrium x x x K = b [HQu ][OH 2 ] x = [Qu] = x 5 2 (6.92 x 10 ) = 3.3 x 10 ( x 10 5 ) 6 2. What is the ph of a 0.20 M solution of ammonia in water (K b = 1.8 x 10 5 )? NH 3 (aq) H 2 O (l) NH 4 (aq) OH (aq) Initial 0.20 M x ---- x x Equilibrium 0.20 x x x Assume that x is small enough to neglect compared with K = b 2 [NH 4 ][OH ] x = [NH ] x = = 1.8 x 10 x x 2 = 3.6 x 10 6 x = [OH ] = 1.89 x 10 3 M (Note that x is negligible compared to 0.20) [H3O ]= x = 5.3 x x 10 ph = log (5.3 x ) =

37 RELATIONSHIP BETWEEN K a AND K b The quantitative relationship between an acid and its conjugate base can be seen in the examples below: NH 4 (aq) H 2 O (l) NH 3 (aq) H 2 O (l) H 3 O (aq) NH 3 (aq) NH 4 (aq) OH (aq) Each of these equilibria can be expressed by an ionization constant: [H3O ][NH 3] [NH 4 ][OH K a = and K b = [NH 4 ] [NH 3] ] Addition of the two equations above results the following: NH 4 (aq) H 2 O (l) H 3 O (aq) NH 3 (aq) K a NH 3 (aq) H 2 O (l) NH 4 (aq) OH (aq) K b 2 H 2 O (l) H 3 O (aq) OH (aq) K w Recall that when two reactions are added to form a third reaction, the equilibrium constant for the third reaction is the products of the two added reactions. Therefore, K w = K a x K b Note that K a and K b are inversely proportional for an acid-base conjugate pair. This is expected, since as the strength of an acid increases (larger K a ), the strength of its conjugate base decreases (smaller K b ). Examples: 1. Hydrofluoric acid (HF) has K a = 6.8 x What is K b for the fluoride ion? K w K b = = Ka 2. Which of the following anions has the largest K b value: NO 3, PO 4 3, HCO 3? 37

38 ACID-BASE PROPERTIES OF SALT SOLUTIONS A salt is an ionic substance resulting from a neutralization reaction: Acid Base Salt Water The aqueous solution of a salt may be: basic, acidic, or neutral Reason: Soluble salts derived from weak acids or weak bases undergo hydrolysis (reaction with H 2 O) in aqueous solution. I. Salt of a strong base (NaOH) and a weak acid (HC 2 H 3 O 2 ) completely NaC 2 H 3 O 2 (aq) Na (aq) dissociated C 2 H 3 O 2 (aq) Na (aq) H 2 O (l) No Reaction H H C 2 H 3 O 2 (aq) H 2 O (l) H C 2 H 3 O 2 (aq) OH (aq) Bronsted Bronsted Bronsted Bronsted base acid acid base NOTE: The resulting solution is basic This reaction is referred to as the Hydrolysis of the acetate (C 2 H 3 O 2 ) ion 38

39 II. Salt of a weak base (NH 4 OH) and a strong acid (HCl) completely NH 4 Cl (aq) NH 4 (aq) dissociated Cl (aq) Cl (aq) H 2 O (l) No Reaction H H NH 4 (aq) H 2 O (l) NH 3 (aq) H 3 O (aq) Bronsted Bronsted Bronsted Bronsted acid base base acid NOTE: The resulting solution is acidic The reaction is referred to as the Hydrolysis of the ammonium(nh 4 )ion III. Salt of a strong base (NaOH) and a strong acid (HCl) completely NaCl (aq) Na (aq) Cl (aq) dissociated Na (aq) H 2 O (l) No Reaction Cl (aq) H 2 O (l) No Reaction The resulting solution is neutral 39

40 IV. Salt of a weak base (NH 4 OH) and a weak acid (HC 2 H 3 O 2 ) Both ions undergo hydrolysis: completely NH 4 C 2 H 3 O 2 (aq) NH 4 (aq) C 2 H 3 O 2 (aq) dissociated C 2 H 3 O 2 (aq) H 2 O (l) HC 2 H 3 O 2 (aq) OH (aq) Is the solution basic? NH 4 (aq) H 2 O (l) NH 3 (aq) H 3 O (aq) Is the solution acidic? Acidity or Basicity of solution depends on the relative acid-base strength of the two ions. K a of NH 4 must be compared with K b of C 2 H 3 O 2 If: K a K b Solution is acidic If: K a K b Solution is basic CONCLUSIONS: Solution of a salt of a strong base and weak acid is basic Solution of a salt of a weak base and a strong acid is acidic Solution of a salt of a strong acid and a strong base is neutral. Solution of a salt of a weak base and a weak acid is: neutral, if: acidic, if: basic, if: K a (cation) K b (anion) K a (cation) K b (anion) K a (cation) K b (anion) 40

41 Examples: 1. For each of the following salts, indicate whether the aqueous solution will be acidic, basic or neutral. Salt Parent Base Parent Acid Solution of salt is: Fe(NO 3 ) 3 Fe(OH) 3 insoluble Ca(CN) 2 Ca(OH) 2 strong base HNO 3 strong acid HCN weak acid acidic Basic Na 2 CO 3 Na 2 S KClO 4 Al(NO 3 ) 3 2. Write hydrolysis equation(s) for each of the following salts: a) Na 2 CO 3 Na 2 CO 3 2 Na CO 3 2 Na H 2 O no reaction (cation of a strong base, NaOH) CO 3 2 H 2 O HCO 3 OH (conjugate base of a weak acid HCO 3 ) b) Na 2 S c) KClO 4 41

42 THE ph OF A SALT SOLUTION What is the ph of a 0.10 M NaCN (sodium cyanide) solution? NOTE: NaCN is the salt of a strong base (NaOH) with a weak acid (HCN). Therefore: we expect the solution to be basic we know beforehand that the ph 7 Consider the hydrolysis of NaCN: NaCN (aq) Na (aq) CN (aq) Na (aq) H 2 O (l) No Reaction CN (aq) H 2 O (l) HCN (aq) OH (aq) K b (CN ) =? (not readily available) Solution: Consider the ionization equilibrium of HCN: HCN (aq) H 2 O (l) H 3 O (aq) CN (aq) K a = 4.9 x K a (HCN) x K b (CN ) = K w K b(cn K 1.0 x 10 K (HCN) 4.9 x 10 w ) = = a = 2.0 x 10 5 Now we can calculate the ph of the solution from equilibrium data: CN (aq) H 2 O (l) HCN(aq) OH (aq) Initial x ---- x x Equilibrium 0.10 x x x K = b 2 [HCN][OH ] x = [CN ] 0.10 = 2.0 x 10 x 5 Consider the usual assumption: 0.10 x 0.10 x x 10 5 = x = [OH ] = 1.4 x poh = log (1.4 x 10 3 ) = 2.85 ph = =

43 Examples: 1. Household bleach is a 5% solution of sodium hypochlorite (NaClO). This corresponds to a molar concentration of about 0.70 M NaClO. K a for HClO (hypochlorous acid) is 3.5 x What is the ph of household bleach? NaClO(aq) Na (aq) ClO (aq) NaClO is the salt of a base with a acid. Therefore the solution of this salt is expected to be ClO (aq) H 2 O (l) HClO (aq) OH (aq) K b =? K b (ClO ) = ClO (aq) H 2 O (l) HClO(aq) OH (aq) Initial x ---- x x Equilibrium 0.70 x x x K = b 2 [HClO][OH ] x = [ClO ] 0.70 = x Consider the usual assumption: 0.70 x 0.70 x 2 K b = = x = [OH ] = 0.70 poh = log [OH ] = ph =