Acids and Bases. Chapter 16

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1 Acids and Bases Chapter 16

2 The Arrhenius Model An acid is any substance that produces hydrogen ions, H +, in an aqueous solution. Example: when hydrogen chloride gas is dissolved in water, the following ions are produced. HCl (g) H + (aq) + Cl - (aq) A base is an substance that produces hydroxide ions, OH -, in an solution. Example: When solid sodium hydroxide is dissolved in water the following ions are produced. NaOH (s) Na + (aq) + OH - (aq) 16-2

3 Another Theory: Brønsted-Lowry Model The Arrhenius model is limiting in its classification of acids and bases, suggesting there is only one kind of acid or base. A more general definition was suggested by a Danish chemist named Johannes Brønsted and an English chemist named Thomas Lowry. The Brønsted-Lowry Model states: An acid is a proton (H + ) donor A base is a proton (H + ) acceptor. 16-3

4 Brønsted-Lowry Model Let s look at a general reaction for an acid (HA) in water. HA (aq) + H 2 O (l) H 3 O + (aq) + A - (aq) In the reaction which reactant was the proton donor (acid)? Which reactant was the proton acceptor (base)? If we look at the products, we see that now we have role reversal. Which product is a proton donor? Which product is a proton acceptor? In this case, the proton donor in the products side is known as the conjugate acid, and the proton acceptor is known as the conjugate base. 16-4

5 Practice In the following reactions identify the acid, base, conjugate acid, and conjugate base. 1. HF (aq) + H 2 O (l) H 3 O + (aq) + F - (aq) 2. H 2 S (aq) + H 2 O (l) H 3 O + (aq) + HS - (aq) 3. NH 3(aq) + H 2 O (aq) NH 4 + (aq) + OH - (aq) 4. HSO 4- + C 2 O 4 2- SO HC 2 O 4-5. Write the conjugate bases for the following acids. a. HClO 4 b. HNO 3 c. HC 2 H 3 O

6 16.2 Acid Strength When we put an acid or base in water, the compound breaks apart into it s respective ions is called dissociation. The degree which a compound dissociates in water determines the strength of the acid or base. Strong acids and bases are substances that are completely ionized, or completely dissociated in solution. Weak acids and bases are substances that only ionize or dissociate partially in solution. 16-6

7 Figure 16.1: Representation of the behavior of acids of different strengths in aqueous solution. 16-7

8 Acids vs Conjugate Bases Strong Acids Acid Strength Increases Weak Acids Acid Conjugate Base HClO 4 ClO - 4 HI I - HBr Br - HCl Cl - H 2 SO 4 HSO - 4 HNO 3 NO - 3 H 3 O + H 2 O HSO - 4 SO 2-4 HF F - HNO 2 NO - 2 HCOOH HCOO - CH 3 COOH CH 3 COO - NH + 4 NH 3 HCN CN - H 2 O OH - Weak Bases Base Strength Increases Strong Bases NH 3 NH

9 16.3 Water as an Acid and a Base A substance is said to be amphoteric if it can behave either as an acid or as a base. Water is the most common amphoteric substance. Let s look at the ionization of 2 water molecules. H 2 O (l) + H 2 O (l) H 3 O + (aq) + OH - (aq) In this reaction one water molecule acts as an acid, and one acts as a base. At 25 C the concentrations of these ions have been calculated repeatedly to be [H 3 O + ]=[OH - ]= 1.0 x 10-7 M 16-9

10 Ion-Product Constant, K w [H 3 O + ][OH - ] = 1.0 x = K w We call this constant, 1.0 x 10-14, K w or the ion-product constant for water. K w = [H + ][OH - ] = 1.0 x In any aqueous solution at 25 C, no matter what it contains, the product of [H + ] and [OH - ] must always equal 1.0 x This means if the [H + ] goes up, the [OH - ] must go down so the product does not change. [H + ] = the concentration of H ions 16-10

11 3 Types of Situations In an aqueous solution, there are 3 possible situations. Each one always maintains a K w =1.0 x A neutral solution, where [H + ]=[OH - ] 2. An acidic solution, where [H + ]>[OH - ] 3. A basic solution, where [H + ]<[OH - ] Example: Calculate the [H + ] and determine the type of solution if you have 1.0 x 10-5 M OH - What information do you know from this equation? K w = [H + ][OH - ] Rearrange to find [H + ] = K w = 1.0 x = [OH - ] 1.0 x 10-5 Is this solution acidic, basic, or neutral? 16-11

12 Practice Problems Determine the [H + ] and [OH - ] and whether the solution is acidic, basic or neutral M H x 10-7 M OH x 10-4 M H x 10-8 M H

13 16.4 The ph Scale Calculating the [H + ] and [OH - ] brings a lot of very small numbers which is rather inconvenient. In 1909 a French chemist, named Soren Sorensen, came up with the ph scale. ph, literally means power of hydrogen When calculating the ph we use the following: ph = - log [H + ] Calculate the ph of a solution with [H + ] = 1.0 x 10-7 M Acidic solution ph < 7.00 Basic solution ph > 7.00 Neutral solution ph=

14 The ph Scale On the ph scale, each increase of 1 unit equals a power of ten change in the [H + ]. A solution with a ph of 3, has a [H + ] = 10-3 M, which is 10 greater than a solution with ph of 4, or [H + ] of 10-4 M and 100 times greater than a ph of 5 or 10-5 =[H + ]. As the [H + ] increases, the ph decreases. The ph scale runs values from 0-14, zero being very acidic, 14 being very basic, and 7 being neutral

15 Figure 16.3: The ph scale. Use the scale to determine if the following are acidic, neutral or basic: a. grapefruit ph=3.2 b. orange juice ph=3.5 c. urine ph= (depends on H 2 O) d. saliva ph= e. milk ph=6.5 f. gastric juice in stomach ph=2.0 g. blood ph= h. tears ph=7.4 i. milk of magnesia ph=10.6 j. ammonia ph=11.4 k. Draino ph=

16 Calculating poh Sometimes we find it easier to calculate the poh, since we are given the [OH - ]. poh = -log [OH - ] Calculate the poh of [OH - ] = 1.0 x 10-2 M The poh is not commonly used to describe a solution, so we must convert the poh to the ph ph + poh = 14 What is the ph of our poh solved at the top of this slide? 16-16

17 Calculating ph and poh Practice Calculate the ph and poh for the following: 1. [H + ] = 1.0 x 10-4 M 2. [OH - ] = 1.0 x 10-3 M 3. [H + ] = 1.0 x M 4. Determine whether each of the solutions above is acidic, basic or neutral. NOTE: When determining sig figs for logs, the number of decimal places for a log must equal the number of significant figures in the original number

18 Calculating the [H + ] from the ph Sometimes the ph is available, but we need to know the [H + ] of the solution. In this case we need to find the inverse log of the ph. [H + ] = inverse log (-ph) On your calculator you may enter the ph, then push the inverse key and then the log. Your calculator may also have a 10 x key, typically above the log key. With this key you would enter your ph, then push the 10 x key. Calculate the [H + ] if the ph is The same sequence is used for [OH - ] and poh

19 Practice Problems The ph of a human blood sample was measured to be What is the [H + ] in this blood? The poh of a liquid drain cleaner was found to be What is the [OH - ] for this cleaner? 16-19

20 The ph curve for the titration of 25.0 ml of M HNO 3 with M NaOH. If you graphed the results from the computer simulation, it would look like this graph

21 Neutralization Reaction Net Ionic Equation Since the acid and base in the reaction are strong, they will both dissociate completely in water. Write the following as an ionic equation: HCl (aq) + NaOH (aq) NaCl (aq) + H 2 O (l) H + (aq) +Cl - (aq) +Na + (aq) +OH - (aq) Na + +Cl - + H 2 O (l) We can then rewrite this as a NIE: H + (aq) + OH - (aq) H 2 O (l) The above is a neutralization reaction. It is called so because if equal amounts of H + and OH - are available for reaction, a neutral solution will result

22 Calculating Volume in Neutralization Reactions What volume of M HCl is needed to neutralize 25.0 ml of a M NaOH solution? M acid V acid (#H + ) + M base V base (#OH - ) (0.100M)?V (1) = 0.35M(.025L)(1) V= (0.35M)(.025L) M V= L or 87.5mL OR YOU CAN CALCULATE USING STOICHIOMETRY! 16-22

23 Step 1: Write a balanced net ionic equation for the reaction. H + (aq) + OH - (aq) H 2 O (l) Step 2: Calculate the moles of reactants ml NaOH x 1L x mol NaOH ml 1 L NaOH = 8.75 x 10-3 mol OH - Step 3: Determine which reactant is limiting. This problem requires the addition of just enough H + ions to react exactly with the OH - ions present. The number of moles of OH - ions present determines the number of moles of H + ions. Therefore the OH - are limiting

24 Step 4: Calculate the moles of H + required x 10-3 mol OH - x 1 mol H + = 8.75 x 10-3 mol H mol OH - Step 5: Calculate the volume of M HCl required. We know that HCl:H + is a 1:1 ratio, so the [H + ] is M. Volume x mol H + = 8.75 x 10-3 mol H + 1L Now we must solve for volume. Volume = 8.75 x 10-2 L or 87.5 ml 16-24

25 Practice Problem 1. Calculate the volume of 0.10 M HNO 3 needed to neutralize 125 ml of M KOH

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