Ionic Equilibria I 離子平衡: Acids and Bases

Ionic Equilibria I 離子平衡: Acids and Bases
18 Ionic Equilibria I 離子平衡: Acids and Bases 維他命C Ascorbic acid 檸檬酸 Citric acid

Chapter Goals A Review of Strong Electrolytes (強電解質)
The Autoionization of Water (水的自身離子化) The pH and pOH Scales (pH值和pOH值) Ionization Constants for Weak Monoprotic Acids and Bases (弱單質子酸及鹼的解離常數) Polyprotic Acids (多質子酸) Solvolysis (溶劑分解) Salts of Strong Bases and Strong Acids (強鹼及強酸形成的鹽) Salts of Strong Bases and Weak Acids (強鹼及弱酸形成的鹽) Salts of Weak Bases and Strong Acids (弱鹼及強酸形成的鹽) Salts of Weak Bases and Weak Acids (弱鹼及弱酸形成的鹽) Salts That Contain Small, Highly Charged Cations

A Review of Strong Electrolytes
This chapter details the equilibria of weak acids and bases. We must distinguish weak acids and bases from strong electrolytes. Weak acids and bases ionize or dissociate partially, much less than 100%. In this chapter we will see that it is often less than 10%! Strong electrolytes ionize or dissociate completely. (強電解質完全解離) Strong electrolytes approach 100% dissociation in aqueous solutions. (強電解質在水溶液中幾近100%解離)

Aqueous Solutions: An Introduction
The reason nonelectrolytes do not conduct electricity is because they do not form ions in solution. ions conduct electricity in solution Acid: a substance that produces hydrogen ions, H+, in aqueous solutions Base: a substance that produces hydroxide ions, OH-, in aqueous solutions Salt: a compound that contains a cation other than H+, and an anion other than hydroxide ion, OH- , or oxide ion, O2- 4 12 12

Classification of solutes strong electrolytes強電解質- conduct electricity extremely well in dilute aqueous solutions Examples of strong electrolytes HCl, HNO3, etc. strong soluble acids NaOH, KOH, etc. strong soluble bases NaCl, KBr, etc. soluble ionic salts ionize in water essentially 100% 檸檬酸 5 13 13

Classification of solutes weak electrolytes - conduct electricity poorly in dilute aqueous solutions CH3COOH, (COOH)2 weak acids NH3, Fe(OH)3 weak bases some soluble covalent salts ionize in water much less than 100% 6 14 14

Strong and Weak Acids Acids are substances that generate H+ in aqueous solutions. Strong acids ionize 100% in water. HCl(g) H+(aq) + Cl-(aq) 100% HNO3 + H2O H3O+(aq) + NO-3(aq) 100% HNO H+(aq) + NO-3(aq) H2O 7 15 15

strong acids--ionize almost 100% 氫氯酸氫溴酸氫碘酸硝酸過氯酸氯酸硫酸 8 16 16

Weak acids--Typically ionize 10% or less! 氫氟酸醋酸氰化氫亞硝酸碳酸亞硫酸磷酸草酸 9 17 17

Strong Bases, Insoluble Bases, and Weak Bases Strong Bases Characteristic of common inorganic bases is that they produce OH- ions in solution. Similarly to strong acids, strong bases ionize 100% in water. NaOH Na+(aq) + OH-(aq) H2O Ba(OH) Ba+2(aq) + 2OH-(aq) H2O 10 19 19

氫氧化鋰氫氧化鈉氫氧化鈣氫氧化鉀氫氧化鍶氫氧化銣氫氧化銫氫氧化鋇 Notice that they are all hydroxides of IA and IIA metals 11

Strong Bases, Insoluble Bases, and Weak Bases Insoluble bases Ionic compounds that are insoluble in water, consequently, not very basic. Cu(OH)2, Zn(OH)2, Fe(OH)2氫氧化亞鐵, Fe(OH)3 Weak bases are covalent compounds that ionize slightly in water. Ammonia is most common weak base-- NH3 NH3(g) + H2O(l) NH4+(aq) + OH-(aq) 12 19 19

Most Water Soluble Salts The solubility guidelines from Chapter 4 will help you remember these salts. NaCl(s) Na+(aq) + Cl-(aq) H2O 100% Ca(NO3) Ca+2(aq) + 2NO3-(aq) H2O 100%

The calculation of ion concentrations in solutions of strong electrolytes is easy. Example 18-1: Calculate the concentrations of ions in M nitric acid, HNO3. HNO3 + H2O H3O+(aq) + NO-3(aq) 100% 0.050M 0.050M 0.050M

Example Calculation of Concentrations of Ions Calculation the molar concentration of Ba2+ and OH- ions in 0.03M barium hydroxide. Ba(OH) Ba2+(aq) + 2OH-(aq) H2O initial 0.03M change -0.03M +0.03M +2x(0.03)M final 0.0M 0.03M +0.06M [Ba2+]=0.03M [OH-]=0.06M Exercise 4 and 6

Example 18-2: Calculate the concentrations of ions in M strontium hydroxide, Sr(OH)2, solution. Sr(OH) Sr2+(aq) + 2OH-(aq) H2O initial 0.02M change -0.02M +0.02M +2x(0.02)M final 0.0M 0.02M +0.04M [Sr2+]=0.02M [OH-]=0.04M

The Autoionization of Water 水的自身離子化[反應]
Pure water ionizes very slightly. The concentration of the ionized water is less than one-millionth molar at room temperature. We can write the autoionization of water as a dissociation reaction similar to those previously done in this chapter. H2O(l) + H2O(l) H3O+(aq) + OH-(aq) Because the activity of pure water is 1, the equilibrium constant for this reaction is: Kc=[H3O+][OH-]

The Autoionization of Water
Experimental measurements have determined that the concentration of each ion is 1.0 x 10-7 M at 25oC. Note that this is at 25oC, not every temperature! We can determine the value of Kc from this information. Kc=[H3O+][OH-] = (1.0x10-7)x(1.0x10-7) = 1.0x10-14

The Autoionization of Water
This particular equilibrium constant is called the ion-product for water and given the symbol Kw. Kw is one of the recurring expressions for the remainder of this chapter and Chapters 19 and 20. Kw=[H3O+][OH-] = 1.0x10-14

HNO3 + H2O H3O+(aq) + NO3- (aq)
Example Calculation of Ions Concentrations Calculation the concentrations of H3O+ and OH- ions in 0.05M HNO3 solution. Strong acid HNO3 + H2O H3O+(aq) + NO3- (aq) initial 0.05M change -0.05M +0.05M +0.05M At equil 0.0M 0.05M +0.05M [H3O+]=[NO3-]=0.05M 2 H2O(l) H3O+(aq) + OH-(aq) initial 0.05M change -2xM +xM +xM At equil (0.05+x)M +xM Kw=[H3O+][OH-] 1.0x10-14 =(0.05+x)(x) (very small number) 1.0x10-14 =(0.05)(x) x=2.0x10-13 M=[OH-] Exercise 14

2 H2O(l) H3O+(aq) + OH-(aq)
Example 18-3: Calculate the concentrations of H3O+ and OH- in M HCl. HCl + H2O H3O+ + Cl- initial 0.05M change -0.05M +0.05M +0.05M At equil 0.0M 0.05M +0.05M [H3O+]=[Cl-]=0.05M 2 H2O(l) H3O+(aq) + OH-(aq) initial 0.05M change -2xM +xM +xM At equil (0.05+x)M +xM Kw=[H3O+][OH-] 1.0x10-14 =(0.05+x)(x) (very small number) 1.0x10-14 =(0.05)(x) x=2.0x10-13 M=[OH-]

The pH and pOH scales pH=-log[H3O+] or [H3O+]=10-pH pOH=-log[OH-]
A convenient way to express the acidity and basicity of a solution is the pH and pOH scales. The pH of an aqueous solution is defined as: pH=-log[H3O+] or [H3O+]=10-pH pOH=-log[OH-] or [OH-]=10-pOH

The pH and pOH scales pH=-log[H3O+] pH=-log(3.0x10-2) pH=1.52
If either the [H3O+] or [OH-] is known, the pH and pOH can be calculated. Example 18-4: Calculate the pH of a solution in which the [H3O+] =0.030 M. pH=-log[H3O+] pH=-log(3.0x10-2) pH=1.52

The pH and pOH scales pH=-log[H3O+] 4.597=-log[H3O+] [H3O+]=10-4.597
Example 18-5: The pH of a solution is What is the concentration of H3O+? pH=-log[H3O+] 4.597=-log[H3O+] [H3O+]= [H3O+]=2.53x10-5M

pH=-log[H3O+] pH=-log(5.0x10-2) pH=1.3 pH=-log[H3O+] 3.301=-log[H3O+]
Example Calculation of pH Calculate the pH of a solution in which the H3O+ concentration is 0.050mol/L. pH=-log[H3O+] pH=-log(5.0x10-2) pH=1.3 Exercise 22 Example Calculation of H3O+ concentration from pH The pH of a solution is What is the concentration of H3O+ in this solution? pH=-log[H3O+] 3.301=-log[H3O+] [H3O+]= [H3O+]=5.0x10-4M Exercise 24

The pH and pOH scales [H3O+][OH-]= 1.0x10-14 log[H3O+]+log[OH-]= -14.0
A convenient relationship between pH and pOH may be derived for all dilute aqueous solutions at 250C. [H3O+][OH-]= 1.0x10-14 Taking the logarithm of both sides of this equation gives: log[H3O+]+log[OH-]= -14.0 Multiplying both sides of this equation by -1 gives: -log[H3O+]+(-log[OH-])= 14.0 Which can be rearranged to this form: pH + pOH = 14.0

The pH and pOH scales [H3O+][OH-]= 1.0x10-14 pH + pOH = 14.0
Remember these two expressions!! They are key to the next three chapters! [H3O+][OH-]= 1.0x10-14 pH + pOH = 14.0

The pH and pOH scales [H3O+]=1.0 M to [H3O+]= 1.0x10-14M
The usual range for the pH scale is 0 to 14. [H3O+]=1.0 M to [H3O+]= 1.0x10-14M pH= to pH=14.0 And for pOH the scale is also 0 to 14 but inverted from pH. pH = 0 has a pOH = 14 and pH = 14 has a pOH = 0. [OH-]= 1.0x up to [OH-]=1.0M pOH= to pOH=0

The pH and pOH scales HNO3 + H2O H3O+(aq) + NO3- (aq) Strong acid
Example 18-6: Calculate the [H3O+], pH, [OH-], and pOH for a M HNO3 solution. Is HNO3 a weak or strong acid? What is the [H3O+] ? Strong acid HNO3 + H2O H3O+(aq) + NO3- (aq) initial 0.02M change -0.02M +0.02M +0.02M At equil 0.0M 0.02M +0.02M [H3O+]= 2x10-2 M pH=-log(2x10-2 M) pH=1.70 Kw=[H3O+][OH-]=1.0x10-14 [OH-]=1.0x10-14/[H3O+] =1.0x10-14/2.0x10-2 =5.0x10-13 M pOH=-log[OH-] =-log(5.0x10-13) =12.30

The pH and pOH scales To help develop familiarity with the pH and pOH scale we can look at a series of solutions in which [H3O+] varies between 1.0 M and 1.0 x M. [H3O+] [OH-] pH pOH 1.0 M 1.0 x M 0.00 14.00 1.0 x 10-3 M 1.0 x M 3.00 11.00 1.0 x 10-7 M 7.00 2.0 x M 5.0 x 10-3 M 11.70 2.30

Ca(OH)2 Ca2+ + 2OH-(aq) initial 0.015M change -0.015M +0.015M
Example Calculations Involving pH and pOH Calculate the [H3O+], pH, [OH-], and pOH for a M Ca(OH)2 solution. Ca(OH) Ca2+ + 2OH-(aq) initial 0.015M change -0.015M +0.015M +2x(0.015)M At equil 0.0M 0.015M +0.03M [OH-]= 3x10-2 M pOH=-log(3x10-2 M) pOH=1.52 pH=14-pOH = =12.48 Kw=[H3O+][OH-]=1.0x10-14 [H3O+]=1.0x10-14/[OH-] =1.0x10-14/3.0x10-2 =3.3x10-13 M Exercise 26 and 37

胃酸炭酸飲料檸檬醋蕃茄啤酒尿液牛奶唾液血漿蛋白鎂乳(瀉藥) 氨水

pH meter Universal indicator Universal pH paper

Ionization Constants for Weak Monoprotic Acids單質子酸 and Bases
Strong acids ionize completely in dilute aqueous, whereas weak acids ionize only slightly Let’s look at the dissolution of acetic acid, a weak acid, in water as an example. The equation for the ionization of acetic acid is: CH3COOH + H2O  H3O+ + CH3COO- The equilibrium constant for this ionization is expressed as: [H3O+] [CH3COO-] Kc= [CH3COOH] [H2O]

Ionization Constants for Weak Monoprotic Acids and Bases
The water concentration in dilute aqueous solutions is very high. 1 L of water contains 55.5 moles of water. Thus in dilute aqueous solutions: [H2O]  55.5M

The water concentration is many orders of magnitude greater than the ion concentrations. Thus the water concentration is essentially that of pure water. Recall that the activity of pure water is 1. Kc= [H3O+] [CH3COO-] [CH3COOH] [H2O] [H3O+] [CH3COO-] [CH3COOH] Kc [H2O] = [H3O+] [CH3COO-] [CH3COOH] K 

We can define a new equilibrium constant for weak acid equilibria that uses the previous definition. This equilibrium constant is called the acid ionization constant 酸解離常數. The symbol for the ionization constant is Ka. [H3O+] [CH3COO-] [CH3COOH] Ka = =1.8x10-5 for acetic acid

In simplified form the dissociation equation and acid ionization expression are written as: CH3COOH  H+ + CH3COO- Ka = [H+] [CH3COO-] [CH3COOH] =1.8x10-5

Ionization Constants for Weak Monoprotic Acids單質子酸 and Bases
Carbonic acid H2CO3 Citric acid C3H5O(COOH)3

From the above table we see that the order of increasing acid strength for these weak acids is: HF > HNO2 > CH3COOH > HClO > HCN The order of increasing base strength of the anions (conjugate bases) of these acids is: F- < NO2- < CH3COO- < ClO- < CN-

Example 18-8: Write the equation for the ionization of the weak acid HCN and the expression for its ionization constant. HCN  H+ + CN- [H+] [CN-] [HCN] Ka = =4.0x10-10

HY  H+ + Y- Ka = Ka = = Ka =3.3x10-4 [H+] [Y-] [HY] [H+] [Y-]
Example 18-9: In a 0.12 M solution of a weak monoprotic acid, HY, the acid is 5.0% ionized. Calculate the ionization constant for the weak acid. HY  H+ + Y- [H+] [Y-] [HY] Ka = Since the weak acid is 5.0% ionized, it is also 95% unionized. Calculate the concentration of all species in solution. [H+]=[Y-]=0.05 x (0.12M) =0.006M =6x10-3M [HY]=0.95 x (0.12M) =0.11M [H+] [Y-] [HY] (6.0x10-3)(6.0x10-3) Ka = = (0.11) Ka =3.3x10-4

HA  H+ + A- Ka = = Ka =1.2x10-5 [H+] [A-] (1.1x10-3)(1.1x10-3) 0.1
Example 18-10: The pH of a 0.10 M solution of a weak monoprotic acid, HA, is found to be What is the value for its ionization constant? pH = 2.97 so [H+]= 10-pH [H+]= [H+]=1.1x 10-3 Use the [H3O+] and the ionization reaction to determine concentrations of all species. HA  H A- At equil ( x10-3)M 1.1x10-3M 1.1x10-3M  0.1M Calculate the ionization constant from this information. [H+] [A-] [HA] (1.1x10-3)(1.1x10-3) 0.1 Ka = = Ka =1.2x10-5

Ka = = Ka =1.4x10-5 pKa =-log(1.4x10-5) =4.85 [H3O+] [A-]
Example Calculations of Ka and pKa from equilibrium Concentrations Nicotinic acid is a weak monoprotic oganic acid that we can represent as HA. A dilute solution of nicotinic acid was found to contain the following concentrations at equilibrium at 25oC. What are the Ka and pKa values? [HA]=0.049M; [H3O+]=[A-]=8.4x10-4M. HA + H2O  H3O A- [H3O+] [A-] [HA] (8.4x10-4)(8.4x10-4) 0.049 Ka = = Ka =1.4x10-5 pKa =-log(1.4x10-5) =4.85 Exercise 30

CH3COOH  H+ + CH3COO- [H+]=[CH3COO-]=0.042 x (0.01M) =4.2x10-4M
Example Calculations of Ka from Percent Ionization In 0.01M solution, acetic acid is 4.2% ionized. Calculate its ionization constant. CH3COOH  H+ + CH3COO- Since the weak acid is 4.2% ionized, it is also 95.8% unionized. Calculate the concentration of all species in solution. [H+]=[CH3COO-]=0.042 x (0.01M) =4.2x10-4M [CH3COOH]=0.958 x (0.01M) =9.58x10-3M [H3O+] [CH3COO-] [CH3COOH] Ka= (4.2x10-4)(4.2x10-4) = (9.58x10-3) = 1.8x10-5 Exercise 38

pH=-log[H3O+] [H3O+]=10-pH=10-1.92=0.012M HA + H2O  H3O+ + A- initial
Example Calculations of Ka from pH The pH of a 0.115M solution of chloroacetic acid, ClCH2COOH, is measure to be Calculate Ka from this weak momoprotic acid. pH=-log[H3O+] [H3O+]=10-pH= =0.012M HA + H2O  H3O A- initial 0.115M change -0.012M +0.012M +0.012M At equil 0.103M 0.012M 0.012M Ka= [H3O+] [A-] [HA] (0.012)(0.012) = (0.103) = 1.4x10-3 Exercise 40

x is small enough to ignore compare to 0.15M Ka= =1.8x10-5 (0.15-x)
Example 18-11: Calculate the concentrations of the various species in 0.15 M acetic acid, CH3COOH, solution. CH3COOH  H+ + CH3COO- initial 0.15M change -xM +xM +xM At equil (0.15-x)M xM xM Ka= [H3O+] [CH3COO-] [CH3COOH] =1.8x10-5 (x)(x) x is small enough to ignore compare to 0.15M Ka= =1.8x10-5 (0.15-x) x2(0.15)x(1.8x10-5) x22.7x10-6 x=1.6x10-3 M=[H3O+]=[CH3COO-] [CH3COOH]= x10-3  0.15M

Let us now calculate the percent ionization for the 0.15 M acetic acid. From Example 18-11, we know the concentration of CH3COOH that ionizes in this solution. The percent ionization of acetic acid is [CH3COOH]ionized [CH3COOH]original % ionization= x100% (1.6x10-3M) % ionization= x100% 0.15M =1.1%

HCN + H2O  H3O+ + CN- initial 0.15M change -xM +xM +xM At equil
Example 18-12: Calculate the concentrations of the species in 0.15 M hydrocyanic acid, HCN, solution. Ka= 4.0 x for HCN HCN + H2O  H3O+ + CN- initial 0.15M change -xM +xM +xM At equil (0.15-x)M xM xM Ka= [H3O+] [CN-] [HCN] (x)(x) =4.0x10-10 = (0.15-x) x2(0.15)x(4.0x10-10) x=7.7x10-6 M=[H3O+]=[CN-] [HCN]=( x10-6)  0.15M x is small enough to ignore compare to 0.15M % ionization= [HCN]ionized [HCN]original x100% =(7.7x10-6/0.15)x100% =0.0051%

Let’s look at the percent ionization of two weak acids as a function of their ionization constants. Examples and will suffice. Solution Ka [H+] pH % ionization 0.15 M acetic acid 1.8 x 10-5 1.6 x 10-3 2.80 1.1 0.15 M HCN 4.0 x 10-10 7.7 x 10-6 5.11 0.0051 Note that the [H+] in 0.15 M acetic acid is 210 times greater than for 0.15 M HCN.

NH3 + H2O  NH4+ + OH- initial 0.15M change -xM +xM +xM At equil
All of the calculations and understanding we have at present can be applied to weak acids and weak bases! Example 18-13: Calculate the concentrations of the various species in 0.15 M aqueous ammonia. NH3 + H2O  NH4+ + OH- initial 0.15M change -xM +xM +xM At equil (0.15-x)M xM xM Kb= [NH4+] [OH-] [NH3] (x)(x) =1.8x10-5 = (0.15-x) x2(0.15)x(1.8x10-5) x=1.6x10-3 M=[NH4+]=[OH-] [NH3]=( x10-3)  0.15M x is small enough to ignore compare to 0.15M % ionization= [NH3]ionized [NH3]original x100% =(1.6x10-3/0.15)x100% =1.1%

pH=11.37 pOH+pH=14 pOH=14-11.37=2.63 [OH-]=10-pOH=10-2.63=2.3x10-3M
Example 18-14: The pH of an aqueous ammonia solution is Calculate the molarity (original concentration) of the aqueous ammonia solution. pH=11.37 pOH+pH=14 pOH= =2.63 [OH-]=10-pOH= =2.3x10-3M NH3 + H2O  NH4+ + OH- initial xM change 2.3x10-3M 2.3x10-3M 2.3x10-3M At equil (x-2.3x10-3)M 2.3x10-3M 2.3x10-3M Kb= [NH4+] [OH-] [NH3] (2.3x10-3)x(2.3x10-3) =1.8x10-5 = (x-2.3x10-3) x(2.3x10-3)(2.3x10-3)/(1.8x10-5) x=0.3 M=[NH3]

HOCl + H2O  H3O+ + OCl- initial 0.10M change -xM +xM +xM At equil
Example Calculation of Concentrations from Ka (a) Calculate the concentration of the various species in 0.10M hypochlorous acid, HOCl.For HOCl, Ka=3.5x10-8. (b) What is the pH of this solution? HOCl + H2O  H3O+ + OCl- initial 0.10M change -xM +xM +xM At equil (0.10-x)M xM xM Ka= [H3O+] [OCl-] [HOCl] (x)(x) = = 3.5x10-8 (0.10-x) x2(0.1)(3.5x10-8) x5.9x10-5=[H3O+]=[OCl-] [HOCl]=( x10-5)=0.10M pH=-log(5.9x10-5) = 4.23 Exercise 42

x=1.3x10-3 M=[H3O+]=[CH3COO-] [CH3COOH]=0.10-1.3x10-3  0.10M
Example Percent Ionization Calculate the percent ionization of a 0.10 M solution of acetic acid. CH3COOH + H2O  H+ + CH3COO- initial 0.10M change -xM +xM +xM At equil (0.10-x)M xM xM Ka= [H3O+] [CH3COO-] [CH3COOH] =1.8x10-5 (x)(x) = (0.10-x) x2(0.1M)x(1.8x10-5) x=1.3x10-3 M=[H3O+]=[CH3COO-] [CH3COOH]= x10-3  0.10M x is small enough to ignore compare to 0.10M [CH3COOH]ionized [CH3COOH]original % ionization= x100% (1.3x10-3M) = x100% =1.3% 0.10M Exercise 48

Increasing acid strength HCN<HOCl<HNO2 (b) Increasing pKa values
Example pKa Value The Ka values for acetic acid and hydrofluoric acid are 1.8x10-5 and 7.2x10-4, respectively. What are their pKa value? For CH3COOH pKa=-logKa =-log(1.8x10-5) =4.74 For HF pKa=-logKa =-log(7.2x10-4) =3.14 Exercise 50 Example Acid Strength and Ka Value Given the following lost of weal acids and their Ka values, arrange the acids in order of (a) increasing acid strength and (b) increasing pKa values. Acid Ka HOCl x10-8 HCN x10-10 HNO x10-4 Increasing acid strength HCN<HOCl<HNO2 (b) Increasing pKa values HNO2<HOCl<HCN Exercise 109

Polyprotic Acids Many weak acids contain two or more acidic hydrogens.
Examples include H3PO4 and H3AsO4. The calculation of equilibria for polyprotic acids is done in a stepwise fashion. There is an ionization constant for each step. Consider arsenic acid, H3AsO4, which has three ionization constants. 1. Ka1 = 2.5 x 10-4 2. Ka2 = 5.6 x 10-8 3. Ka3 = 3.0 x 10-13

Polyprotic Acids H3AsO4  H+ + H2AsO4- H2AsO4-  H+ + HAsO42-
The first ionization step for arsenic acid is: H3AsO4  H+ + H2AsO4- Ka1= [H+] [H2AsO4-] [H3AsO4] =2.5x10-4 The second ionization step for arsenic acid is: H2AsO4-  H+ + HAsO42- Ka2= [H+] [HAsO42-] [H2AsO4-] =5.6x10-8

Polyprotic Acids HAsO42-  H+ + AsO43- [H+] [AsO43-] Ka3= =3.0x10-13
The third ionization step for arsenic acid is: HAsO42-  H+ + AsO43- Ka3= [H+] [AsO43-] [H2AsO42-] =3.0x10-13 Notice that the ionization constants vary in the following fashion: Ka1>Ka2>Ka3 This is a general relationship. For weak polyprotic acids the Ka1 is always > Ka2, etc.

Polyprotic Acids H3AsO4  H+ + H2AsO4- Ka1= [H+] [H2AsO4-] [H3AsO4] =
Example 18-15: Calculate the concentration of all species in M arsenic acid, H3AsO4, solution. H3AsO4  H+ + H2AsO4- At equil (0.10-x)M xM xM Ka1= [H+] [H2AsO4-] [H3AsO4] = (x)(x) (0.10-x) =2.5x10-4 X2+2.5x10-4X-2.5x10-5=0 -2.5x10-4  (2.5x10-4)2-4(1)(-2.5x10-5) x= 2x1 x=-5.1x10-3 and x=4.9x10-3 [H+]=[H2AsO4-]=4.9x10-3M [H3AsO4]=( x10-3) M =0.095M

Polyprotic Acids H2AsO4-  H+ + HAsO4-2 [H+] [HAsO4-2] Ka2= =
Next, write the equation for the second step ionization and represent the concentrations. H2AsO4-  H+ + HAsO4-2 [] from 1st step (4.9x10-3)M (4.9x10-3)M At equil (4.9x10-3-y)M yM yM Ka2= [H+] [HAsO4-2] [H2AsO4-] = (4.9x10-3-y) (4.9x10-3+y)(y) =5.6x10-8 y<<4.9x10-3 Thus (4.9x10-3 –y)  4.9x10-3 Ka2= (4.9x10-3) (4.9x10-3)(y) =5.6x10-8 y=5.6x10-8=[H+]2nd=[HAsO42-] Note the [H+]1st >>[H+]2nd

Polyprotic Acids HAsO42-  H+ + AsO43- [H+] [AsO4-3] Ka3= = [HAsO4-]
Finally, repeat the entire procedure for the third ionization step. HAsO42-  H AsO43- [] from 1st and 2nd step (5.6x10-8)M (4.9x x10-8)M At equil (5.6x10-8-z)M zM zM Ka3= [H+] [AsO4-3] [HAsO4-] = (5.6x10-8-z) (4.9x x10-8+z)(z) =3.0x10-13 z<<5.6x10-8 Thus (5.6x10-8 –z)  5.6x10-8 Ka3= (5.6x10-8) (4.9x10-3)(z) =3.0x10-13 z=3.4x10-18=[H+]3rd=[AsO43-] Use Kw to calculate the [OH-] in the M H3AsO4 solution. [H+][OH-] =1.0x10-14 (4.9x10-3)x[OH-]=1.0x10-14 [OH-]=2.0x10-12

Polyprotic Acids A comparison of the various species in M H3AsO4 solution follows. Species Concentration H3AsO4 0.095 M H+ M H2AsO4- HAsO42- 5.6 x 10-8 M AsO43- 3.4 x M OH- 2.0 x M

H2SO4 + H2O  H3O+ + HSO4- 100% ionization 0.10M 0.10M 0.10M
Example Solutions Strong Polyprotic Acid Calculate concentration of all species present in 0.10M H2SO4. Ka2=1.2x10-2. H2SO4 + H2O  H3O+ + HSO4- 100% ionization 0.10M 0.10M 0.10M HSO4- + H2O  H3O+ + SO4-2 0.10M 0.10M -xM +xM +xM (0.10-x)M (0.10+x)M xM Ka2= [H3O+] [SO4-2] [HSO4-] = (0.1+x)(x) (0.1-x) = 1.2x10-2 x2+0.1x= 1.2x10-3 – (1.2x10-2)x x=0.01 or x=-0.12 [H3O+]2nd=[SO42-]=0.01M [H2SO42-]  0M [HSO4-]= M=0.09 M [H3O+]= M=0.11 M [OH-]=1.0x10-14/0.11=9.1x10-14 M Exercise 38

Solvolysis溶劑分解[作用] This reaction process is the most difficult concept in this chapter. Solvolysis is the reaction of a substance with the solvent in which it is dissolved. Hydrolysis refers to the reaction of a substance with water or its ions. Combination of the anion of a weak acid with H3O+ ions from water to form nonionized weak acid molecules.

Solvolysis A- + H3O+  HA + H2O Recall H2O + H2O  H3O+ + OH-
Hydrolysis refers to the reaction of a substance with water or its ions. Hydrolysis is solvolysis in aqueous solutions. The combination of a weak acid’s anion with H3O+ ions, from water, to form nonionized weak acid molecules is a form of hydrolysis. A- + H3O+  HA + H2O Recall H2O + H2O  H3O+ + OH- The reaction of the anion of a weak monoprotic acid with water is commonly represented as: A- + H2O  HA + OH- The removal of H3O+ upsets the water equilibrium

Solvolysis Recall that at 25oC in neutral solutions:
[H3O+] = 1.0 x 10-7 M = [OH-] in basic solutions: [H3O+] < 1.0 x 10-7 M and [OH-] > 1.0 x 10-7 M in acidic solutions: [OH-] < 1.0 x 10-7 M and [H3O+] > 1.0 x 10-7 M

Solvolysis Remember from BrØnsted-Lowry acid-base theory:
The conjugate base of a strong acid is a very weak base. The conjugate base of a weak acid is a stronger base.

Solvolysis HCl +H2O H3O++Cl-
Hydrochloric acid, a typical strong acid, is essentially completely ionized in dilute aqueous solutions. HCl +H2O H3O++Cl- ~100% The conjugate base of HCl, the Cl- ion, is a very weak base. The chloride ion is such a weak base that it will not react with the hydronium ion. Cl- + H3O+ → No rxn. In dilute aqueous solutions This fact is true for all strong acids and their anions.

Solvolysis HF +H2O H3O++F- H3O++F- HF +H2O
HF, a weak acid, is only slightly ionized in dilute aqueous solutions. Its conjugate base, the F- ion, is a much stronger base than the Cl- ion. The F- ions combine with H3O+ ions to form nonionized HF. Two competing equilibria are established. HF +H2O H3O++F- Only slightly H3O++F HF +H2O Nearly completely

Solvolysis Dilute aqueous solutions of salts that contain no free acid or base come in four types: 1. Salts of Strong Bases and Strong Acids 2. Salts of Strong Bases and Weak Acids 3. Salts of Weak Bases and Strong Acids 4. Salts of Weak Bases and Weak Acids

Salts of Strong Bases and Weak Acids
Salts made from strong acids and strong soluble bases form neutral aqueous solutions. An example is potassium nitrate, KNO3, made from nitric acid and potassium hydroxide. KNO3(s) K++NO3- ~100% in H2O H2O +H2O H3O++OH- The ions that are in solution The KOH and HNO3 are present in equal amounts There is no reaction to upset [H3O+][OH-] Thus the solution is neutral HNO3 KOH

Salts made from strong soluble bases and weak acids hydrolyze to form basic solutions. Anions of weak acids (strong conjugate bases) react with water to form hydroxide ions. An example is sodium hypochlorite, NaClO, made from sodium hydroxide and hypochlorous acid. NaClO(s) Na++ClO- ~100% in H2O H2O +H2O H3O++OH- The ions that are in solution Which is the stronger acid or base? HClO NaOH

NaClO(s) Na++ClO- ~100% in H2O H2O +H2O H3O++OH- ClO- + H3O HClO + H2O We can combine these last two equations into one single equation that represents the total reaction. ClO- + H2O HClO + OH- The equilibrium constant for this reaction, called the hydrolysis constant, is written as: Kb= [HClO] [OH-] [ClO-]

Algebraic manipulation of the previous expression give us a very useful form of the expression. Multiply the expression by one written as [H+]/ [H+]. H+/H+ = 1 Kb= [HClO][OH-] [ClO-] x [H+] Kb= [HClO] [ClO-][H+] x [H+][OH-] 1 Kb= 1 Ka for HClO = Kw Ka for HClO = 1x10-14 3.5x10-8 x Kw Kb= [HClO] [OH-] [ClO-] = 2.9x10-7

This same method can be applied to the anion of any weak monoprotic acid. A- + H2O HA + OH- Kb= [HA] [OH-] [A-] = Kw Ka for HA Kw=KaKb

Example 18-16: Calculate the hydrolysis constants for the following anions of weak acids. The fluoride ion, F-, the anion of hydrofluoric acid, HF. For HF, Ka=7.2 x 10-4. F- + H2O HF + OH- Kb= [HF] [OH-] [F-] = Kw Ka for HF Kb= 1.0x10-14 7.2x10-4 =1.4x10-11

The cyanide ion, CN-, the anion of hydrocyanic acid, HCN. For HCN, Ka = 4.0 x CN- + H2O HCN + OH- Kb= [HCN] [OH-] [CN-] = Kw Ka for HCN Kb= 1.0x10-14 4.0x10-10 =2.5x10-5

NaClO(s) Na++ ClO- ClO- + H2O HClO + OH- Kb= [HClO] [OH-] [ClO-]
Example 18-17: Calculate [OH-], pH and percent hydrolysis for the hypochlorite ion in 0.10 M sodium hypochlorite, NaClO, solution. “Clorox”, “Purex”, etc., are 5% sodium hypochlorite solutions. NaClO(s) Na++ ClO- ~100% in H2O 0.10M 0.10M 0.10M ClO- + H2O HClO + OH- initial 0.10M change -xM +xM +xM At equil (0.10-x)M xM xM Kb= [HClO] [OH-] [ClO-] (x)(x) =2.9x10-7= (0.1-x) x=1.7x10-4M=[ClO-]=[OH-] pOH=-log(1.7x10-4)=3.77 pH= =10.23 x<<0.1 so (0.1-x)0.1 [ClO-]hydrolyzed % hydrolysis= x100% [ClO-]original 1.7x10-4 % hydrolysis= x100% =0.17% 0.10M

If a similar calculation is performed for 0.10 M NaF solution and the results from 0.10 M sodium fluoride and 0.10 M sodium hypochlorite compared, the following table can be constructed. Solution Ka Kb [OH-] (M) pH % hydrolysis NaF 7.2 x 10-4 1.4 x 10-11 1.2 x 10-6 8.08 0.0012 NaClO 3.5 x 10-8 2.9 x 10-7 1.7 x 10-4 10.23 0.17

Salts of Weak Bases and Strong Acids
Salts made from weak bases and strong acids form acidic aqueous solutions. An example is ammonium bromide, NH4Br, made from ammonia and hydrobromic acid. NH4Br (s) NH4++ Br- ~100% in H2O H2O +H2O OH- + H3O+ The ions that are in solution Which is the stronger acid or base? NH3 HBr

The relatively strong acid, NH4+, reacts with the OH- ion removing it from solution leaving excess H3O+ NH4+ + OH NH3 + H2O generates excess H3O+ The reaction may be more simply represented as: NH4+ + H2O NH3 + H3O+ NH NH3 + H+ more simply as: The hydrolysis constant expression for this process is: Ka= [NH3][H3O+] [NH4+] [NH3][H+] [NH4+] or Ka=

Multiplication of the hydrolysis constant expression by [OH-]/ [OH-] gives: Ka= [NH3][H3O+] [NH4-] x [OH-] Ka= [NH3] [NH4+][OH-] x [H3O+][OH-] 1 Ka= 1 Kb for NH3 = Kw Kb for NH3 = 1x10-14 1.8x10-5 x Kw Ka= [NH3] [H3O+] [NH4-] = 5.6x10-10

In its simplest form for this hydrolysis: NH NH3 + H+ Ka= [NH3] [H+] [NH4-] = 5.6x10-10

NH4+ NH3 + H+ Ka= [NH3] [H+] [NH4+] = (x)(x) (0.1-x) = 5.6x10-10
Example 18-18: Calculate [H+], pH, and percent hydrolysis for the ammonium ion in 0.10 M ammonium bromide, NH4Br, solution. NH NH3 + H+ initial 0.10M change -xM +xM +xM At equil (0.10-x)M xM xM Ka= [NH3] [H+] [NH4+] = (x)(x) (0.1-x) = 5.6x10-10 x2=5.6x10-11M X=7.5x10-6M=[NH3]=[H+] pH=-log(7.5x10-6)=5.12 x<<0.1 so (0.1-x)0.1 [NH4+]hydrolyzed % hydrolysis= x100% [NH4+]original 7.5x10-6 % hydrolysis= x100% =0.0075% 0.10M

Salts of Weak Bases and Weak Acids
Salts made from weak acids and weak bases can form neutral, acidic or basic aqueous solutions. The pH of the solution depends on the relative values of the ionization constant of the weak acids and bases. Salts of weak bases and weak acids for which parent Kbase =Kacid make neutral solutions. An example is ammonium acetate, NH4CH3COO, made from aqueous ammonia, NH3,and acetic acid, CH3COOH. Ka for acetic acid = Kb for ammonia = 1.8 x 10-5.

The ammonium ion hydrolyzes to produce H+ ions. Its hydrolysis constant is: NH NH3 + H+ Ka= [NH3] [H+] [NH4-] = 5.6x10-10 The acetate ion hydrolyzes to produce OH- ions. Its hydrolysis constant is: CH3COO- + H2O  CH3COOH + OH- Ka= [OH-] [CH3COOH] [CH3COO-] =5.6x10-10

Because the hydrolysis constants for both ions are equal, their aqueous solutions are neutral. Equal numbers of H+ and OH- ions are produced. NH4+CH3COO NH4++CH3COO- ~100% in H2O H2O +H2O OH- + H3O+ The ions that are in solution A weak acid and base are formed in solution NH4OH CH3COOH

Salts of weak bases and weak acids for which parent Kbase > Kacid make basic solutions. An example is ammonium hypochlorite, NH4ClO, made from aqueous ammonia, NH3,and hypochlorous acid, HClO. Kb for NH3 = 1.8 x 10-5 > Ka for HClO = 3.5x10-8

Salts of weak bases and weak acids for which parent Kbase < Kacid make acidic solutions. An example is trimethylammonium fluoride,(CH3)3NHF, made from trimethylamine, (CH3)3N,and hydrofluoric acid acid, HF. Kb for (CH3)3N = 7.4 x 10-5 < Ka for HF = 7.2 x 10-4

Summary of the major points of hydrolysis up to now. The reactions of anions of weak monoprotic acids (from a salt) with water to form free molecular acids and OH-. A- + H2O HA + OH- Kb= [HA] [OH-] [A-] = Kw Ka for HA

The reactions of anions of weak monoprotic acids (from a salt) with water to form free molecular acids and OH-. BH+ + H2O B + H3O+ Ka= Kw Kb (waek base)

Aqueous solutions of salts of strong acids and strong bases are neutral. Aqueous solutions of salts of strong bases and weak acids are basic. Aqueous solutions of salts of weak bases and strong acids are acidic. Aqueous solutions of salts of weak bases and weak acids can be neutral, basic or acidic. The values of Ka and Kb determine the pH.

Ionic Equilibria I 離子平衡: Acids and Bases

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Ionic Equilibria I 離子平衡: Acids and Bases

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