Wednesday, February 25, Acid and Base Reactions

Acid and Base Reactions

Dilute aqueous solution of acetic acid, C3COO Aqueous solution of sodium hydroxide, NaO

The role of the ydrogen Ion Cl (aq) Æ + (aq) + Cl - (aq) What does the neutral atom consist of? What must + ion then be? A PROTON!

Definitions of acids and bases Svante Arrhenius (Sweden) 1859-1927 Johannes Bronsted (Denmark) 1879-1947 G. N. Lewis (U.S.) 1875-1946

Arrhenius definitions of acids and bases An acid dissolves in water to yield protons X + (aq) + X (aq) A base dissolves in water to yield hydroxide ions Y O Y + (aq) + O (aq)

Bronsted Definition An acid is a proton donor A base is a proton acceptor

Proton transfer from Cl to water. hydronium ion + O Cl O C Cl

A base is a proton acceptor O Cl O Cl C

Ammonia is a Bronsted base N O N + A Weak Base O hydroxide ion

Example of Acid/Base Neutralization Cl(aq) + NaO(aq) NaCl(aq)??? + 2 O (l ) Acid + Base Salt + Water Complete ionic equation Na + (aq) + O (aq) + + (aq) + Cl (aq) Na + (aq) + Cl (aq) + 2 O(l) Cancel Spectator ions Net ionic equation O (aq) + + (aq) 2 O(l )

Bronsted Definition An acid is a proton donor A base is a proton acceptor Conjugate acid-base pair

Consider the transfer of a proton from Cl to water. + O Cl O base acid Conjugate acid Cl Conjugate base

Acetic acid O C 3 CO O + 2 O C 3 CO + 3 O + acid base Conjugate base Conjugate acid

Practice Exercise N C You complete this reaction and identify the conjugate acid-base pairs. O N C O base acid Conjugate acid Conjugate base

Equilibrium constant for proton transfer + O Cl O Check cheat sheet and write down the chemical reaction that goes with its Ka. Ka = K a = [ 3 O + ][Cl - ] [Cl] [ + ][Cl - ] [Cl] Cl

The Acid - Base Properties of Water

Water as an Acid and a Base O O + O O base acid Conjugate acid Conjugate base

Ion Product of Water ( K w ) The equilibrium constant for the ionization of water into + and O - ions

Ion Product of Water ( K w ) base O acid O + O Conjugate acid 2 2 O (l) 3 O + (aq) + O - (aq) O Conjugate base K = K = [ 3 O + ] [ O - w ] = [ + ] [ O - ] = 1.0 x 10-14 at 25 ºC

Practice Exercise Calculate the concentration of O - ions in a hydrochloric acid solution whose hydrogen ion concentration is 1.3 M. K w = 1 x 10-14 = [ 3 O + ] [O - ] 1 x 10-14 = ( 1.3 ) [O - ] 7.7 x 10-15 M = [O - ]

p A Measure of Acidity

The p Scale p = - log 10 [ + ]

Log Review Given log10(x) = y Then y is the exponent which, when you raise 10 to that power, you get x. Or x = 10 y. solving for y log(10) = log(100) = log(0.0001) = log(1) = 1 2-4 0 solving for x log(x) = 3 what is x? x = 10 3 = 1000

p of a Neutral solution - just water Strategy Find [ + ] from the autoionization of water and then take -log of it. recall K w = [ + ] [ O - ] = 1.0 x 10-14 I C E 2 2 O (l) 3 O + (aq) + O - (aq) 0 M 0 M +x +x +x +x K x = 1 x 10-7 M = [ + ] = [O - w = x 2 = 1 x 10-14 ] p = -log[ + ] = -log(1 x 10-7 ) = 7

The p of acidic and basic solutions K w = [ + ] [O - ] Acidic solution [ + ] > [ O - ] [ + ] > 1 x 10-7 M (higher [ + ] conc. than neutral) Ex p = - log ( 10-6 ) = acidic solutions have p < 7 Basic Ex p = - log ( 10-8 ) = = 1.0 x 10-14 solution [ + ] < [ O - ] basic solutions have p > 7 6 [ + ] < 1 x 10-7 M (lower [ + ] conc. than neutral) 8

the p scale is useful in that it lets us express acidity by numbers that are not exponentials but we must keep in mind that a difference of 1 p unit is equivalent to a factor of 10 in [ + ] concentration. thus, a solution with a p of 2 is not twice as concentrated in [ + ] as one with a p of 4, it is 10 2 (or 100) times as concentrated

p values of common fluids

Practice Exercise The p of a certain fruit juice is 3.33 Calculate [ + ]. p < 7 so solution is acidic/basic? p = 3.33 = - log 10 [ + ] log [ + ] = -3.33 [ + ] = 10-3.33 [ + ] = 4.7 x 10-4 M [ + ] > 10-7 so solution is acidic/basic?

Variation on K w K w = 1 x 10-14 = [ 3 O + ] [ O - ] log(k w ) = log(1 x 10-14 ) = log([ 3 O + ] [ O - ]) log(1) + log(10-14 ) = log([ 3 O + ]) + log([ O - ]) 0-14 = log([ 3 O + ]) + log([ O - ]) 14 = - log([ 3 O + ]) - log([ O - ]) 14 = p + po on cheat sheet?

Practice Exercise The O- concentration of a blood sample is 2.5 x 10-7 M. What is its p? roadmap a) convert [O - ] to po via -log([o - ]) b) convert po to p via p + po = 14 po = - log 10 [ O - ] = - log [ 2.5 x 10-7 ] po = 6.60 p = 14 - po p = 7.40 p + po = 14 perfectly fine alternative method a) convert [O - ] to [ + ] via Kw b) take -log([ + ])

Strength of Acid and Bases

there are relatively few strong acids the vast majority of acids are weak

The strength of acids and bases A strong acid is one that is completely ionized in water. A weak acid is one that partially ionizes in water to some extent less than 100%.

+ + + +

Strong acid Ex NO3, 1st + of 2SO4, Cl, Br, I, etc. Weak acid Ex acetic C3COO, all organic acids (R-COO), all other acids

Some strong acids Acid is strong ClO 4 Cl, Br, I NO 3 2 SO 4 Conjugate base is weak ClO - 4 Cl - Br - I - NO 3 - SO 4 -

Some weak acids Acid is weak F O (acetic C CO acid) 3 N 4 + Conjugate base is stronger F - C 3 COO - N 3

Some strong bases Base is strong (ydroxides of group 1A) NaO, KO,... (ydroxides of group 2A) Ba(O) 2

Practice Exercise Calculate the p of a 1.8 x 10-2 M Ba(O) 2 solution. Strategy For a strong base, use straight stoichiometry to find [O - ], then [ + ], then take -log of [ + ]. Ba(O) 2 Ba 2+ + 2 O - I 1.8 x 10-2 M 0 0 E 0 1.8 x 10-2 M 2(1.8 x 10-2 M) [ + ] [ O - ] = 1 x 10-14 [ + ] x 2( 1.8 x 10-2 ) = 1 x 10-14 p = 12.56 [ + ] = 1 x 10-14 3.6 x 10-2 = 2.8 x 10-13 M

Generalizations If an acid is strong (dissociates its + 100%), its conjugate base has no measurable strength (the weakest of the weak) because, essentially by definition, the strong acid it would form by gaining an + would immediately dissociate. The stronger the weak acid (the more completely its + dissociate), the weaker the conjugate base (the harder it is for the conjugate base to rip an + off water). 3 O + is the strongest of the weak acids. O - is the strongest base.

Relative strengths of conjugate acid-base pairs chloric

Relative strengths of conjugate acid-base pairs

The position of equilibrium lies to the side of the weaker acid. Why? O F + O F weaker acid Stronger acid The weaker acid by definition holds onto its + more tightly.

Practice Exercise In which direction does the equilibrium lie? Is there mostly NO 2 or CN floating around? Equilibrium Stronger acid lies to the right weaker acid NO + CN - 2 NO - 2 + CN K a = 4.0x10-4 K a = 6.2x10-10 First identify the acid on each side Second specify which acid is stronger The weaker acid by definition holds onto its + more tightly.

Practice Exercise In which direction does the equilibrium lie? Is there mostly C3COO or COO floating around? Weaker acid C 3 COO Stronger Equilibrium lies to the left acid + COO - 3 CCOO - + COO K a = 1.8x10-5 K a = 1.8x10-4 First identify the acid on each side Second specify which acid is stronger The weaker acid by definition holds onto its + more tightly.

Molecular Structure of Acids A + + A - Major factors affecting acid strength are Polarity of the bond to Strength of the bond to

Acidity of Binary Acids Increases going down a column strongest bond F weakest acid Cl weakest bond Br I strongest acid Bond strength is the major factor

Acidity of Binary Acids Increases going across a period C 4 N 3 2 O F weakest strongest acid acid Polarity of bond the major factor

Oxoacids Atom or group O Acidity increases with the ability of the group attached to oxygen to attract electrons toward itself

Oxoacids Atom in a high oxidation state δ O δ+ Atom in a high oxidation state O - +

Nitric Acid Nitrous Acid +5 oxidation state O +3 O O N + O N O - Strong Acid NO 3 Weak Acid NO 2

Compare Acid perchloric ClO 4 OClO 3 chloric ClO 3 OClO 2 chlorous +7 ClO 2 +7 +5 +5 +3 +3 OClO Ka ~10 9 1.0x10-1 1.2x10-2 hypochlorous +1 ClO +1 OCl 3.0x10-8

Group 1A or 2A Metal + - O these compounds are ionic and are strong bases

Lewis Acids and Bases

Lewis definitions X Y an acid is an electronpair acceptor a base is an electronpair donor X Y X Y

Example F 3 N + BF 3 N B F Lewis base Lewis acid F

Weak Acids and Acid Ionization Constants Acid ionization constants are a particular kind of equilibrium constant

Equilibrium constant for proton transfer + O + A O + A A (aq ) + 2 O 3 O + (aq ) + A - (aq ) K a [ 3 O + ][A - ] = [A ] Acid ionization constant,acid + WATER

The Base Ionization Constant K b Negatively charged base + + B - 2 O B O - base acid Conjugate acid Conjugate base K b = [B ] [O - ] [B - ] base ionization constant, BASE + WATER

The Base Ionization Constant K b neutral base + + B 2 O B + O - base acid Conjugate acid Conjugate base K b = [B + ] [O - ] [B] base ionization constant check cheat sheet. Which form of Kb is there?

Practice Exercise What is the p of a 0.122 M solution of a weak monoprotic acid A that has K a = 5.7 x 10-4 I C E A (aq) + (aq) + A - (aq) 0.122 M 0.00M 0.00 M - x + x +x 0.122 - x + x +x K a = [ + ] [A - ] [A] = (x) 2 (0.122 - x) = 5.7 x 10-4

[ + ] [F - ] (x) 2 [F] = (0.122 - x) Approximation.122 - x.122 = 5.7 x 10-4 Small K a x 2 0.122 = 5.7 x 10-4 p = -log [ + ] x = [ + ] =.008 M p = -log (.008 ) p = 2.1

Weak Bases and Base Ionization Constants

Ammonia is an example of a neutral molecule that is a weak base 3 N + base acid O + 3 N Conjugate acid - + O Conjugate base [N + 4 ] [O - ] K b = = 1.8 x 10-5 [ 3 N]

Example Methylamine + C 3 N 2 + O C 3 N 3 - + O K b = 4.4 x 10-4 Conjugate acid K a = 2.3 x 10-11

Practice Exercise K b = 1.8 x 10-5 What is the p of a 0.400 M ammonia solution N 3 (aq) N 4 + (aq) + O - (aq) I C E K b = 0.400 M 0.00M 0.00 M - x + x +x 0.400- x + x +x [N 4+ ] [O - ] [N 3 ] = (x) 2 = 1.8 x 10-5 (0.400 - x)

K a = [N 4+ ] [O - ] [N 3 ] = (x) 2 (0.400 - x) = 1.8 x 10-5 Approximation.400 - x.400 x 2 0.400 = 1.8 x 10-5 po = -log [ O - ] po = 2.57 x = [O - ] = 2.7 x 10-3 M p = 14.00-2.57 p = 11.43

The Relationship Between Conjugate Acid-Base Equilibrium Constants

For the two equilibria that involve a conjugate acid-base pair in aqueous solution K a A + + A - K b A - + 2 O A + O - 2 O + + O - Recall when adding two equilibria, multiply their equilibrium constants. K a K w K b = K w At 25 C K a K b = 1.0 x 10-14

K a K b = K w If K a is big; K b is small and vice versa the stronger the acid, the weaker its conjugate base the stronger the base, the weaker its conjugate acid

Relationships between K a and K b

Percent Ionization A (aq) + (aq) + A- (aq) percent ionization = [ A-] (at equilibrium) [ A] (original) x 100

Percent Ionization consider 1.00 M OAc K a = 1.8 x 10-5 OAc (aq) AcO - (aq) + (aq) + I C E 1.00 M 0.00M 0.00 M - x + x +x 1 - x + x +x K a = [ + ] [A - ] [OAc] = (x) 2 (1.00 - x) = 1.8 x 10-5

(x) 2 = 1.8 x 10-5 (1.00 - x) x = 0.0042 M [ AcO - ] % ionization = 100 x [ OAc ]o [ AcO - ] = 0.0042 M [OAc ] o = 1.00 M % ionization = 100 (0.0042) / 1.0 ) % ionization = 0.42%

Percent Ionization Is greater in more dilute solution acetic acid K a = 1.8 x 10-5 [acetic acid ] [ + ] %dissoc 1.00 M 0.0042 M 0.42% 0.100 M 0.0013 M 1.3%

Percent Ionization consider 3.00 M A and a % dissociation of 12% K a =? A (aq) + (aq) + A - (aq) I C 3.00 M 0.00M 0.00 M - x + x +x E 3.00 - x + x +x x 0.12 = x = 0.36 M 3.00 M K a = (0.36) (0.36) (3.00-0.36) = 4.9x10-2

Diprotic and Polyprotic Acids

Important Points When an acid has more than one ionizable proton, the protons are lost in successive reactions.

Phosphoric Acid 3 PO 4 (aq ) + (aq ) + 2 PO 4 -(aq ) 2 PO 4 -(aq ) + (aq ) + PO 4 2-(aq ) PO 4 2-(aq ) + (aq ) + PO 4 3-(aq )

Important Points When an acid has more than one ionizable proton, the protons are lost in successive reactions. Each ionization is characterized by a separate K a value. Each successive K a is smaller than the preceding one.

Phosphoric Acid 7.5 x 10-3 3 PO 4 (aq ) + (aq ) + 2 PO 4 -(aq ) 6.2 x 10-8 2 PO 4 -(aq ) + (aq ) + PO 4 2-(aq ) 4.8 x 10-13 PO 4 2-(aq ) + (aq ) + PO 4 3-(aq )

Sulfuric Acid 2 SO 4 is a strong acid in its first ionization; but a weak acid in its second. K a = very large O SO 2 O O SO 2 O - + + K a = 0.013 - - O SO 2 O O SO 2 O - + +

Calculate the p of a 0.056 M solution of 2 SO 4 2 SO 4 SO 4 -+ + I 0.056 M 0 0 For first ionization use stoichiometry E 0 0.056 M 0.056 M SO 4 - SO 4 2- + + I E 0.056 M 0 0.056 M 0.056 M - x x 0.056 M + x

SO 4 - SO 4 2- + +.056 - x x.056 + x K a = [SO 4 2-] [ + ] [SO 4 -] 0.013 = ( x )(.056) (.056).013 = ( x (.056 ) + x ) (.056 - x ) x =.013.013 100 x (.056) assume x is small compared 0.056 = 23%

SO - 2 SO 2-4 + +.056 - x x.056 + x K a =.013 = [SO 4 2-] [ + ] [SO 4 -] ( x (.056 ) + x ) (.056 - x ) x =.056 ; so the assumption is invalid

SO - 4 SO 2-4 + + K a =.013 =.056 - x x.056 + x [SO 4 2-] [ + ] [SO 4 -] ( x (.056 ) + x ) (.056 - x ) x -.069 + - = x 2 +.069 x - 7.28 x 10-4 = 0.069 2-4(1)(- 7.28 x 10-4 ) 2(1) x = 0.0093

SO 4 - SO 4 2- + +.056 - x x.056 + x K a =.013 = [SO 4 2-] [ + ] [SO 4 -] ( x (.056 ) + x ) (.056 - x ) x = 0.0093 [ + ] = (.056 + x ) = 0.063 M p = 1.19

Acid-Base Properties of Salts

Definition A salt is an ionic compound formed by the reaction of an acid and a base. Salts sometimes react with water. When this occurs, we speak of the hydrolysis of a cation, an anion, or both. Salt hydrolysis usually affects the p of a solution. Demos w/ notes

Salt ydrolysis Demos w/ Notes Write this table down and fill in the first 2 blank columns. Salt the 2 ions* predicted p observed p NaCl N4Cl Na3PO4 AlCl3 NaCO3 *circle ones that might react with water

NaCl - why was its p the way it was? Na + is the conjugate of a strong base (NaO) so it will not react with water. Cl - is the conjugate of a strong acid (Cl) so it will not react with water. NaCl(aq) should therefore be neutral.

N4Cl - why was its p the way it was? Ammonium ion is a weak acid N + (aq) N (aq) + + (aq) 4 3 K a = 5.6 x 10-10 = [N 3 ] [N 4 +] [ + ]

What is the p of a 0.10 M N 4 Cl solution? N + (aq) N (aq) + + (aq) 4 3 0.10 - x + x + x x 2 5.6 x 10-10 = 0. 1 - x p = 5.12 x = [ + ] = 7.5 x 10-6 M

Na3PO4 - why was its p the way it was? Phosphate ion is a weak base PO 3- (aq) PO4 2- +O - 4 + 2O (aq) (aq)

AlCl3 - why was its p the way it was? Salts of highly charged metal ions give acidic solutions AlCl 3 (s) + 2 O (l) Al( 2 O) 6 3+ (aq) 2 O Al( 2 O) 5 3+ O Al( 2 O) 5 2+ + +

Li + Be 2+ Na + Mg 2+ Al 3+ O 2 F S 2 Cl K + Ca 2+ Se 2 Br

δ δ δ 3+ δ δ δ

δ δ δ δ+ δ δ δ

δ δ δ δ+ δ δ + δ

NaCO3 Why was its p the way it was? Salts in which both the Cation and Anion ydrolyze or one is Amphoteric

General rules If K b for the anion is greater than K a for the cation, the solution is basic.

Example ammonium cyanide (N 4 CN) dissolves in water to give a basic solution K a = 5.5 x 10-10 N 4 + N 3 + + CN - + 2 O CN + O - K b = 1.6 x 10-5

General rules If K b for the anion is greater than K a for the cation, the solution is basic. If K b for the anion is less than K a for the cation, the solution is acidic. If K a and K b are similar, the solution is close to neutral.

NaCO3 Why was its p the way it was? CO3 - is amphoteric. CO3 - could gain (2CO3 ) or lose (CO3 2- ) an + when reacting with water. Ka for CO3 - = 5.6x10-11 With such a small Ka, the amphoteric CO3 - is terrible at donating + s. So it s better at accepting an + and forms a basic solution.

ClO 4 (aq ) Reactions of Strong Acids with + NaO(aq ) Strong Bases What should the final p be? NaClO 4 (aq ) + 2 O(l ) acid base salt water Ask yourself ow good are ClO4 - and Na + at being acids or bases? Will they react with water to donate or accept a +? ClO4 - and Na + = weakest/poorest bases & acids so they re spectators. So solution is neutral, p = 7 + (aq ) + O - (aq ) 2 O(l )

Reactions of Weak Acids with Strong Bases What should the final p be? CN(aq ) + NaO(aq ) NaCN(aq ) + 2 O(l ) acid base salt water Ask yourself ow good are CN - and Na + at being acids or bases? Will they react with water to donate or accept a +? Na + = spectator but CN - is a (weak) base! So solution is basic, p > 7 CN - (aq) + 2O (l) CN(aq) + O - (aq)

Reactions of Strong Acids with Weak Bases Br (aq ) + N 3 (aq ) N 4 Br(aq ) acid base salt Ask yourself ow good are Br - and N4 + at being acids or bases? Will they react with water to donate or accept a +? Br - = spectator (conjugate base of a strong N4 + (aq) + What should the final p be? acid) but N4 + is a (weak) acid! So solution is acidic, p < 7 2O (l) 3O + (aq) + N3(aq)

Reactions of Weak Acids with Weak Bases What should the final p be? CN(aq ) + N 3 (aq ) N 4 CN(aq ) acid base salt Ask yourself ow good are CN - and N4 + at being acids or bases? Will they react with water to donate or accept a +? Both react with water! p depends on the relative strengths of N 4 + and CN -. N4 + (aq) + CN - (aq) CN(aq) + N3(aq)

Acid-Base Properties of Oxides and ydroxides

Acidic, Basic, and Amphoteric Oxides Metallic oxides give basic solutions in water Oxides of nonmetals give acidic solutions in water Amphoteric oxides also exist

Acidic, Basic, and Amphoteric Oxides Metallic oxides give basic solutions in water Na O (s) + O(l ) 2NaO(aq) 2 2 BaO (s) + O(l ) Ba(O) (aq) 2 2

Acidic, Basic, and Amphoteric Oxides Metallic oxides give basic solutions in water react with acids BaO (s) + 2NO (aq) 3 Ba(NO 3 ) 2 (aq) + 2 O (l ) The BaO first turned into Ba(O) 2 and then reacted with the acid.

Acidic, Basic, and Amphoteric Oxides Oxides of nonmetals give acidic solutions in water. demo w/ CO2 CO (g) + O(l ) CO (aq) 2 2 2 3 What would Le Chatelier say will happen to this equilibrium with the water in our oceans if the CO 2 partial pressure (concentration) increases in our atmosphere?

Acidic, Basic, and Amphoteric Oxides Oxides of nonmetals give acidic solutions in water. SO 3 (g) (from burning coal w/ S in it) + O(l) SO 2 2 4 (rain) (aq) (acid rain) P O (s) + 6 O(l ) 4 PO (aq) 4 10 2 3 4

Acidic, Basic, and Amphoteric Oxides Oxides of nonmetals give acidic solutions in water react with bases CO (g) + 2NaO (aq ) 2 Na CO (aq) + O (l ) 2 3 2

Acidic, Basic, and Amphoteric Oxides Amphoteric oxides react with both acids and bases Al O (s) + 6Cl (aq ) 2 3 2AlCl 3 (aq) + 3 2 O (l ) Al O (s) + 2NaO(aq ) 2 3 + 3 2 O (l ) 2NaAl(O) 4 (aq)