1 Chapter 17 Acids and Bases - we are all familiar with 'acids' - depicted on television as burning liquids - from foods (i.e. vinegar) - taste "sour" or "tart' - less familiar with 'bases' - taste "bitter" - for example, beer or coffee - several different definitions have been used for the terms "acid" and "base" over the years
2 17.1 The Arrhenius Theory Svante Arrhenius was interested in "electrolytes" and "electrolysis" and he devised much of the early theory to explain ions in solution - according to Arrhenius, acids dissolve in water to give protons while bases dissolve to give hydroxide ions and these react to give water and a salt - this reaction is called "neutralization": acid + base salt + water
3 for example, hydrochloric acid yields a proton in solution sodium hydroxide generates hydroxide ions hence: HCl (aq) + NaOH (aq) NaCl (aq) + H 2 O in ionic form: H + (aq) + Cl - (aq) + Na + (aq) + OH - (aq) Na + (aq) + Cl - (aq) + H 2 O or in "net" ionic form: H + (aq) + OH - (aq) H 2 O
4 - from the Arrhenius point of view, HCl is an acid because it participates in a neutralization reaction that gives water and a salt (in this case, "table salt") and NaOH is a base because it also participates in a neutralization reaction that is, the reason that we know that 'HCl' and 'NaOH' are an acid and a base, respectively, is because they give us water - hence, they must yield up a proton and a hydroxide ion this was a good "first step" but it was not sufficient - for example, it does not explain "ammonia"
5 7.2 Bronsted-Lowry Theory redefines acids and bases in terms of their ability to donate or accept protons an 'acid' is a 'proton donor' a 'base' is a 'proton acceptor' hence, anything that increases the proton content of a solution is, by definition, an acid and anything that decreases it - by accepting a proton - is a base - explains ammonia which accepts protons
6 but this definition is also slightly flawed because bases don't "accept" protons - they steal them! - arrow pushing is how we show this:
7 Bronsted-Lowry gives us the concept of "conjugate acids and bases" NH 4 + (aq) + OH - (aq) NH 3 (aq) + H 2 O acid base conj. conj. base acid NH 4+ /NH 3 and OH - /H 2 O both form acid/base pairs and the partner of an acid in such a pair is called "the conjugate base" (and vice versa for a base) Note: this all depends upon how the reaction is written!
8 also, most acid/base reactions are actually equilibria for example, acetic acid CH 3 COOH (aq) + H 2 O W CH 3 COO - (aq) + H 3 O + (aq) acid base base acid conjugates we can write an equilibrium expression for this reaction
9 K a = [CH 3 COO - ][H 3 O + ] / [CH 3 COOH] where [H 2 O] is treated as a constant and incorporated into the value of K a, the "acid ionization constant" could equally well write an expression for K b, the "base ionization constant" Bronsted-Lowry is better than Arrhenius at accounting for substances that can both donate and accept protons such as water. Such substances are called "amphiprotic"
10 17.3 The (Auto)Ionization of Water and ph - water is one of many substances that are capable of donating a proton to itself that is, H 2 O + H 2 O W H 3 O + (aq) + OH - (aq) acid base conj. conj. acid base this is more commonly written as, H 2 O W H + (aq) + OH - (aq)
11 as an equilibrium, we can write the equilibrium constant expression, K = [H + (aq) ][OH- (aq) ] / [H 2 O] but, since the concentration of water is a constant, this is more commonly expressed as: K w = [H + ][OH - ] simply put, the concentrations of [H + ] and [OH - ] in any and all aqueous solutions are inversely related
12 furthermore, K w = 1.0 x = [H + ][OH - ] or, in pure water, [H + ] = [OH - ] = (1.0 x ) ½ = 1.0 x 10-7 This is our definition of "neutral". In pure water with no other species present, the concentrations of both H + (aq) and OH - (aq) are the same. Note: K w is an "ion product" as it is only concerned with the ionic components of the equilibrium.
13 Measuring the concentration of either the protons or hydroxide ions in a solution results in a measurement of the other. We could use molarity and scientific notation to express the concentration of protons and hydroxide ions (for example, [H + ] = 2.34 x10-5 M) but we more frequently use a "p-function" By definition, a p-function is the "negative log 10 " of the quantity in question. ph = -log [H + ] i.e. ph = -log (2.34 x10-5 ) = 4.631
14 note that p-functions are not restricted to simply measuring ph poh = - log [OH - ] pk a = - log K a pk b = - log K b po 2 = - log [O 2 ] etcetera The purpose of the p-functions is to change to a logarithmic scale but not to change the answers.
15 Also, ph is not actually -log[h + ] but rather it is the negative logarithm of the hydrogen ion activity. That is, ph = -log(ά H+ ) The "activity" is the amount of a substance that is actually available to react - which is not necessarily equal to the amount present. This is a subtle point and not one that we will pursue further in this course.
16 Note: K w = [H + ][OH - ] = 1.0 x10-14 pk w = ph + poh = 14 - as ph decreases, poh increases and vice versa
17 Acid Strength - the strength of an acid is not necessarily correlated with its concentration - that is, a dilute solution of a strong acid might be more reactive than a concentrated solution of a dilute acid - acid strength is a measure of how freely an acid will donate it proton - for example, perchloric acid (HClO 4 ) is a strong acid because it will readily give up its proton to any other base - hence, the perchlorate ion (ClO 4- ) is a lousy base
18 - the stronger the acid, the weaker the conjugate base - that is, a strong acid will lose its proton to any other base and is incapable of stealing the proton back - the strongest acid in water is the hydronium ion (H 3 O + ) - strong acids, when dissolved in water, will donate their proton to water i.e. HCl + H 2 O Cl - + H 3 O + HClO 4 + H 2 O ClO H 3 O + HNO 3 + H 2 O NO H 3 O +
19 - but any acid (or base) can only donate to their concentration - that is, if the concentration of hydrochloric acid in an aqueous solution is M, then the maximum concentration of H + is M. (and it can only react with M of anything else!) - hence, a dilute solution of a strong acid is not particularly reactive but a concentrated solution is!
20 17.4 Strong Acids and Strong Bases - a strong acid or a strong base will completely dissociate in aqueous solution - giving the concentration of H + and OH - directly - generally, can ignore the contribution of water as it will be very small compared to the contribution from the strong acid or base - except when the concentration of the strong acid or base is close to or less than 1 x10-7 M
21 - hence, calculating the ph, the poh, or the pcl for the dissolution of 0.23M HCl is aqueous solution is straightforward: 0.23M HCl 0.23M H M Cl - ph = -log(0.23) = since, [H + ] = [Cl - ], pcl = ph = poh = 14 - ph = =
22 Table 17-2 has the common "Strong acids and Bases"
23 "Are you wondering.." These are sections of the book that seem a little tangental at times. However, the ones on pgs. 676 and 682 are important. pg. 676 How to calculate [H 3 O + ] in an extremely dilute solution of a strong acid. Example: What is the ph of a solution of 1.0 x10-8 HCl? If we assume [H + ] total = 1.0 x10-8, then the answer would be ph = -log (1.0 x 10-8 ) = 8 - but clearly, an acid should have a ph more acidic than neutral
24 - must take auto-ionization of water into account H 2 O + H 2 O W H 3 O + + OH - HCl + H 2 O W H 3 O + + Cl - [H 3 O + ] total = 1.0 x x (from water) - hence, K w = [H 3 O + ][OH - ] = 1.0 x10-14 (x x10-8 )(x) = 1.0 x10-14 x x10-8 x x10-14 = 0
25 solving the quadratic equation gives, x = -(1.0 x10-8 ) ± [(1.0 x10-8 ) 2 2(1) - 4(1)(1.0 x10-14 )] ½ = -1.0 x10-8 ± x x = x10-8 and [H 3 O + ] total = 1.0 x x = 1.0 x x10-8 = x10-7 ph = 6.978
26 17.5 Weak acids and Weak bases - the difference between a "strong acid" and a "weak acid" is that a weak acid is only partially dissociated - some of the "parent" acid will be present in solution along with the conjugate base i.e. strong acid: HClO 4 + H 2 O W H 3 O + + ClO 4 - K eq > 10 6 weak acid: CH 3 COOH + H 2 O W CH 3 COO - + H 3 O + K eq = 1.8 x10-5
27 - the size of the equilibrium constants tell us that in the strong acid, the concentrations of H 3 O + and ClO 4- are very large compared to the concentration of HClO 4 whereas in the weak acid, the concentrations of H 3 O + and CH 3 COO - are small compared to CH 3 COOH. - why does this happen? the covalent bond between the proton and the acetate ion is much stronger than the one holding the proton to the perchlorate ion Note: even the strongest covalent bond will eventually break, so even methane (CH 4 ) is an acid - just a very, very weak acid!
28 - to talk about weak acids, we use pk a 's as a method of presenting the acid equilibrium constant
29 working through questions follows the same steps that we used in solving other equilibrium problems: example: determine the K a of a weak acid if the total concentration of acid is M and the ph is 2.46 step 1) write out equilibrium HA W H + + A - step 2) write out equilibrium constant expression K a = [H + ][A - ] / [HA]
30 step 3) ICE table HA W H + + A - initial M 0.0 M 0.0 M change - x + x + x equilibrium x x x but, in this case, we know "x" because we know the final ph - hence, x = [H + ] = = x10-3 M hence, [H + ] = [A - ] = M; [HA] = M K a = ( )( )/(0.2965) = x10-5
31 Example: What is the ph of a 1.0 M solution of a weak acid (K a = x10-5 )? step 1) equation: HA W H + + A - step 2) equilibrium: K a = [H + ][A - ] / [HA] step 3) ICE: HA W H + + A - initial change - x + x + x equilibrium x x x
32 step 4) K a = (x)(x) / (1.0 - x) = x10-5 step 5) x 2 = x10-5 (1.0 - x) = 4.05 x x10-5 x x x10-5 x x10-5 = 0 step 6) x = -(4.054 x10-5 ) ± [(4.054 x10-5 ) 2-4(1)(4.054x10-5 )] ½ 2(1) x = x10-5 ± x10-2 x =
33 step 7) [H + ] = x = ph = -log( ) = step 8) can't really check the ph but we can check the equilibrium: K a = ( ) 2 /( ) = x10-5 Also, does a ph of sound reasonable? Yes. In the last problem, an acid concentration of M resulted in a ph of It makes sense that an increase in the acid concentration should result in a decrease in ph (an increase in the number of H + ions in solution).
34 Example: A solution of vinegar is typically about 3% acetic acid in water, which corresponds to about M. At this concentration, with K a = 1.8 x10-5, what is the ph of vinegar? step 1)
37 What about weak bases? Example: What is the ph of a 2.3 x10 2 M solution of Ammonia (K b = 1.8 x10-5 )? step 1)
40 book advocates using a simplifying assumption - that the amount of dissociation is small compared to the overall concentration of acid in acetic acid example, we had K a = (x)(x) / ( x) = 1.8 x10-5 but if x << then ( x) and the equation would be: x 2 / = 1.8 x10-5 x 2 = 0.90 x10-5 x = M
41 this is reasonably close to the value that we got using the quadratic, 2.99 x10-3 M, and certainly within rounding errors - hence, we could have saved some time this way however, consider the example of a weak base such as ammonia after simplifying, K b = (x)(x) / ( x) = 1.8 x10-5 x 2 / = 1.8 x10-5 x 2 = 4.14 x10-5 x = x10-4 compared to x10-4
42 - while this is close, they are not exactly the same and could lead to errors. Simplification does not always work. It is necessary to know the long route, no matter what. As an approach, you can always try the short route first but if x is more than 5% of the concentration, then do the quadratic. Note: quadratic formula always works but is slower. simplification doesn't always work but is fast.
43 Percent Ionization - the percent ionization gives the proportion of ionized molecules (or dissociated molecules) % ionization = molarity of H + from HA x 100% molarity of HA - strong acids - percent ionization is virtually 100% - weak acids - percent ionization is less than 100% but it changes with concentration
44 - the more dilute the acid, the greater the % ionization i.e. for acetic acid % ionization is 0.6% at M % ionization is 4% at M % ionization is 100% at M - this gives us a clue as to what sort of answer to expect as the more dilute the acid, the closer the concentration of H + will be to the concentration of the acid that is, [H + ] goes to [HA] as the acid gets more dilute
45 Were you wondering.. on page math is obtuse and it is not something that we are going to deal with overtly at this point, however, it is a reminder that water is a good acid and a good base under dilute conditions - also, the iterative process is a stepwise approach used in a lot of computer programming - we will need to consider this when we talk about ions as acids and bases, and more importantly, the titration of weak acids and bases - but we will find a mathematically less obtuse answer!
46 17.6 Polyprotic Acids - most of the acids that we have dealt with so far are "mono-protic" - they only give up a single proton upon dissociation (and the bases only accept a single proton) - some acids are "polyprotic" as are some bases: polyprotic acids: H 3 PO 4, H 2 SO 4, H 2 SO 3, H 2 CO 3 also the organic acids: oxalic, maleic, succinic polyprotic bases: diaminoethane, morphine, DNA, RNA
47 - as examples, we will deal with two that are of equal importance to industry - phosphoric acid and sulfuric acid Phosphoric acid, H 3 PO 4, has three acid equilibria: 1) H 3 PO 4 W H + + H 2 PO - 4 K a1 = [H + ][H 2 PO 4- ] / [H 3 PO 4 ] = 7.1 x10-3 2) H 2 PO 4- W H + + HPO 2-4 K a2 = [H + ][HPO 2-4 ] / [H 2 PO 4- ] = 6.3 x10-8 3) HPO 2-4 W H + + PO 3-4 K a3 = [H + ][PO 3-4 ] / [HPO 2-4 ] = 4.2 x10-13
48 as you might expect, K a1 > K a2 > K a3 - it gets harder to remove the positively charged proton as the species gets more negatively charged One way to illustrate what is going on is with a "speciation curve" - a plot of ph versus concentration:
49 Working questions with polyprotic acids means that we have to take each deprotonation into account. Example, What is the ph of a 0.10M solution of phosphoric acid? - need to do this step-wise first reaction: step 1) H 3 PO 4 W H + + H 2 PO 4 - step 2) K a = [H + ][H 2 PO 4- ] / [H 3 PO 4 ] = 7.1 x10-3
50 step 3) ICE table H 3 PO 4 W H + + H 2 PO - 4 initial 0.10 M change - x x x equilibrium x x x step 4) K a = x 2 /( x) = 7.1 x10-3 check simplification: x 2 / 0.10 = 7.1 x10-3 x = (7.1 x10-3 x 0.10) ½ = M which is not small compared to 0.10 M
51 step 5) go the quadratic route x 2 = 7.1 x10-3 ( x) x x10-3 x x10-4 = 0 step 6) x = -(0.0071) ± [(0.0071) 2-4(1)(-7.1 x10-4 )] ½ x = ± x = step 7) [H + ] = [H 2 PO 4- ] = M; [H 3 PO 4 ] = M step 8) check the equilibrium - it works
52 but what about the second step? does the second deprotonation now contribute? step 1) write out equation: H 2 PO 4- W H + + HPO 4 2- step 2) K a = [H + ][HPO 4 2- ] / [H 2 PO 4- ] = 6.3 x10-8 Note that we already have [H + ] = M and [HPO 4- ] = M from the first equilibrium. This means that we are not starting from "scratch" here and the ICE table must reflect this.
53 step 3) ICE table H 2 PO - 4 W H + + HPO 2-4 initial M M 0.0M change - x + x + x equlibrium x x x step 4) re-write K a K a = ( x)(x) / ( x) = 6.3 x10-8 if we assume that 'x' is small compared to , then we get: ( )(x) / ( ) 6.3 x10-8 or: x = 6.3 x10-8 M
54 In this instance, the simplification works as 6.3 x10-8 M is very much smaller than M. Hence, [H + ] = x M and ph = -log( ) = Note that since K a3 is even smaller, it would make a very small contribution to [H + ] - one that we can safely ignore.
55 Example: What is the ph of a 0.10M solution of sulfuric acid? - for sulfuric acid, the first deprotonation proceeds 100% as it is a "strong acid" and hence, we don't have to calculate this as the concentration of protons will equal the concentration of the acid. That is, H 2 SO 4 H + + HSO M 0.10M Note that these are the initial concentrations for these species in the second deprotonation, so once again, for the second deprotonation we are not starting from 'scratch'.
56 the second deprotonation: step 1) HSO 4 - W H + + SO 4 2- step 2) K a = [H + ][SO 4 2- ] / [HSO 4- ] = step 3) ICE table: HSO - 4 W H + + SO 2-4 initial change - x x x equilibrium x x x step 4) K a = ( x)(x) / ( x) = 0.013
57 step 5) x x = 0.013( x) = x x x = 0 step 6) x = -(0.113) ± [(0.113) 2-4(1)( )] ½ 2(1) = ( ± 0.134)/2 x = or x = step 7) [H + ] e = x = = M ph = -log(0.1105) = less than 1, as expected, because 0.10M H + = ph 1
58 17.7 Ions as Acids and Bases - ionic species that are components of acid/base equilibria will change the ph of a solution when dissolved - further, "the product of the ionization constant of an acid and its conjugate base equals the ion product of water" - that is, K a x K b = K w for example, NaF in water gives F - which binds H + because it is the conjugate base of a weak acid F - + H + W HF K a -1 = [HF]/[H + ][F - ]
59 Hydrolysis - dissolving NaCl in water gives Na + (aq) and Cl - (aq) but no change in ph as Cl - is a weak base and doesn't pull protons from water and Na + is a weak acid - dissolving NH 4 Cl in water gives NH 4 + (aq) and Cl - (aq) which undergoes a change in ph due to the hydrolysis of the ammonium ion NH 4+ + H 2 O W NH 3 + H 3 O + - the ionic component of a weak acid or base, when added to water as a salt, will result in the breaking up of water molecules and a shift in the ph from neutrality
60 Example: What is the ph of a 0.10M solution of NaF? step 1) F - + H 2 O W HF + OH - step 2) K b = [HF][OH - ] / [F - ] = K w /K a = 1.52 x10-11 step 3) ICE: F - W HF + OH - initial 0.10M change - x x x equilibrium x x x
61 step 4) K b = (x)(x) / ( x) = 1.52 x10-11 step 5) in this case, it is safe to assume that 'x' is going to be very much less than 0.10, hence, x 2 / 0.10 = 1.52 x10-11 x = [(1.52 x10-11 )(0.10)] ½ = 1.23 x10-6 step 7) [OH - ] e = x = 1.26 x10-6 poh = ph = = the fluoride ion shifts the ph from neutral to basic
62 17.8 Molecular Structure and Acid-Base Behaviour - this section deals with questions around why some acids are strong and some are weak - and the simple answer is the strength of the bond holding the proton - however, general trends are: acidity increases going down a group HF K a = 6.6 x10-4 HCl K a = 1.3 x10 6 HBr K a = 10 8 HI K a = 10 9
63 oxoacids - acidity increases with the number of oxygens K HClO 2 < K HClO 3 < K HClO 4 - decreases for different central atoms as one goes down a group K HClO 4 > K HBrO 4 > K HIO 4 strength of organic acids - the length of the organic chain has little effect - presence of an electron withdrawing group has a major effect on K a strength of amine bases - leave for organic chemistry
64 17.9 Lewis Acids and Bases - this is the most useful and general definition of acids and bases - it encompasses all of the other definitions and then some - a "Lewis acid" is an electron pair acceptor - a "Lewis base" is an electron pair donor i.e. H + accepts a pair of electrons from OH - O 2 donates a pair of electrons to the iron in hemoglobin
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Chemistry 12 Acid-Base Equilibrium II Name: Date: Block: 1. Strengths of Acids and Bases 2. K a, K b 3. Ionization of Water 4. Relative Strengths of Brønsted-Lowry Acids and Bases Strengths of Acids and
As you work through the chapter, you should be able to: Advanced Placement Chemistry Chapters 14 16 Syllabus Chapter 14 Acids and Bases 1. Describe acid and bases using the Bronsted-Lowry, Arrhenius, and
Chapter 16 - Acids and Bases 16.1 Acids and Bases: The Brønsted Lowry Model 16.2 ph and the Autoionization of Water 16.3 Calculations Involving ph, K a and K b 16.4 Polyprotic Acids 16.1 Acids and Bases:
Acids and Bases Chapter 15 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1 Acids Have a sour taste. Vinegar owes its taste to acetic acid. Citrus fruits contain
Principles of Reactivity: The Chemistry of Acids and Bases **a lot of calculations in this chapter will be done on the chalkboard Do not rely on these notes for all the material** Acids, Bases and Arrhenius
Chapter 6 Acids and Bases Introduction Brønsted acid-base reactions are proton transfer reactions. Acids donate protons to bases. In the process, the acid is converted into its conjugate base and the base
Acid-Base Equilibria 1. Classify each of the following species as an acid, a base, or amphoteric in aqueous solution: (a) H 2 O; (b) CH 3 CH 2 ; (c) PO 4 3 ; (d) C 6 H 5 NH 3 2. Write the proton transfer
Chapter 9 Aqueous Solutions and Chemical Equilibria At equilibrium, the rate of a forward process or reaction and that of the reverse process are equal. 9A The chemical composition of aqueous solutions
Properties of Acids and Bases Chapter 14 Acids and Bases Svante Arrhenius (1859-1927) First to develop a theory for acids and bases in aqueous solution Arrhenius Acids Compounds which dissolve (dissociate)
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
Talk n Acids & Bases... Lady Dog! Definitions So far in this course, we have looked at processes in chemistry that deal with, or are best explained by, ionic salts or molecules. Now we will turn our attention
16.1 Acids and Bases: A Brief Review Arrhenius concept of acids and bases: an acid increases [H + ] and a base increases [OH ]. 16.2 BrønstedLowry Acids and Bases In the BrønstedLowry system, a BrønstedLowry
Slide 1 Chapter 14 Aqueous Equilibria: Acids and Bases Slide 2 Acid Base Concepts 01 Arrhenius Acid: A substance which dissociates to form hydrogen ions (H + ) in solution. HA(aq) H + (aq) + A (aq) Arrhenius
Aqueous Equilibria, Part 1 AP Chemistry Lecture Outline Name: Acids and Bases Arrhenius...acids increase the when dissolved in H 2 O....bases increase the when dissolved in H 2 O. e.g., HCl and NaOH Bronsted-Lowry
Chpt 16 Acids and Bases Defining Acids Arrhenius: Acid: Substances when dissolved in water increase the concentration of H+. Base: Substances when dissolved in water increase the concentration of OH- Brønsted-Lowry:
Molecular Definitions of Acids and Bases: Arrhenius Definition: An acid produces hydrogen ions (H + ) in an aqueous solution. For example: Consider hydrochloric acid, HCl, which when dissolved in water
Acids and Bases Properties, Reactions, ph, and Titration C-19 2017 Properties of acids 1. Taste Sour (don t try this except with foods). 2. Are electrolytes (conduct electricity). Some are strong, some
Lecture Presentation Chapter 16 Acid Base Equilibria John D. Bookstaver St. Charles Community College Cottleville, MO Some Definitions Arrhenius An acid is a substance that, when dissolved in water, increases
Chapter 14 Acids and Bases I. Bronsted Lowry Acids and Bases a. According to Brønsted- Lowry, an acid is a proton donor and a base is a proton acceptor. Therefore, in an acid- base reaction, a proton (H
CHAPTER 14 ACIDS AND BASES Topics Definition of acids and bases Bronsted-Lowry Concept Dissociation constant of weak acids Acid strength Calculating ph for strong and weak acids and bases Polyprotic acids
Chem 106 Thursday, March 10, 2011 Chapter 17 Acids and Bases K a and acid strength Acid + base reactions: Four types (s +s, s + w, w + s, and w + w) Determining K from concentrations and ph ph of aqueous
Chapter 10 - Acids & Bases 10.1-Acids & Bases: Definitions Arrhenius Definitions Acids: substances that produce hydrogen ions when dissolved in H 2 O Common Strong Acids: Common Weak acids: Organic carboxylic
Chemistry, The Central Science, 11th edition Theodore L. Brown, H. Eugene LeMay, Jr., Bruce E. Bursten Chapter 16 John D. Bookstaver St. Charles Community College Cottleville, MO Some Definitions Arrhenius
P a g e 1 Chapter 16 ACID-BASE EQUILIBRIA Nature of Acids and Bases Before we formally define acids and bases, let s examine their properties. Properties of Acids Sour taste Ability to dissolve many metals
I. A/B General A. Definitions 1. Arrhenius 2. B-L 3. Lewis *H+=H3O+ B. Properties 1. Acid 2. Base II. ACIDS A. Types 1. Binary 2. Ternary 3. Acid Anhydride B. Protic 1. mono 2. di 3. tri C. Strengths 1.
Acids and Bases Acid/Base Definitions Arrhenius Model Acids produce hydrogen ions in aqueous solutions Bases produce hydroxide ions in aqueous solutions Bronsted-Lowry Model Acids are proton donors Bases
Unit 12: Acids & Bases Aim: What are the definitions and properties of an acid and a base? Mar 23 12:08 PM Properties of an Acid 3. Are electrolytes. (Dissociate and conduct electricity when aq) 2. Turns
CHAPTER 14 THE CHEMISTRY OF ACIDS AND BASES "ACID"--Latin word acidus, meaning sour. (lemon) "ALKALI"--Arabic word for the ashes that come from burning certain plants; water solutions feel slippery and
Unit 4 Toxins, Section IV, L17-22 Lesson 17 Heartburn Lesson 18 Pass the Proton Lesson 19 phooey! Lesson 20 Watered Down Lesson 21 Neutral Territory Lesson 22 Drip Drop Acids and Bases What are the properties
Assessment Chapter Test B Chapter: Acids and Bases PART I In the space provided, write the letter of the term or phrase that best completes each statement or best answers each question. 1. Which of the