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Acid and Base Equilibrium Chapter 16 Brown LeMay.

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Presentation on theme: "Acid and Base Equilibrium Chapter 16 Brown LeMay."— Presentation transcript:

1 Acid and Base Equilibrium Chapter 16 Brown LeMay

2 Basic Concepts Acids – sour or tart taste, electrolytes, described by Arrhenius as sub that inc. the H + conc. Bases – bitter to the taste, slippery, electrolytes, describes by Arrhenius as sub that inc. the OH - conc.

3 Dissociation of Water Pure water exists almost entirely of water molecules. It is essentially a non- electrolyte. Water ionizes to a small extent – auto- ionization The equilibrium expression is H 2 0 (l) H + (aq) + OH - (aq) Kw = [H+] [OH-]

4 Since the H+ ion does not exist alone in water Kw is often expressed Kw = [H 3 O + ] [OH - ] because water conc is constant it does not appear in the expression The proton in water

5 Values for Kw and [H+] and [OH-] Kw = [H 3 O + ] [OH - ] = 1.0 x x = [X] [X] 1.0 x = X 2 X = 1.0 X = [H 3 O + ] = [OH - ]

6 The Bronsted – Lowry definition holds true for situations not involving water Acids donate protons Bases accept protons

7 HCl (g) + NH3 (g)  NH 4 + (g) + Cl - (aq)\ donates H+ accepts H+ conj acid conj base BL-Acid BL-Base notice that the reaction doesn’t happen in water and that the OH - concentration has not increased

8 Conjugate Acids & Bases and Amphoteric Substances HNO 2(aq) + H 2 O (l)  NO 2 - (aq) +H 3 O + (aq) conjugate acid base base acid donates - accepts protons note HNO 2 is considered a Arrenius acid H+ inc in H20 also a BL because donates a Proton all Arrenius acids and bases are also BL acids and bases – however all BL acids and bases may or may not be arrenius acids or bases water is not an Arrenius base in this example

9 every acid has a conjugate base formed from the removal of a proton from that acid every base has a conjugate acid formed from the addition of a proton to that base

10 Amphotheric Substances NH 3(aq) +H 2 0 (l)  NH 4 + (aq) + OH - (aq) Base Acid Conj Acid Conj Base p-acc p-donar note water is acting as an acid in this reaction and a base in the previous one that makes it a amphortic substance

11 Strengths of Acid, Bases The stronger the acid is the weaker its conjugate base (weaker acid  stronger conj base) The stronger the base the weaker its conjugate acid (weaker base  stronger conj acid) Stronger acids and bases ionize to a greater extent than do weak acids and bases.

12 Strong Acids – dissociate completely into ions HNO 3 (aq)  H + (aq) + NO 3 - (aq) the production of H + ions from the acid dominates – ignore the H+ donated from the water it is insignificant The  equilibrium lies so far to the right because HNO 3 doesn’t reform The neg log of the H + from the acid determines pH. Strong bases also dissociate completely and the conc of OH - from the base is the only factor considered when calculating pH.

13 Acids and Bases Strong Acids HClO 4 prechloric H 2 SO 4 sulfuric HI hydroiodic HCl hydochloric HBr hydrobromic HNO 3 nitric Strong Bases GI hydroxides ex. NaOH,KOH G2 Hydroxides Sr(OH) 2 GI Oxides ex. Na 2 O, K 2 O GI,II Amides ex. KNH 2, Ca(NH 2 ) 2

14 The pH scale pH – is defined as the neg log (base-10) of the H + ion concentration pH = -log [H + ] What is the pH of a neutral solution [H + ] = 1.0 x 10^-7 pH = -log [1.0 x 10^-7 pH = 7

15 Strong acids and the pH scale An acidic solution must have a [H+] conc greater than 1.0 X ex 1.0 X log [1.0 X 10 -6] = pH 6 What is the pH of a basic solution? A basic solution is one in which the [OH - ] is greater than 1X10 -7.

16 Calc. pH of strong basic solutions Calculate the pH of a sol that has a [OH - ] con. Of 1.0 X10 -5 Kw = 1x = [H + ] [OH - ] Kw = 1x = [H + ] [1.0X10 -5 ] [H + ] = 1x = 1.0 X X10 -5 pH = -log 1.0 x pH = 9

17 The “p” Scale The negative log of a quantity is labeled p (quantity) Ex: we could reference the quantity of OH - directly: pOH = -log[OH - ] From the definition of Kw -log Kw =(-log [H + ]) (-log [OH - ])= -log 1x10 -4 Kw = pH + pOH = 14

18 Calc. pH using the p scale Ex. OH- conc = 1.0X log 1.0X10 -5 = 5 = pOH pH + pOH = 14 pH + 5 = 14 pH = 9

19 Weak Acids partially ionize in aqueous solution mixture of ions and un-ionized acid in sol. WA are in equilibrium (H 2 0 is left out because it a pure liquid) HA (aq) + H20 (l)  H (aq) + A - (aq) Ka is the acid dissociation constant Ka = [H 3 0] [A - ] = [H + ] [A - ] [HA] [HA]

20 Acid Dissociation Constant The larger the Ka value the stronger the acid is – more product is in solution

21 Weak Bases Weak bases in water react to release a hydroxide (OH-) ion and their conjugate acid: Weak Base(aq) + H2O(l) Conjugate Acid(aq) + OH-(aq)

22 A common weak base is ammonia NH 3 (aq) + H 2 O(l)  NH 4 + (aq) + OH - (aq) Since H 2 O is a pure liquid it is not expressed in the equilibrium Kb expression Kb = [NH 4 +] [OH - ] (base dissociation [NH 3 ] constant) Kb always refers to the equilibrium in which a base reacts with H2O to form the conjugate acid and OH-

23 Lewis Acids and Bases Review - An Arrhenius acid reacts in water to release a proton - base reacts in water to release a hydroxide ion In the Bronstead-Lowry description of acids and bases: acid reacts to donate a proton - a base accepts a proton

24 G.N.Lewis defination - Lewis acid is defined as an electron- pair acceptor Lewis base is defined as an electron- pair donor

25 In the example with ammonia, the ammonia is acting as a Lewis base (donates a pair of electrons), and the proton is a Lewis acid (accepts a pair of electrons)

26 Lewis is consistent with the description by Arrhenius, and with the definition by Bronstead- Lowry. However, the Lewis description, a base is not restricted in donating its electrons to a proton, it can donate them to any molecule that can accept them.

27 Calculating the pH of a Weak Acid What is the pH of an aq sol that is M pyruvic acid HC 2 H 3 P 3 ? Ka = 1.4x10 -4 at 25 o C HC 2 H 3 P 3  H + + C 2 H 3 P 3 - I C -X X X E X +X +X

28 Ka = [H + ] [C 2 H 3 P 3 - ] pluggin in the values [HC 2 H 3 P 3 ] from the table 1.4x10 -4 = X 2 ( X) 1.4x10 -4 (0.0030) = X 2 4.2x x10 -4 x = X 2 X x10 -4 x-4.2x10 -7 = 0 a quadratic ignore the neg sol x = 5.82 x10 -4 pH = -log 5.82 x10 -4 pH = 3.24

29 Learning Check What is the pH at 25 o C of a solution made by dissolving a 5.00 grain tablet of aspirin (acetylsalicylic acid) in liters of water? The tablet contains 0.325g of the acid HC 9 H 7 O 4. Ka = 3.3x10-4 mm = 180.2g/l H + = 9.4x10 -4 pH = 3.03

30 Buffers A solution that resists changes in pH when a limited amount of an acid or base is added to it. Buffers contain either a weak acid and it’s conj. base or a weak base and it’s conj. acid.

31 Examples Ex. Weak acid and conj. base equal molar amounts Strong Acid added H + + A -  HA conj. base weak acid the conj base interacts with the H+ ions from the strong acid changing them to a weak acid

32 Strong base added OH + HA  HOH + A - weak acid conj base the weak acid interacts with the OH - ion from the base to form water and the conj. base If the concentration of A - and HA are large and the amount of H+ or OH- is small the solution will be buffered or the change in pH will be minimized.

33 Buffering capacity – the amount of acid or base a buffer can react with before a significant change in pH occurs Ratio of acid to conj base – unless the ratio is close to 1 ( between 1:10 and 10:1) will be too low to be useful.

34 Calculating the pH of a buffer Note: a solution of 0.10 M acedic acid and its conj base 0.20 M acetate from sodium acetate is a buffer solution pH = 5.07 Ex. Calc. the pH of a buffer by mixing 60.0 ml of M NH 3 with 40.0ml of M NH 4 Cl.

35 1 St cal the conc. of each species M = moles/liters mol of NH M = X/0.060 l = mol mol of NH M = X/0.040 l = mol [NH 3 ] = mol/ l = M [NH 4 ] = mol/ l = M NH 3 + H2O  NH4 + + OH - I 0.060M O C -X +X +X E x x X

36 Kb = [NH 4 + ] [OH - ] ( X)X 1.8X10 -5 [NH 3 ] (0.060-X) Ignore X 1.8X10 -5 = 0.040X X = 2.7X log (2.7x10-5) = 4.6 pOH pH = 9.4 Or using Henderson - Hasselbalch equation pOH = pKb + log [conj acid] = log ( [0.04] [B] [0.06]) = 4.6 pOH pH= 9.4

37 What is the pH of a buffer prepared by adding 30.0ml of.15M HC 2 H 3 O 2 to 70ml of.2M NaC 2 H 3 O 2 ? [HC 2 H 3 O 2 ] =.15M = x/.03 =.0045/.01 =.045 [C 2 H 3 O 2 ] =.20M =x/.07 =.0140/.01 =.140 ka = 1.7x10-5 HHeq pH = pKa + log [conj base] [acid] pH = log(.140 = )

38 Adding an acid or a base to a buffer Calc the ph of 75ml of the buffer solution of(0.1M HC 2 H 3 O 2 and 0.2M NaC 2 H 3 O 2 ) to which 9.5 ml of 0.10M HCl has been added. Compare the change to that of adding HCl to pure water. H + + C 2 H 3 O 2 -  HC 2 H 3 O 2 H + = 0.10M = n/.0095l = moles C 2 H 3 O 2 - = 0.2M = n/.075 = moles HC 2 H 3 O 2 = 0.10M = n/.075 = moles

39 Neutralization Reaction C 2 H 3 O 2 = C 2 H 3 O 2 moles – H+ moles n n = HC 2 H 3 O 2 = Orginial Conc. + Conc Contributed by reaction moles = mol [C 2 H 3 O 2 ] = 0.014mol/0.085l = 0.16M [HC 2 H 3 O 2 ] = /0.085 = 0.10M pH = log (.16/.10) = 4.96


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