Properties of Acids Sour taste Ability to dissolve many metals Ability to neutralize bases Change blue litmus paper to red

Properties of Bases Taste bitter alkaloids = plant product that is alkaline  often poisonous Feel slippery Ability to turn red litmus paper blue Ability to neutralize acids

Some Definitions Arrhenius acids and bases –An acid is a substance that, when dissolved in water, increases the concentration of hydrogen ions. –A base is a substance that, when dissolved in water, increases the concentration of hydroxide ions. © 2012 Pearson Education, Inc.

Arrhenius Theory Acids: produce H + ions in aqueous solution. HCl(aq) → H + (aq) + Cl − (aq)

Hydronium Ion The H + ions produced by the acid are so reactive they cannot exist in water. –H + ions are protons! Instead, they react with water molecules to produce complex ions, mainly hydronium ion, H 3 O +.H + + H 2 O  H 3 O + –There are also minor amounts of H + with multiple water molecules, H(H 2 O) n +.

Arrhenius Theory Bases: produce OH − ions in aqueous solution. NaOH(aq) → Na + (aq) + OH(aq)

Some Definitions Brønsted–Lowry –An acid is a proton donor. –…must have a removable (acidic) proton –A base is a proton acceptor. –…must have a pair of nonbonding electrons © 2012 Pearson Education, Inc.

If it can be either… …it is amphiprotic or amphoteric. HCO 3  HSO 4  H2OH2O © 2012 Pearson Education, Inc.

A.Both substances act as Brønsted-Lowry bases. B.Neither substance acts as a Brønsted-Lowry base. C.H 2 S(aq) D.CH 3 NH 2 (aq)

Conjugate Acids and Bases The term conjugate comes from the Latin word “conjugare,” meaning “to join together.” Reactions between acids and bases always yield their conjugate bases and acids. © 2012 Pearson Education, Inc.

Which of the following is the conjugate acid of SO 4 2– ? a.H 2 SO 4 b.HSO 4 1– c.SO 3 2– d.H 3 SO Add an H, take away an electron

Which of the following is the conjugate base of HPO 4 2– ? a.H 3 PO 4 b.H 2 PO 4 1– c.PO 4 3– d.HPO 3 2– - Take away the H, add an electron

Acid and Base Strength Strong acids are completely dissociated in water. –Their conjugate bases are negligible. Weak acids only dissociate partially in water. –Their conjugate bases are weak bases. © 2012 Pearson Education, Inc.

Acid and Base Strength Substances with negligible acidity do not dissociate in water. –Their conjugate bases are exceedingly strong. © 2012 Pearson Education, Inc.

Strong or Weak A strong acid is a strong electrolyte. –Practically all the acid molecules ionize. A strong base is a strong electrolyte. –Practically all the base molecules form OH – ions, either through dissociation or reaction with water. A weak acid is a weak electrolyte. –Only a small percentage of the molecules ionize, . A weak base is a weak electrolyte –only a small percentage of the base molecules form OH – ions, either through dissociation or reaction with water, .

Strong Acids Strong acids donate practically all their H’s. 100% ionized in water Strong electrolyte [H 3 O + ] = [strong acid] [X] means the molarity of X

Weak Acids Weak acids donate a small fraction of their H’s. Most of the weak acid molecules do not donate H to water. Much less than 1% ionized in water [H 3 O + ] << [weak acid]

Ionic Attraction and Acid Strength

A.Strong base B.Weak base C.Negligible basicity D.Never a base

Acid and Base Strength In any acid–base reaction, the equilibrium will favor the reaction that moves the proton to the stronger base. HCl (aq) + H 2 O (l)  H 3 O + (aq) + Cl  (aq) H 2 O is a much stronger base than Cl , so the equilibrium lies so far to the right that K is not measured (K >> 1). Product favored. © 2012 Pearson Education, Inc.

Acid and Base Strength In any acid–base reaction, the equilibrium will favor the reaction that moves the proton to the stronger base: Acetate is a stronger base than H 2 O, so the equilibrium lies to the left side (K < 1). Reactant favored. Stop here CH 3 CO 2 H (aq) + H 2 O (l) H 3 O + (aq) + CH 3 CO 2  (aq) © 2012 Pearson Education, Inc.

BELL QUESTIONS: For the following proton-transfer reaction use your chart to predict whether the equilibrium lies to the left (K c 1) being product favored: The CO 3 2  ion appears lower in the right-hand column and is therefore a stronger base than SO 4 2 . CO 3 2 , therefore, will get the proton preferentially to become HCO 3 , while SO 4 2  will remain mostly unprotonated. The resulting equilibrium lies to the right, favoring products (that is, K c > 1):

A Brønsted-Lowry base is a.a proton donor. b.a proton acceptor. c.an electron-pair donor. d.an electron-pair acceptor.

A Brønsted-Lowry acid is a.a proton donor. b.a proton acceptor. c.an electron-pair donor. d.an electron-pair acceptor.

A.Na + and OH – B.H + and OH – CNa + and Cl – D.H + and Cl –

Autoionization of Water water is amphiprotic. a few molecules act as bases and a few act as acids. This is referred to as autoionization. © 2012 Pearson Education, Inc.

Ion Product Constant-Kw The equilibrium expression for this process is K c = [H + ] [OH  ] This special equilibrium constant is referred to as the ion product constant for water, K w. At 25  C, K w = 1.0  10  14 © 2012 Pearson Education, Inc.

pH pH is defined as the negative base-10 logarithm of the concentration of hydronium ion: pH =  log [H + ] © 2012 Pearson Education, Inc.

pH In pure water, K w = [H + ] [OH  ] = 1.0  10  14 Since in pure water, [H + ] = [OH  ], [H + ] =  1.0  10  14 = 1.0  10  7 © 2012 Pearson Education, Inc.

pH Therefore, in pure water, pH =  log (1.0  10  7 ) = 7.00 An acid has a higher [H + ] than pure water, so its pH is <7. A base has a lower [H + ] than pure water, so its pH is >7. © 2012 Pearson Education, Inc.

pH © 2012 Pearson Education, Inc.

Other “p” Scales The “p” in pH tells us to take the negative base-10 logarithm of the quantity (in this case, hydronium ions). Some similar examples are –pOH:  log [OH  ] –pK w :  log K w –pK a :  log K a –pK b :  log K b © 2012 Pearson Education, Inc.

Watch This! Because [H + ] [OH  ] = K w = 1.0  10  14 we know that  log [H + ] +  log [OH  ] =  log K w = or, in other words, pH + pOH = pK w = © 2012 Pearson Education, Inc.

Calculating [H + ] from [OH  ] Calculate the concentration of H + (aq) in (a) a solution in which [OH  ] is M, (b) a solution in which [OH  ] is 1.8  10  9 M. Note: In this problem and all that follow, we assume, unless stated otherwise, that the temperature is 25  C. [OH  ] > [H + ] [H + ] > [OH  ]

Calculating pH from [H + ] Calculate the pH values for the two previous solutions: (a) In the first instance we found [H + ] to be 1.0  10  12 M, so that pH =  log(1.0  10  12 ) =  (  12.00) = Because 1.0  10  12 has two significant figures, the pH has two decimal places, (b) For the second solution, [H + ] = 5.6  10  6 M. Before performing the calculation, it is helpful to estimate the pH. To do so, we note that [H + ] lies between 1  10  6 and 1  10  5. Thus, we expect the pH to lie between 6.0 and 5.0. pH =  log(5.6  10  6 ) = 5.25

Calculating [H + ] from pOH A sample of freshly pressed apple juice has a pOH of Calculate [H + ]. pH =  pOH pH = – = 3.76 Next we use: pH =  log [H + ] = 3.76 Thus, log [H + ] =  3.76 To find [H+], we need to determine the antilogarithm of  Scientific calculators have an antilogarithm function (sometimes labeled INV log or 10 x ) that allows us to perform the calculation: [H + ] = antilog (  3.76) = 10  3.76 = 1.7 × 10  4 M

Calculating the pH of a Strong Acid What is the pH of a M solution of HClO 4 ? Because HClO 4 is a strong acid, it is completely ionized, giving [H + ] = [ClO 4  ] = M. Solve pH =  log(0.040) = 1.40 Check Because [H + ] lies between 1  10  2 and 1  10  1, the pH will be between 2.0 and 1.0. Our calculated pH falls within the estimated range. Furthermore, because the concentration has two significant figures, the pH has two decimal places.

Calculating the pH of a Strong Base What is the pH of (a) a M solution of NaOH, (b) a M solution of Ca(OH) 2 ? (a) NaOH dissociates in water to give one OH  ion per formula unit. Therefore, the OH  concentration for the solution in (a) equals the stated concentration of NaOH, M. Method 1: Method 2: pOH =  log(0.028) = 1.55 pH = – pOH = 12.45

Continued (b) Ca(OH) 2 is a strong base that dissociates in water to give two OH  ions per formula unit. Thus, the concentration of OH  (aq) for the solution in part (b) is 2  ( M) = M. Method 1: Method 2: pOH =  log(0.0022) = 2.66 pH =  pOH = Stop here, do problems as class.

How Do We Measure pH? For less accurate measurements, one can use –Litmus paper “Red” paper turns blue above ~pH = 8 “Blue” paper turns red below ~pH = 5 –Or an indicator. © 2012 Pearson Education, Inc.

How Do We Measure pH? For more accurate measurements, one uses a pH meter, which measures the voltage in the solution. © 2012 Pearson Education, Inc.

Which of these indicators is best suited to distinguish between a solution that is slightly acidic and one that is slightly basic? A.Thymol blue B.Bromothymol blue C.Phenophthalein D.Yellow blue

What is the pH of a M aqueous solution of HBr? a.1.00 b.1.70 c.2.30 d.12.30

What is the pH of a M aqueous solution of KOH? a b c.4.00 d.1.40

A M solution of Ba(OH) 2 has the higher pH. B M solution of NaOH has the higher pH. C.Both solutions have the same pH as both have the same concentration.

A.O B.H C.N D.C

Dissociation Constants For a generalized acid dissociation, the equilibrium expression would be This equilibrium constant is called the acid-dissociation constant, K a. [H 3 O + ] [A  ] [HA] Ka = HA (aq) + H 2 O (l) A  (aq) + H 3 O + (aq) © 2012 Pearson Education, Inc.

Dissociation Constants The greater the value of K a, the stronger the acid. Dissoc constants are needed to calculate ph of weak acids/bases. © 2012 Pearson Education, Inc.

Calculating K a from the pH of a Weak Acid The pH of a 0.10 M solution of formic acid, HCOOH, at 25  C is Calculate K a for formic acid at this temperature. We know that [H + ] [HCOO  ] [HCOOH] K a = © 2012 Pearson Education, Inc.

Calculating K a from the pH The pH of a 0.10 M solution of formic acid, HCOOH, at 25  C is Calculate K a for formic acid at this temperature. To calculate K a, we need the equilibrium concentrations of all three things. We can find [H + ], which is the same as [HCOO  ], from the pH. © 2012 Pearson Education, Inc.

Calculating K a from the pH pH=  log [H + ] 2.38=  log [H + ]  2.38= log [H + ] 10  2.38 = 10 log [H + ] = [H + ] 4.2  10  3 = [H + ] = [HCOO  ] © 2012 Pearson Education, Inc.

Calculating K a from pH [HCOOH], M[H + ], M[HCOO  ], M Initially Change At equilibrium Now we can set up a table… © 2012 Pearson Education, Inc.

Calculating K a from pH [HCOOH], M[H + ], M[HCOO  ], M Initially Change  4.2  10   10  3 At equilibrium Now we can set up a table… © 2012 Pearson Education, Inc.

Calculating K a from pH [HCOOH], M[H + ], M[HCOO  ], M Initially Change  4.2  10   10  3 At equilibrium 0.10  4.2  10  3 = =  10  3 Now we can set up a table… © 2012 Pearson Education, Inc.

Calculating K a from pH [4.2  10  3 ] [0.10] K a = = 1.8  10  4 © 2012 Pearson Education, Inc.

Calculating Percent Ionization – this is also the verification of the small approximation Percent ionization =  100 In this example, [H + ] eq = 4.2  10  3 M [HCOOH] initial = 0.10 M [H + ] eq [HA] initial Percent ionization =   10  = 4.2% STOPPED HERE © 2012 Pearson Education, Inc.

Weak Acid - Calculating pH from K a Calculate the pH of a 0.30 M solution of acetic acid, HC 2 H 3 O 2, at 25  C. HC 2 H 3 O 2 (aq) + H 2 O (l) H 3 O + (aq) + C 2 H 3 O 2  (aq) K a for acetic acid at 25  C is 1.8  10  5. © 2012 Pearson Education, Inc.

Calculating pH from K a The equilibrium constant expression is [H + ] [C 2 H 3 O 2  ] [HC 2 H 3 O 2 ] K a = © 2012 Pearson Education, Inc.

Calculating pH from K a [C 2 H 3 O 2 ], M[H 3 O + ], M[C 2 H 3 O 2  ], M Initially Change At equilibrium We next set up a table… We are assuming that x will be very small compared to 0.30 and can, therefore, be ignored. © 2012 Pearson Education, Inc.

Calculating pH from K a [C 2 H 3 O 2 ], M[H 3 O + ], M[C 2 H 3 O 2  ], M Initially Change xx +x+x+x+x At equilibrium We next set up a table… We are assuming that x will be very small compared to 0.30 and can, therefore, be ignored. © 2012 Pearson Education, Inc.

Calculating pH from K a [C 2 H 3 O 2 ], M[H 3 O + ], M[C 2 H 3 O 2  ], M Initially Change xx +x+x+x+x At equilibrium 0.30  x  0.30 xx We next set up a table… We are assuming that x will be very small compared to 0.30 and can, therefore, be ignored. © 2012 Pearson Education, Inc.

Calculating pH from K a Now, (x) 2 (0.30) 1.8  10  5 = (1.8  10  5 ) (0.30) = x  10  6 = x  10  3 = x © 2012 Pearson Education, Inc.

Calculating pH from K a pH =  log [H 3 O + ] pH =  log (2.3  10  3 ) pH = 2.64 © 2012 Pearson Education, Inc. RULE *always check your assumption!!! If eq conc/initial conc x 100= more than 5% you will need to go back and solve the quadratic!!!!!

A.Because K a for weak acids slightly less than 1 and therefore, the extent of ionization is very small and can be neglected. B.Because most weak acids of low to moderate concentration undergo very little ionization, often less than 1% in solution, and thus the extent of ionization can be neglected. C.Because most weak acids of high or low concentration will form minimal conjugate base and thus the extent of ionization can be neglected. DBecause for each weak acid dissociating in solution, a weak acid molecule is formed with minimal conjugate base forming and thus the extent of ionization can be neglected.

Polyprotic Acids Polyprotic acids have more than one acidic proton. If the difference between the K a for the first dissociation and subsequent K a values is 10 3 or more, the pH generally depends only on the first dissociation. See page 731 in your text for an example. © 2012 Pearson Education, Inc.

Weak Bases Bases react with water to produce hydroxide ion. © 2012 Pearson Education, Inc.

Weak Bases The equilibrium constant expression for this reaction is [HB] [OH  ] [B  ] K b = where K b is the base-dissociation constant. © 2012 Pearson Education, Inc.

Weak Bases K b can be used to find [OH  ] and, through it, pH. © 2012 Pearson Education, Inc. *You already know how to get from [OH] to ph!!!

Weak Base - pH of Basic Solutions What is the pH of a 0.15 M solution of NH 3 ? [NH 4 + ] [OH  ] [NH 3 ] K b = = 1.8  10  5 NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH  (aq) © 2012 Pearson Education, Inc.

pH of Basic Solutions [NH 3 ], M[NH 4 + ], M[OH  ], M Initially At equilibrium Tabulate the data. © 2012 Pearson Education, Inc.

pH of Basic Solutions [NH 3 ], M[NH 4 + ], M[OH  ], M Initially At equilibrium 0.15  x  0.15 xx Tabulate the data. © 2012 Pearson Education, Inc.

pH of Basic Solutions (1.8  10  5 ) (0.15) = x  10  6 = x  10  3 = x 2 (x) 2 (0.15) 1.8  10  5 = © 2012 Pearson Education, Inc.

pH of Basic Solutions Therefore, [OH  ] = 1.6  10  3 M pOH =  log (1.6  10  3 ) pOH = 2.80 pH =  2.80 pH = © 2012 Pearson Education, Inc.

What is the pH of a M aqueous solution of NH 3 ? The K b of NH 3 is 1.8  10 –5. a.1.00 b.4.74 c.9.36 d.11.13

K a and K b K a and K b are related in this way: K a  K b = K w Therefore, if you know one of them, you can calculate the other. Stop here day 3 or 4. © 2012 Pearson Education, Inc.

For a conjugate acid- base pair, K w = K a ___ K b. a.+ b.– c.  d./

A few notes on pK….. A way of expressing the strength of an acid or base is pK. pK a = −log(K a ), K a = 10 −pKa pK b = −log(K b ), K b = 10 −pKb The stronger the acid, the smaller the pK a. –Larger K a = smaller pK a –Because it is the –log The stronger the base, the smaller the pK b. –Larger K b = smaller pK b

Reactions of Anions with Water Anions are bases. As such, they can react with water in a hydrolysis reaction to form OH  and the conjugate acid: X  (aq) + H 2 O (l) HX (aq) + OH  (aq) © 2012 Pearson Education, Inc.

Reactions of Cations with Water Cations with acidic protons (like NH 4 + ) will lower the pH of a solution. Most metal cations that are hydrated in solution also lower the pH of the solution. © 2012 Pearson Education, Inc.

Reactions of Cations with Water Attraction between nonbonding electrons on oxygen and the metal causes a shift of the electron density in water. This makes the O–H bond more polar and the water more acidic. © 2012 Pearson Education, Inc.

Reactions of Cations with Water Greater charge and smaller size make a cation more acidic. © 2012 Pearson Education, Inc.

Effect of Cations and Anions 1.An anion that is the conjugate base of a strong acid will not affect the pH. 2.An anion that is the conjugate base of a weak acid will increase the pH. 3.A cation that is the conjugate acid of a weak base will decrease the pH. © 2012 Pearson Education, Inc.

Effect of Cations and Anions 4.Cations of the strong Arrhenius bases will not affect the pH. 5.Other metal ions will cause a decrease in pH. 6.When a solution contains both a weak acid and a weak base, the affect on pH depends on the K a and K b values. Stop here. © 2012 Pearson Education, Inc.

Factors Affecting Acid Strength The more polar the H–X bond and/or the weaker the H–X bond, the more acidic the compound due to size. So acidity increases from left to right across a row and from top to bottom down a group. © 2012 Pearson Education, Inc.

Factors Affecting Acid Strength In oxyacids, in which an –OH is bonded to another atom, Y, the more electronegative Y is, the more acidic the acid. © 2012 Pearson Education, Inc.

Factors Affecting Acid Strength For a series of oxyacids, acidity increases with the number of oxygens. © 2012 Pearson Education, Inc.

Factors Affecting Acid Strength Resonance in the conjugate bases of carboxylic acids stabilizes the base and makes the conjugate acid more acidic. © 2012 Pearson Education, Inc.

Lewis Acids Lewis acids are defined as electron-pair acceptors or electrophiles. Atoms with an empty valence orbital can be Lewis acids. © 2012 Pearson Education, Inc.

Lewis Bases Lewis bases are defined as electron-pair donors or nucleophiles. Anything that could be a Brønsted–Lowry base is a Lewis base. Lewis bases can interact with things other than protons, however. © 2012 Pearson Education, Inc.