Chemistry 102(060) Summer 2015 Instructor: Dr. Upali Siriwardane e-mail: upali@latech.edu Office: CTH 311 Phone 257-4941 Office Hours: M,Tu, W,Th,F 9:00-11:00 am or by appointment..; Test Dates: July 20, 2015 (Test 1): Chapter 13 July 27, 2015 (Test 2): Chapter 14 August 4, 2015 (Test 3): Chapter 15 August 12, 2015 (Test 4): Chapter 17 August 13, 2015 (Make-up test) comprehensive: Chapters 13-17

Chapter 15. Acids and Bases 15.1 Heartburn 659 15.2 The Nature of Acids and Bases 660 15.3 Definitions of Acids and Bases 662 15.4 Acid Strength and the Acid Dissociation Constant (Ka) 665 15.5 Autoionization of Water and pH 668 15.6 Finding the and pH of Strong and Weak Acid Solutions 673 15.7 Base Solutions 682 15.8 The Acid–Base Properties of Ions and Salts 685 15.9 Polyprotic Acids 693 15.1 0 Acid Strength and Molecular Structure 698 15.1 1 Lewis Acids and Bases 700 15.1 2 Acid Rain 701

Types of Reactions a) Precipitation Reactions. Reactions of ionic compounds or salts b) Acid/base Reactions. Reactions of acids and bases c) Redox Reactions. reactions of oxidizing & reducing agents

What are Acids &Bases? Definition? a) Arrhenius b) Bronsted-Lowry c) Lewis

Arrhenius Definitions Arrhenius, Svante August (1859-1927), Swedish chemist, 1903 Nobel Prize in chemistry Acid Anything that produces hydrogen ions in a water solution. HCl (aq) H+ ( aq) + Cl- ( aq) Base Anything that producs hydroxide ions in a water solution. NaOH (aq) Na+ ( aq) + OH- ( aq) Arrhenius definitions are limited proton acids and hydroxide bases to aqueous solutions.

Brønsted-Lowry definitions Expands the Arrhenius definitions to include many bases other than hydroxides and gas phase reactions Acid Proton donor Base Proton acceptor This definition explains how substances like ammonia can act as bases. Eg. HCl(g) + NH3(g) ------> NH4Cl(s) HCl (acid), NH3 (base). NH3(g) + H2O(l) NH4+ + OH-

Lewis Definition G.N. Lewis was successful in including acid and bases without proton or hydroxyl ions. Lewis Acid: A substance that accepts an electron pair. Lewis base: A substance that donates an electron pair. E.g. BF3(g) + :NH3(g) F3B:NH3(s) the base donates a pair of electrons to the acid forming a coordinate covalent bond common to coordination compounds. Lewis acids/bases will be discussed later in detail

1) Acids and bases can be defined in several ways 1) Acids and bases can be defined in several ways. Which definitions of the bases that fits the description below? a) a compound that produces more OH- ions in water: b) a proton acceptor: c) an electron pair donor:

Types of Acids and Bases Binary acids Oxyacid Organic acids Acidic oxides Basic oxides Amine Polyprotic acids

Types of Acids and Bases Binary acids: HCl, HBr, HI, H2S More than two elements: HCN Oxyacid: HNO3, H2SO4, H3PO4 Polyprotic acids: H2SO4, H3PO4 Organic acids: R-COOH, R= CH3-, CH3CH2- Acidic oxides: SO3, NO2, CO2, Basic oxides: Na2O, CaO Amine: NH3. R-NH2, R= CH3-, CH3CH2- : primary R2-NH : secondary, R3-N: tertiary Lewis acids & bases: BF3 and NH3

These are usually oxides of non-metallic elements such as P, S and N. Acidic Oxides These are usually oxides of non-metallic elements such as P, S and N. E.g. NO2, SO2, SO3, CO2 They produce oxyacids when dissolved in water SO3 + H2O ---> H2SO4 CO2 + H2O ---> H2CO3 NO2 + H2O ---> HNO3

Basic Oxides Oxides oxides of metallic elements such as Na, K, Ca. They produce hydroxyl bases when dissolved in water. e.g. CaO + H2O ---> Ca(OH)2 Na2O + H2O ---> 2 NaOH

Protic Acids Monoprotic Acids: The form protic refers to acidity due to protons. Monoprotic acids have only one acidic proton. e.g. HCl. Polyprotic Acids: They have more than one acidic proton. e.g. H2SO4 - diprotic acid H3PO4 - triprotic acid.

Polyprotic Acids acids where more than one hydrogen per molecule is released

Amines Class of organic bases derived from ammonia NH3 by replacing hydrogen by organic groups. They are defined as bases similar to NH3 by Bronsted-Lowery or Lewis acid/base definitions.

Amines

2) Identify types of acids/bases as: binary acids, oxy acids, organic acids, acidic oxides, basic oxides, amine and polyprotic acids.   a) HF b) HBr c) H3PO4 d) H2SO4 e) HNO3 f) R-COOH g) NO2 h) SO3 j)CaO k) R-NH2

Influence of Molecular Structure on Acid Strength Binary Hydrides hydrogen & one other element Bond Strengths weaker the bond, the stronger the acid Stability of Anion higher the electronegativity, stronger the acid

Binary Acids Compounds containing acidic protons bonded to a more electronegative atom. e.g. HF, HCl, HBr, HI, H2S The acidity of the haloacid (HX; X = Cl, Br, I, F) Series increase in the following order: HF < HCl < HBr < HI

Oxyacids Compounds containing acidic - OH groups in the molecule. Acidity of H2SO4 is greater than H2SO3 because of the extra O (oxygens) The order of acidity of oxyacids from the a halogen (Cl, Br, or I) shows a similar trend. HClO4 > HClO3 > HClO2 > HClO perchloric chloric chlorus hyphochlorus

Influence of Molecular Structure on Acid Strength Oxyacids hydrogen, oxygen, & one other element H-O-E higher the electronegativity on E, stronger the acid as this weakens the bond between the O and H

Oxo Acid < < < <

3) Which of the following is stronger acid or base: H2SO4 or H2SO3: HCl or HI: HClO or HClO3: H2S or HF: CF3COOH or CH3COOH: CH3COOH or CH3CH2COOH

3) Which of the following is stronger acid or base: H2SO4 or H2SO3: Ka H2SO4>> 1 ; H2SO3= 1.4 x 10-2 HCl or HI: HClO or HClO3: H2S or HF: H2S: Ka:H2S=6.3 x 10-8 ; HF= 6.3 x 10-4 CF3COOH or CH3COOH: 2.5 x 10-3 and 1.8 x 10-5 CH3COOH or CH3CH2COOH: 1.8 x 10-5 and 1.2 x 10-5

Dissociation Strong Acids: HCl(aq) + H2O(l) H3+O(aq) + Cl-(aq) H2SO4(aq) + H2O(l) H3+O(aq) + HSO4-(aq) Dissociation Equilibrium Weak Acid/base: H2O(l) + H2O(l) H3+O(aq) + OH-(aq) This dissociation is called autoionization of water. HC2H3O2(aq) + H2O(l) H3+O(aq) + C2H3O2-(aq) NH3 (aq) + H2O(l) NH4+ + OH-(aq) Equilibrium constants: Ka, Kb and Kw

4) Write equations for the dissociation equilibrium reactions for the following acids and bases in water. Which of these are acid or dissociations? a) HCl: b) H2SO4 :

4) Write equations for the dissociation equilibrium reactions for the following acids and bases in water. Which of these are acid or dissociations? c) H2O (autoionization): What is auto ionization? d) HC2H3O(acetic acid): e) NH3:

Bronsted acid/conjugate base and base/conjugate acid pairs in acid/base equilibria HCl(aq) + H2O(l) H3+O(aq) + Cl-(aq) HCl(aq): acid H2O(l): base H3+O(aq): conjugate acid Cl-(aq): conjugate base H2O/ H3+O: base/conjugate acid pair HCl/Cl-: acid/conjugate base pair

Brønsted-Lowry Definitions Conjugate acid-base pairs. Acids and bases that are related by loss or gain of H+ as H3O+ and H2O. Examples. Acid Base H3O + H2O HC2H3O2 C2H3O2- NH4 + NH3 H2SO4 HSO4- HSO4- SO42-

Select acid, base, acid/conjugate base pair, base/conjugate acid pair H2SO4(aq) + H2O(l) H 3+O(aq) + HSO4-(aq) acid base conjugate acid conjugate base base/conjugate acid pair acid/conjugate base pair

5) For HOCl write: a) Dissociation equilibrium reaction for the HOCl: b) Identify the acid/conjugate base pair: c) Identify the base/conjugate acid pair: d) The equilibrium constant expression:

Strong Acid vs. Weak Acids completely ionized Hydrioidic HI Ka ~ 1011 pKa = -11 Hydrobromic HBr Ka ~ 109 pKa = -9 Perchloric HClO4 Ka ~ 107 pKa = -7 Hyrdrochloric HCl Ka ~ 107 pKa = -7 Chloric HClO3 Ka ~ 103 pKa = -3 Sulfuric H2SO4 Ka ~ 102 pKa = -2 Nitric HNO3 Ka ~ 20 pKa = -1.3 Weak acid partially ionized Hydrofluoric acid HF Ka = 6.6x10-4 pKa = 3.18 Formic acid HCOOH Ka = 1.77x10-4 pKa = 3.75 Acetic acid CH3COOH Ka = 1.76x10-5 pKa = 4.75 Nitrous acid HNO2 Ka = 4.6x10-4 pKa = 3.34 Acetyl Salicylic acid C9H8O4 Ka = 3x10-4 pKa = 3.52 Hydrocyanic acid HCN Ka = 6.17x10-10 pKa = 9.21

Strong Base vs. Weak Base completely ionized Lithium hydroxide LiOH Sodium hydroxide NaOH Potassium hydroxide KOH Kb~ 102-103 Rubidium hydroxide RbOH Cesium hydroxide CsOH Boarder-line Bases Magnesium hydroxide Mg(OH)2 Calcium hydroxide Ca(OH)2 Strotium hydroxide Sr(OH)2 Kb~ 0.01 to0.1 Barium hydroxide Ba(OH)2 Weak Base partially ionized Ammonia NH3 Kb=1.79x10-5 pKb = 4.74 Ethyl amine CH3CH2NH2 Kb=5.6x10-4 pKb = 3.25

Acid and Base Strength Strong acids Ionize completely in water. HCl, HBr, HI, HClO3, HNO3, HClO4, H2SO4. Weak acids Partially ionize in water. Most acids are weak. Strong bases Ionize completely in water. Strong bases are metal hydroxides - NaOH, KOH Weak bases Partially ionize in water.

Common Acids and Bases Acids Formula Molarity* nitric HNO3 16 hydrochloric HCl 12 sulfuric H2SO4 18 acetic HC2H3O2 18 Bases ammonia NH3(aq) 15 sodium hydroxide NaOH solid *undiluted.

6. Identify stronger and weaker acids:

Autoionization of Water Autoionization When water molecules react with one another to form ions. Acids and bases alter the dissociation equilibrium of water based on Le Chaterlier’s principle Kw = [ H3O+ ] [ OH- ] = 1.0 x 10-14 at 25oC Note: [H2O] is constant and is included in Kw. H2O(l) + H2O(l) H3O+(aq) + OH-(aq) (10-7M) (10-7M) ion product of water

What is pH? Kw = [H3+O][OH-] = 1 x 10-14 [H3+O][OH-] = 10-7 x 10-7 Extreme cases: Basic medium [H3+O][OH-] = 10-14 x 100 Acidic medium [H3+O][OH-] = 100 x 10-14 pH value is -log[H+] spans only 0-14 in water.

pH and other “p” scales Substance pH 1 M HCl 0.0 Gastric juices 1.0 - 3.0 Lemon juice 2.2 - 2.4 Classic Coke 2.5 Coffee 5.0 Pure Water 7.0 Blood 7.35 - 7.45 Milk of Magnesia 10.5 Household ammonia 12.0 1M NaOH 14.0 We need to measure and use acids and bases over a very large concentration range. pH and pOH are systems to keep track of these very large ranges. pH = -log[H3O+] pOH = -log[OH-] pH + pOH = 14

pH scale A logarithmic scale used to keep track of the large changes in [H+]. 0 7 14 10 0 M 10-7 M 10-14 M Very Neutral Very acidic Basic When you add an acid to, the pH gets smaller. When you add a base to, the pH gets larger.

pH of Aqueous Solutions

pH, pKw and pOH The relation of pH, Kw and pOH Kw = [H+][OH-] log Kw = log [H+] + log [OH-] -log Kw= -log [H+] -log [OH-] ; previous equation multiplied by -1 pKw = pH + pOH; pKw = 14 since Kw =1 x 10-14 14 = pH + pOH pH = 14 - pOH pOH = 14 - pH

Measuring pH Arnold Beckman inventor of the pH meter father of electronic instrumentation

7. Identify the following as acidic/basic/neutral and estimate/calculate pH. Solution Acidic/basic/neutral pH of the solution a) [ H+] > [OH-] and [H+] > 1.0 x 10-7 M : : b) [H+] < [OH-] and [H+] < 1.0 x 10-7 M : : c) [H+] = [OH-] = 1.0 x 10-7 M : : d) [H+] > [OH-] = 1.0 × 10-14 M : : e) [H+]< [OH-] = 1.0 x 10+7 M : :

pH and pOH calculations of acid and base solutions a) Strong acids/bases dissociation is complete for strong acid such as HNO3 or base NaOH [H+] is calculated from molarity (M) of the solution b) weak acids/bases needs Ka , Kb or percent(%)dissociation

pH of 0.5 M H2SO4 Solution HSO4-(aq) + H2O(l) H3+O(aq) + SO42-(aq) H2SO4(aq) + H2O(l) H3+O(aq) + HSO4-(aq) HSO4-(aq) + H2O(l) H3+O(aq) + SO42-(aq) [H3+O][HSO4-] H2SO4 ; Ka1 = ------------------- [H2SO4] [H3+O][SO42-] H2SO4 ; Ka2 = ------------------- ; Ka2 ignored [HSO4-]

pH of 0.5 M H2SO4 Solution pH = -log(0.5) pH = 0.30 H2SO4(aq) + H2O(l) H3+O(aq) + HSO4-(aq) the moles of H+ ions in the solution is equal to moles of H2SO4 at the beginning. [H2SO4] = [H+] = 0.5 mole/L pH = -log [H+] pH = -log(0.5) pH = 0.30

1.5 x 10-2 M NaOH. 1.5 x 10-2 M NaOH. NaOH is also a strong base dissociates completely in water. [NaOH] = [HO- ] = 1.5 x 10-2 mole/L pOH = -log[HO-]= -log(1.5 x 10-2) pOH = 1.82 As defined and derived previously: pKw= pH + pOH; pKw= 14 pH = pKw + pOH pH = 14 - pOH pH = 14 - 1.82 ; pH = 12.18

How many OH- are in the compound? 8) For a 0.10 M solution of Ba(OH)2.Is it a strong base? How many OH- are in the compound? b) Calculate the [OH-] and [H+]: c) pH of the solution:

b) Is it polyprotic acid? c) Dissociation equilibria: 9) Calculate the pH of the strong acid 0.2 M H2SO4. a) Is it a strong acid? b) Is it polyprotic acid? c) Dissociation equilibria: d) Why second dissociation equilibria is not considered for [H+] concentration? e) Calculate the [H+] f) pH of the solution:

pH of Mixtures of Strong and Weak Acids the presence of the strong acid retards the dissociation of the weak acid The pH of the solution is mainly based on the strong acid Eg. 1.0 M HCl and 1.0 HC2H3O2 HCl(aq) + H2O(l)  H3+O(aq) + Cl-(aq) HC2H3O2(aq) + H2O(l) H3+O(aq) + C2H3O2-(aq)

pH of Mixtures of Acids and Bases The pH of the solution is mainly based on the excess acid or base present Eg. 10 mL of 1.0 M HCl and 20 mL 1.0 NaOH Moles of excess NaOH ( M x L) = 1.0 x 0.001= 0.001 Mixed together volume = 30 mL = 0.030 L Molarity of excess NaOH = 0.001/0.030= 0.030 Calculate pOH and then pH

Equilibrium, Constant, Ka & Kb Ka: Acid dissociation constant for a equilibrium reaction. Kb: Base dissociation constant for a equilibrium reaction. Acid: HA + H2O H3+O + A- Base: BOH + H2O B+ + OH- [H3+O][ A-] [B+ ][OH-] Ka = --------------- ; Kb = ----------------- [HA] [BOH]

Acid Dissociation Constant HCl(aq) + H2O(l) H3+O(aq) + Cl-(aq) [H3+O][Cl-] Ka= ----------------- [HCl] [H+][Cl-]

Base Dissociation Constant NH3 + H2O NH4+ + OH- [NH4+][OH-] K = [NH3]

Comparing Kw and Ka & Kb Any compound with a Ka value greater than Kw of water will be a an acid in water. Any compound with a Kb value greater than Kw of water will be a base in water.

Ionization Constants for Acids

WEAKER/STRONGER Acids and Bases & Ka and Kb values A larger value of Ka or Kb indicates an equilibrium favoring product side. Acidity and basicity increase with increasing Ka or Kb. pKa = - log Ka and pKb = - log Kb Acidity and basicity decrease with increasing pKa or pKb.

Which is weaker? a. HNO2 ; Ka= 4.0 x 10-4. b. HOCl2 ; Ka= 1.2 x 10-2. c. HOCl     ;  Ka= 3.5 x 10-8. d. HCN      ;  Ka= 4.9 x 10-10.

What is Ka1 and Ka2? H2SO4(aq) + H2O(l) H3+O(aq) + HSO4-(aq) HSO4-(aq) + H2O(l) H3+O(aq) + SO42-(aq)

Ka Examples H2SO4(aq) + H2O(l) H3+O(aq) + HSO4-(aq) HSO4-(aq) + H2O(l) H3+O(aq) + SO42-(aq) [H3+O][HSO4-] H2SO4 ; Ka1 = ------------------- [H2SO4] [H3+O][SO42-] H2SO4 ; Ka2 = ------------------- [HSO4-]

Ka Examples HC2H3O2(aq) + H2O(l) H3+O(aq) + C2H3O2-(aq) [H+][C2H3O2-] H C2H3O2; Ka= ------------------ [H C2H3O2] NH3 (aq) + H2O(l) NH4+ + OH-(aq) [NH4+][OH-] NH3; Kb= -------------- [ NH3]

% Dissociation gives x (amount dissociated) need for pH calculation % Dissoc. = ------------------------- x 100 Initial amount/con. x % Dissoc. = --------------------------- x 100 concentration

How do you calculate pH of weak acids/bases? From % dissociation From Ka or Kb What is % dissociation Amount dissociated % Dissoc. = ------------------------- x 100 Initial amount

How do you calculate % dissociation from Ka or Kb 1.00 M solution of HCN; Ka = 4.9 x 10-10 What is the % dissociation for the acid?

1.00 M solution of HCN; Ka = 4.9 x 10-10 First write the dissociation equilibrium equation: HCN(aq) + H 2O(l) <===> H 3+O(aq) + CN-(aq) [HCN] [H+ ] [CN- ] Ini. Con. 1.00 M 0.0 M 0.00 M Cha. Con -x x x Eq. Con. 1.0 - x x x [H 3+O ][CN-] x2 Ka = ------------------- = ---------------- [HCN] 1.0 - x

1.00 M solution of HCN; Ka = 4.9 x 10-10 1.0 - x ~ 1.00 since x is small x2 Ka = -----------; Ka = 4.9 x 10-10 = x2 1.0 x = 4.9 x 10-10 = 2.21 x 10 -5 Amount disso. 2.21 x 10 -5 ----------------- x 100 =- ------------- x 100 Ini. amount 1.00 % Diss. =2.21 x 10 -5 x 100 = 0.00221 %

a) Dissociation equilibria: b) ICE setup: c) Amount dissociated: 10) Calculate the % dissociation of 2.00 M solutions of HCN (Ka= 4.9 x 10-10) a) Dissociation equilibria: b) ICE setup: c) Amount dissociated: d) % dissociation::

Calculate the pH of a weak acid from % dissociation 1 M HF, 2.7% dissociated Notice the conversion of % dissociation to a fraction (x): 2.7/100=0.027) x=0.027

Calculate the pH of a weak acid from % dissociation HF(aq) + H 2O(l) <===> H 3+O(aq) + F-(aq) [H+][F-] Ka = ----------- [HF] [HF] [H+ ] [F- ] Ini. Con. 1.00 M 0.0 M 0.00 M Chg. Con -x x x Eq.Con. 1.0-0.027 0.027 0.027 pH = -log [H+] pH = -log(0.027) pH = 1.57

11) Calculate the Ka of if 5. 0 M HF, 2 11) Calculate the Ka of if 5.0 M HF, 2.7% dissociated: a) Dissociation equilibrium: b) ICE setup: c) Amount dissociated: d) Ka :

HBz(aq) + H2O(l) H3O+(aq) + Bz-(aq) Weak acid Equilibria Example Determine the pH of a 0.10 M benzoic acid solution at 25 oC if Ka = 6.28 x 10-5 HBz(aq) + H2O(l) H3O+(aq) + Bz-(aq) The first step is to write the equilibrium expression Ka = [H3O+][Bz-] [HBz]

Weak acid Equilibria HBz H3O+ Bz- Initial conc., M 0.10 0.00 0.00 Change, DM -x x x Eq. Conc., M 0.10 - x x x [H3O+] = [Bz-] = x We’ll assume that [Bz-] is negligible compared to [HBz]. The contribution of H3O+ from water is also negligible.

Weak Acid Equilibria Solve the equilibrium equation in terms of x Ka = 6.28 x 10-5 = x = (6.28 x 10-5 )(0.10) H3O+ = 0.0025 M pH = 2.60 x2 0.10

pH from Ka or Kb 1.00 M solution of HCN; Ka = 4.9 x 10-10 First write the dissociation equilibrium equation: HCN(aq) + H 2O(l) H 3+O(aq) + CN-(aq) [HCN] [H+ ] [CN- ] Ini. Con. 1.00 M 0.0 M 0.00 M Chg. Con -x x x Eq. Con. 1.0 - x x x

Weak Acid Equilibria [H 3+O ][CN-] x2 [HCN] 1.0 - x 1.0 - x ~ 1.00 since x is small x2 Ka = -----------; Ka = 4.9 x 10-10 = x2 1.0 x = 4.9 x 10-10 = 2.21 x 10 -5 pH = -log [H+] pH = -log(2.21 x 10-5) pH = 4.65

The Conjugate Partners of Strong Acids and Bases The conjugate acid/base of a strong base/acid has no net effect on the pH of a solution The conjugate base of a weak acid hydrolyze in water and basic or pH of a solution > 7.00 E.g. Na+C2H3O2- sodium acetate The conjugate acid of a weak base hydrolyze in water and acidic or pH of a solution < 7.00 E.g NH4Cl

12) Calculate the [H+], [OH-] and pH of 0. 90 M HC2H3O2; Ka= 1 12) Calculate the [H+], [OH-] and pH of 0.90 M HC2H3O2; Ka= 1.8 x 10-5. a) Dissociation equilibria: b) ICE setup: c) [H+] and [OH-]: d) pH:

13) Calculate the [H+], [OH-] pOH and pH 5. 0 M NH3; Kb = 1 13) Calculate the [H+], [OH-] pOH and pH 5.0 M NH3; Kb = 1.8 x 10-5 a) Dissociation equilibria: b) ICE setup: c) [H+] and [OH-]: d) pOH and pH:

14) Calculate the pH of a 0. 015 M solution of lactic acid 14) Calculate the pH of a 0.015 M solution of lactic acid. The Ka for lactic acid is 1.4 x 10-4.

15) Calculate the pH of a 0. 14 M solution of an acid with Ka = 6 15) Calculate the pH of a 0.14 M solution of an acid with Ka = 6.2 x 10-8 (pH = 4.03)

Acid-Base Properties of Typical Ions

Hydrolysis Reaction of a basic anion or acidic cation with water is an ordinary Brønsted-Lowry acid-base reaction. CH3COO-(aq) + H2O(l) CH3COOH(aq) + OH-(aq) NH4+(aq) + H2O(l) NH3 (aq) + H3O+(aq) This type of reaction is given a special name. Hydrolysis The reaction of an anion with water to produce the conjugate acid and OH-. The reaction of a cation with water to produce the conjugate base and H3O+.

What salt solutions would be acidic, basic and neutral? 1) strong acid + strong base = neutral 2) weak acid + strong base = basic 3) strong acid + weak base = acidic weak acid + weak base = neutral, basic or an acidic solution depending on the relative strengths of the acid and the base.

What pH? Neutral, basic or acidic? a)NaCl neutral b) NaC2H3O2 basic c) NaHSO4 acidic d) NH4Cl

1) If the following substance is dissolved in pure water, will the solution be acidic, neutral, or basic? a) Solid sodium carbonate-(Na2CO3): b) Sodium chloride- (NaCl): c) Sodium acetate- (NaC2H3O2): d) Ammonium sulfate-((NH4)2SO4):

How do you calculate pH of a salt solution? Find out the pH, acidic or basic? If acidic it should be a salt of weak base If basic it should be a salt of weak acid if acidic calculate Ka from Ka= Kw/Kb if basic calculate Kb from Kb= Kw/Ka Do a calculation similar to pH of a weak acid or base

What is the pH of 0.5 M NH4Cl salt solution? (NH 3; Kb = 1.8 x 10-5) Find out the pH, acidic if acidic calculate Ka from Ka= Kw/Kb Ka= Kw/Kb = 1 x 10-14 /1.8 x 10-5) Ka= 5.56. X 10-10 Do a calculation similar to pH of a weak acid

Continued NH4+ + H2O H 3+O + NH3 [NH4+] [H3+O ] [NH3 ] Ini. Con. 0.5 M 0.0 M 0.00 M Change -x x x Eq. Con. 0.5 - x x x [H 3+O ] [NH3 ] Ka(NH4+) = -------------------- = [NH 4+] x2 ---------------- ; appro.:0.5 - x . 0.5 (0.5 - x)

Continued pH of 0.5 M NH4Cl solution is 4.77 (acidic) x2 Ka(NH4+) = ----------- = 5.56 x 10 -10 0. 5 x2 = 5.56 x 10 -10 x 0.5 = 2.78 x 10 -10 x= 2.78 x 10 -10 = 1.66 x 10-5 [H+ ] = x = 1.66 x 10-5 M pH = -log [H+ ] = - log 1.66 x 10-5 pH = 4.77 pH of 0.5 M NH4Cl solution is 4.77 (acidic)