Corrosive Base Equilibria

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Corrosive Base Equilibria. 18.1 Acids and Bases in Water . 18.2 Autoionization of Water and the pH Scale. 18.3 Proton Transfer and the Br

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Section 18 Acid-Base Equilibria

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Acid-Base Equilibria 18.1 Acids and Bases in Water 18.2 Autoionization of Water and the pH Scale 18.3 Proton Transfer and the Brønsted-Lowry Acid-Base Definition 18.4 Solving Problems Involving Weak-Acid Equilibria 18.5 Weak Bases and Their Relations to Weak Acids 18.6 Molecular Properties and Acid Strength 18.7 Acid-Base Properties of Salt Solutions 18.8 Generalizing the Brønsted-Lowry Concept: The Leveling Effect 18.9 Electron-Pair Donation and the Lewis Acid-Base Definition

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The Nature of Acids and Bases: Acids: Acids taste harsh. Respond with metals and create H gas. turns blue litmus red pH < 7 * Bases: Bases taste severe. They are elusive. turns red litmus blue. pH >7

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Arrhenius idea Acids deliver H+ particles. Bases create OH-particles. This idea is restricted on the grounds that it applies to just watery arrangement and characterizes just OH containing bases.

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Neutralization: corrosive + base _______salt + water.

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Acid Dissociation Constant (Ka) Write Ka expression for solid and powerless acids.

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Strong corrosive: HA( g or l ) + H 2 O( l ) H 2 O + ( aq ) + A - ( aq ) The degree of separation for solid acids. Figure 18.1

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Weak corrosive: HA( aq ) + H 2 O( l ) H 2 O + ( aq ) + A - ( aq ) The degree of separation for frail acids. Figure 18.2

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1M HCl( aq ) 1M CH 3 COOH( aq ) Reaction of zinc with a solid and a feeble corrosive. Figure 18.3

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HA( g or l ) + H 2 O( l ) H 3 O + ( aq ) + A - ( aq ) HA( aq ) + H 2 O( l ) H 3 O + ( aq ) + A - ( aq ) [H 3 O + ][A - ] more grounded corrosive higher [H 3 O + ] K c = [H 2 O][HA] bigger K a [H 3 O + ][A - ] K c [H 2 O] = K a = [HA] littler K a lower [H 3 O + ] weaker corrosive Strong acids separate totally into particles in water. K c >> 1 Weak acids separate marginally into particles in water. K c << 1 The Acid-Dissociation Constant

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Classifying the Relative Strengths of acids and Bases: Strong acids: HCl, HBr, HI Oxo acids. HNO3, H2SO4, HClO4 Weak acids: HF HCN , H2S (H not clung to O or halogen) Oxo acids. HClO, HNO2, H3PO4 Carboxylic acids. CH3COOH

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Classifying the Relative Strengths of acids and Bases: Strong bases: M2O, MOH M= (assemble 1A metal) MO , M(OH)2 M=group 2A metal Weak bases: (N molecule and solitary combine of electrons) NH3 Amines.

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ACID STRENGTH

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PROBLEM: Classify each of the accompanying mixes as a solid corrosive, feeble corrosive, solid base, or powerless base. Arrange: Pay regard for the content meanings of acids and bases. Take a gander at O for acids and also the - COOH gather; look for amine gatherings and cations in bases. (a) Strong corrosive - H 2 SeO 4 - the quantity of O molecules surpasses the quantity of ionizable protons by 2. (b) Weak corrosive - (CH 3 ) 2 CHCOOH is a natural corrosive having a - COOH assemble. (c) Strong base - KOH is a Group 1A(1) hydroxide. (d) Weak base - (CH 3 ) 2 CHNH 2 has a solitary match of electrons on the N and is an amine. Test PROBLEM 18.1: Classifying Acid and Base Strength from the Chemical Formula (a) H 2 SeO 4 (b) (CH 3 ) 2 CHCOOH (c) KOH (d) (CH 3 ) 2 CHNH 2 SOLUTION:

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Autoionization of water and the pH scale Water separates into its particles.

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H 2 O( l ) H 2 O( l ) OH - ( aq ) H 3 O + ( aq ) Autoionization of Water and the pH Scale +

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H 2 O( l ) + H 2 O( l ) H 3 O + ( aq ) + OH - ( aq ) [H 3 O + ][OH - ] K c = [H 2 O] 2 The Ion-Product Constant for Water K c [H 2 O] 2 = K w = [H 3 O + ][OH - ] = 1.0 x 10 - 14 at 25 0 C A change in [H 3 O + ] causes a backwards change in [OH - ]. In an acidic arrangement, [H 3 O + ] > [OH - ] In an essential arrangement, [H 3 O + ] < [OH - ] In a nonpartisan arrangement, [H 3 O + ] = [OH - ]

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Divide into K w [H 3 O + ] > [OH - ] [H 3 O + ] = [OH - ] [H 3 O + ] < [OH - ] The relationship between [H 3 O + ] and [OH - ] and the relative causticity of arrangements. Figure 18.4 [H 3 O + ] [OH - ] ACIDIC SOLUTION BASIC SOLUTION NEUTRAL SOLUTION

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SAMPLE PROBLEM 18.2: Calculating [H 3 O + ] and [OH - ] in an Aqueous Solution PROBLEM: An exploration scientist includes a deliberate measure of HCl gas to immaculate water at 25 0 C and gets an answer with [H 3 O + ] = 3.0x10 - 4 M. Ascertain [OH - ]. Is the arrangement nonpartisan, acidic, or essential? Arrange: Use the K w at 25 0 C and the [H 3 O + ] to locate the relating [OH - ]. Arrangement: K w = 1.0x10 - 14 = [H 3 O + ] [OH - ] so [OH - ] = K w/[H 3 O + ] = 1.0x10 - 14/3.0x10 - 4 = 3.3x10 - 11 M [H 3 O + ] is > [OH - ] and the arrangement is acidic.

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pH scale pH   log[H+] pH in water ranges from 0 to 14. K w = 1.00  10  14 = [H+] [OH  ] p K w = 14.00 = pH + pOH As pH rises, pOH falls (aggregate = 14.00). pH=7-nonpartisan, pH>7-essential, pH<7-acidic

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pH scale Acidic solns have a higher pOH. pK=-log K condition comes to equ, generally items present, low pK (high K) Reverse of the above explanation is likewise valid. pH is measure utilizing pH meter, pH paper or corrosive base marker.

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pH (marker) paper pH meter Figure 18.7 Methods for measuring the pH of a watery arrangement

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Figure 18.5 The pH estimations of some recognizable fluid arrangements pH = - log [H 3 O + ]

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Table 18.3 The Relationship Between K an and pK an Acid Name (Formula) K an at 25 0 C pK a 1.02x10 - 2 Hydrogen sulfate particle (HSO 4 - ) 1.991 3.15 7.1x10 - 4 Nitrous corrosive (HNO 2 ) 4.74 1.8x10 - 5 Acetic corrosive (CH 3 COOH) 8.64 2.3x10 - 9 Hypobromous corrosive (HBrO) 1.0x10 - 10 Phenol (C 6 H 5 OH) 10.00

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Figure 18.6 The relations among [H 3 O + ], pH, [OH - ], and pOH.

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Problems P.3 .Calculate pH and pOH at 25 C for: 1.0 M H+ P.4 .pH=6.88, Calculate [ H+] and [ OH-] for this example. P.5 .Calculate pH of 0.10 M HNO3 P.6. Ascertain pH of 1.0 x 10-10 HCl.

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Bronsted-Lowry demonstrate : A corrosive is a proton (H+ ) giver. A base is a proton acceptor. * A conjugate corrosive base match just contrasts by one H. HCl + H2O _________ H3O+ Cl-Acid base conj corrosive conj base HA( aq ) + H2O ( l )  H3O+ ( aq ) +A-( aq ) CA1 CB2 CA2 CB1

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Bronsted-Lowry display : conjugate base: everything that remaining parts of the corrosive atom after a proton is lost. Has one H less and one more less charge than the corrosive. conjugate corrosive: framed when the proton is exchanged to the base. Has one more H and one less charge than the base. Do Follow up issue 18.4-Page.779

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A corrosive is a proton contributor, any animal groups which gives a H + . A base is a proton acceptor, any animal varieties which acknowledges a H + . Brønsted-Lowry Acid-Base Definition A corrosive base response can now be seen from the point of view of the reactants AND the items. A corrosive reactant will deliver a base item and the two will constitute a corrosive base conjugate combine.

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Acid + Base + Acid Reaction 1 HF + H 2 O F - + H 3 O + Reaction 2 HCOOH + CN - HCOO - + HCN Reaction 3 NH 4 + CO 3 2-NH 3 + HCO 3 - Reaction 4 H 2 PO 4 - + OH - HPO 4 2-+ H 2 O Reaction 5 H 2 SO 4 + N 2 H 5 + HSO 4 - + N 2 H 6 2+ Reaction 6 HPO 4 2-+ SO 3 2-PO 4 3-+ HSO 3 - Table 18.4 The Conjugate Pairs in Some Acid-Base Reactions Conjugate Pair Conjugate Pair

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Lone match ties H + HCl H 2 O Cl - H 3 O + Lone combine ties H + NH 3 H 2 O NH 4 + OH - Proton exchange as the basic component of a Brønsted-Lowry corrosive base response. Figure 18.8 (corrosive, H + contributor) (base, H + acceptor) (base, H + acceptor) (corrosive, H + giver)

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PROBLEM: The accompanying responses are imperative ecological procedures. Distinguish the conjugate corrosive base sets. (a) H 2 PO 4 - ( aq ) + CO 3 2-( aq ) HPO 4 2-( aq ) + HCO 3 - ( aq ) (a) H 2 PO 4 - ( aq ) + CO 3 2-( aq ) HPO 4 2-( aq ) + HCO 3 - ( aq ) PLAN: Identify proton benefactors (acids) and proton acceptors (bases). (b) H 2 O( l ) + SO 3 2-( aq ) OH - ( aq ) + HSO 3 - ( aq ) (b) H 2 O( l ) + SO 3 2-( aq ) OH - ( aq ) + HSO 3 - ( aq ) SAMPLE PROBLEM 18.4: Identifying Conjugate Acid-Base Pairs conjugate match 2 conjugate combine 1 SOLUTION: proton benefactor proton acceptor proton acceptor proton giver conjugate combine 2 conjugate match 1 proton contributor proton acceptor proton acceptor proton giver

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PROBLEM: Predict the net course and whether K an is more prominent or under 1 for each of the accompanying responses (expect approach beginning convergences of all species): (a) H 2 PO 4 - ( aq ) + NH 3 ( aq ) HPO 4 2-( aq ) + NH 4 + ( aq ) (a) H 2 PO 4 - ( aq ) + NH 3 ( aq ) HPO 4 2-( aq ) + NH 4 + ( aq ) PLAN: Identify the conjugate corrosive base combines and afterward counsel Figure 18.10 (catch) to decide the relative quality of each. The more grounded the species, the more prevalent its conjugate. (b) H 2 O( l ) + HS - ( aq ) OH - ( aq ) + H 2 S( aq ) (b) H 2 O( l ) + HS - ( aq ) OH - ( aq ) + H 2 S( aq ) more grounded corrosive more grounded base weaker base weaker corrosive weaker corrosive weaker base more grounded base more grounded corrosive SAMPLE PROBLEM 18.5: Predicting the Net Direction of an Acid-Base Reaction SOLUTION: Net bearing is to one side with K c > 1. Net bearing is to one side with K c < 1.

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Figure 18.9 Strengths of conjugate corrosive base sets

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Solving Problems Involving Weak-Acid Equilibria. Compose adjusted condition and Ka expression. Make an ICE table. Make required presumptions. Substitute values and explain for x. Check suspicions by ascertaining % mistake.

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Problems P.7. Ascertain the pH of 1.00 M fluid arrangement of HF. Ka=7.2 x 10-4 P.8 Calculate the pH of 0.100 M aq arrangement of HOCl. Ka=3.5 x 10-8

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