Section 14.1

Nature of Acids and Bases

Arrhenius DefinitionArrhenius Definition Acids produce hydrogen ions in Acids produce hydrogen ions in

aqueous solution.aqueous solution.– HCl (aq) H+(aq) + Cl-(aq)

Bases produce hydroxide ions Bases produce hydroxide ions when dissolved in water.when dissolved in water.– NaOH (aq) Na+(aq) + OH-(aq)

Limits what can be considered Limits what can be considered bases (as we’ll see with other bases (as we’ll see with other definitions).definitions).

Lewis DefinitionLewis Definition An acid is an electron pair An acid is an electron pair

acceptor.acceptor. A base is an electron pair donor.A base is an electron pair donor. Easy to see if Lewis structures are Easy to see if Lewis structures are

drawn:drawn:

e- pair acceptor e- pair

donor

Bronsted-Lowry DefinitionBronsted-Lowry Definition This is the one we’ll focus on!This is the one we’ll focus on! An acid is a proton (HAn acid is a proton (H++) donor and a ) donor and a

base is a proton acceptor.base is a proton acceptor. HCl is an acid.HCl is an acid. When it dissolves in water it gives its When it dissolves in water it gives its

proton to water.proton to water. HCl(g) + HHCl(g) + H22O(O(ll) ) HH33OO++ + Cl + Cl-- Water is a base since it accepts the HWater is a base since it accepts the H++..

– In the Arrhenius definition water would not be considered a base!

Conjugate Acid-Base PairsConjugate Acid-Base Pairs General equation General equation HAHA(aq) + H(aq) + H22O(O(ll) ) H H33OO++(aq) + (aq) + AA--(aq)(aq) acidacid + base conjugate acid + + base conjugate acid + conjugate conjugate basebase This is an equilibrium.This is an equilibrium. Equilibrium favors the side with the weaker acid Equilibrium favors the side with the weaker acid

and base. and base. Refer to the handout for acid and base strength.Refer to the handout for acid and base strength. Note: conjugate bases of strong acids are weak, Note: conjugate bases of strong acids are weak,

and conjugate bases of weak acids are strong. and conjugate bases of weak acids are strong. In other words, the stronger the acid/base, the In other words, the stronger the acid/base, the weaker the conjugate base/acid and vice-versa.weaker the conjugate base/acid and vice-versa.

Acid dissociation constant KAcid dissociation constant Kaa Recall strong vs. weak acids/bases!Recall strong vs. weak acids/bases!– Strong = essentially completely dissociate

in water (no equilibrium).– Weak = partially dissociate in water.

Equilibrium is therefore present! For a weak acid:For a weak acid: HA(aq) + HHA(aq) + H22O(O(ll) ) H H33OO++(aq) + (aq) +

AA--(aq)(aq) Water is often left out:Water is often left out: HA (aq) HHA (aq) H++(aq) + A(aq) + A--(aq) (aq)

Acid dissociation constant KAcid dissociation constant Kaa Since weak acids result in equilibrium, Since weak acids result in equilibrium, an equilibrium expression can be an equilibrium expression can be written:written:

HA(aq) + HHA(aq) + H22O(O(ll) ) H H33OO++(aq) + A(aq) + A--

(aq)(aq) KKaa = [H = [H33OO++][A][A--]] [HA][HA] Acid dissociation constantAcid dissociation constant = K = Kaa.. Note: technically all acids have a KNote: technically all acids have a Kaa

value. Size of Kvalue. Size of Kaa indicates strength! indicates strength!

Acid dissociation constant KAcid dissociation constant Kaa If water is left out:If water is left out: HA(aq) HA(aq) HH++(aq) + A(aq) + A--(aq)(aq) KKaa = [H = [H++][A][A--]]

[HA] [HA]Shown without H3O+

Section 14.2

Acid Strength

Types of AcidsTypes of Acids Polyprotic Acids- more than 1 acidic Polyprotic Acids- more than 1 acidic

hydrogen (diprotic, triprotic).hydrogen (diprotic, triprotic).– Ex: H2SO4, H3PO4

Oxyacids - Proton is attached to the Oxyacids - Proton is attached to the oxygen of an ion.oxygen of an ion.– Ex: H2SO4, HNO3

Organic acids contain the carboxyl Organic acids contain the carboxyl group -COOH (acidic Hgroup -COOH (acidic H++ attached to O). attached to O).– Ex: CH3COOH (acetic acid)

Generally very weak.Generally very weak.

KKaa of Polyprotic Acids of Polyprotic Acids The first KThe first Kaa value is much larger than value is much larger than

the second, third, etc.the second, third, etc. The second KThe second Kaa value is much larger value is much larger

than the third Kthan the third Kaa value, etc. value, etc. Since the first KSince the first Kaa value is by far the value is by far the

largest, second, third, etc. Klargest, second, third, etc. Kaa values values can be ‘ignored’ and the first Kcan be ‘ignored’ and the first Kaa value value can usually be used for the acid can usually be used for the acid dissociation constant of the acid.dissociation constant of the acid.

AmphotericAmphoteric Can behave as either an acid or a base.Can behave as either an acid or a base. Water undergoes autonionization (it Water undergoes autonionization (it

autoionizes).autoionizes). 2H2H22O(O(ll) ) H H33OO++(aq) + OH(aq) + OH--(aq)(aq) KKWW= [H= [H33OO++][OH][OH--]=[H]=[H++][OH][OH--]] At 25ºC KAt 25ºC KWW = 1.0 x10 = 1.0 x10-14-14 Occurs in Occurs in EVERYEVERY aqueous solution. aqueous solution. Neutral solution [HNeutral solution [H++] = [OH] = [OH--]= 1.0 x10]= 1.0 x10-7-7 Acidic solution [HAcidic solution [H++] > [OH] > [OH--]] Basic solution [HBasic solution [H++] < [OH] < [OH--]]

Notice it’s very small!

Amphoteric ContinuedAmphoteric Continued Note: although water dissociation Note: although water dissociation

occurs in all solutions and occurs in all solutions and contributes to [Hcontributes to [H++], it is small in ], it is small in comparison and can be ignored comparison and can be ignored when calculating [Hwhen calculating [H++].].

Adding other species to water can Adding other species to water can also allow water to act as an acid also allow water to act as an acid or a base.or a base.

Strong AcidsStrong Acids HBr, HI, HCl, HNOHBr, HI, HCl, HNO33, H, H22SOSO44, HClO, HClO44 Completely dissociatedCompletely dissociated [H[H++] = [HA]] = [HA] Ex: What is [HEx: What is [H++] in a 0.10M HNO] in a 0.10M HNO33

solution?solution?

HNOHNO33 H H++ + NO + NO33--

*Mole ratio is 1:1 for HNO*Mole ratio is 1:1 for HNO33:H:H++

*Thus [H*Thus [H++] = 0.10M] = 0.10M

Weak AcidsWeak Acids Ka will be small.Ka will be small. It will be an equilibrium problem to It will be an equilibrium problem to

find [Hfind [H++].]. Determine whether most of the HDetermine whether most of the H++

will come from the acid or the water.will come from the acid or the water. Compare Ka and Kw.Compare Ka and Kw.

– Whichever is larger is the one that will donate more H+ (this is usually the acid).

Rest is just like last chapter.Rest is just like last chapter.

Weak AcidsWeak Acids Because youBecause you’’re dealing with weak re dealing with weak

acids, it can be assumed that the acids, it can be assumed that the acid wonacid won’’t dissociate much. t dissociate much.

Lets you make assumptions and Lets you make assumptions and simplify terms when solving for x. simplify terms when solving for x. (If you forget and don’t simplify (If you forget and don’t simplify that’s OK- you will just need to use that’s OK- you will just need to use the quadratic formula).the quadratic formula).

Then calculate the [HThen calculate the [H++].].

Weak AcidsWeak Acids For weak acids, the concentrations of For weak acids, the concentrations of

H H++ and A and A-- may be found if both the may be found if both the initial concentration of the acid and initial concentration of the acid and KKaa are known. are known.

Ex: What is the [HEx: What is the [H++] in 0.300M acetic ] in 0.300M acetic acid solution? Kacid solution? Kaa = 1.8 x 10 = 1.8 x 10-5-5

Remember- this is an equilibrium Remember- this is an equilibrium problem! Set up your ICE table, then problem! Set up your ICE table, then solve for x.solve for x.– Answer: x = [H+] = 2.3 x 10-3 M

Practice Problem #1Practice Problem #1 Pg. 675 #55:Pg. 675 #55: A 0.0560g sample of acetic acid is A 0.0560g sample of acetic acid is

added to enough water to make added to enough water to make 50.00mL of solution. Calculate the 50.00mL of solution. Calculate the [H[H++], ],

[CH[CH33COOCOO--], and [CH], and [CH33COOH]. The KCOOH]. The Kaa for acetic acid = 1.8 x 10for acetic acid = 1.8 x 10-5-5..

Summary: Strong vs. WeakSummary: Strong vs. Weak Remember- strong acids: [HA] = Remember- strong acids: [HA] =

[H[H++] and weak acids = equilibrium ] and weak acids = equilibrium problem to find [Hproblem to find [H++]. ].

Otherwise, once the [HOtherwise, once the [H++] is known, ] is known, pH problems are the same. pH problems are the same.

Section 1 HomeworkSection 1 Homework Pg. 672 # 19(a&c), 21, 24, 28, 29, Pg. 672 # 19(a&c), 21, 24, 28, 29,

3232

pH ScalepH Scale pH= -log[HpH= -log[H++] or pH = -log [H] or pH = -log [H33OO++]] pH of pure water = -log(1.0 x 10pH of pure water = -log(1.0 x 10-7-7) = ) =

7.007.00– In other words, this is neutral.– pH < 7.00 is acidic; pH > 7.00 is basic– Notice that as pH decreases, [H+]

increases Sig figs: only the digits after the Sig figs: only the digits after the

decimal place of a pH are significantdecimal place of a pH are significant [H[H++] = 1.0 x 10] = 1.0 x 10-8-8 pH= 8.00 2 sig figs pH= 8.00 2 sig figs

pH Scale ContinuedpH Scale Continued The pH of a solution can be The pH of a solution can be

estimated if the [Hestimated if the [H++] is known.] is known.– Look at the exponent to estimate the

pH– Ex: [H+] = 1 x 10-5; pH = 5

pOH can also be calculated:pOH can also be calculated: pOH = -log[OH-]

Basic

Acidic Neutral

100

10-1

10-3

10-5

10-7

10-9

10-11

10-13

10-14

[H+]

0 1 3 5 7 9 11

13

14

pH

100

10-1

10-3

10-5

10-7

10-9

10-11

10-13

10-14

[OH-]

01357911

13

14

pOH

Important RelationshipsImportant Relationships KKWW = 1.0 x 10 = 1.0 x 10-14-14 = [H = [H++][OH][OH--]] pKpKWW = 14.00 = pH + pOH = 14.00 = pH + pOH pH = -log[HpH = -log[H++]] pOH = -log[OHpOH = -log[OH--]] [H[H++],[OH],[OH--],pH and pOH],pH and pOH

Given any one of these we can Given any one of these we can find the other three!find the other three!

There are also often multiple ways There are also often multiple ways to correctly solve these problems!to correctly solve these problems!

Example #1Example #1 The [HThe [H++] = 1.00 x 10] = 1.00 x 10-5 -5 M. What is the M. What is the

pOH for this solution?pOH for this solution? Step 1: Find pH.Step 1: Find pH. Step 2: Then use Step 2: Then use

14 = pH + pOH to solve for pOH. 14 = pH + pOH to solve for pOH. Can you think of another approach?Can you think of another approach? Find [OHFind [OH--] using: 1.0x10] using: 1.0x10-14-14 = [H = [H++]]

[OH[OH--]. Then use: pOH = -log[OH]. Then use: pOH = -log[OH--].].Answer: Either way you get pOH = 9.000

Note: if [H+] is given, no need to worry if acid is strong or weak.

Example #2Example #2 Calculate the pH of a 1.00M solution of HF, Calculate the pH of a 1.00M solution of HF,

whose Kwhose Kaa = 7.2x10 = 7.2x10-4-4..(1) (1) Small KSmall Kaa, so it, so it’’s a weak acid.s a weak acid.(2) (2) Major species: HF and HMajor species: HF and H22O.O.(3) (3) Which provides the most HWhich provides the most H++ ions? ions?KKaa = 7.2 x 10 = 7.2 x 10-4-4 and K and Kww = 1.0 x 10 = 1.0 x 10-14-14 So HF provides more HSo HF provides more H++ ions. ions.(4) (4) Equilibrium weEquilibrium we’’re investigating:re investigating:HF (aq) HHF (aq) H++ (aq) + F (aq) + F-- (aq) (aq)So, KSo, Kaa = 7.2 x 10 = 7.2 x 10-4-4 = [H = [H++][F][F--]/[HF] ]/[HF]

[H+] is not given, so now we need to decide if the acid is strong or weak, and then solve for [H+].

Example #2Example #2(5) (5) Set up ICE table:Set up ICE table: HF (aq) HHF (aq) H++ (aq) + F (aq) + F-- (aq) (aq)I 1.00 0 0I 1.00 0 0C -x +x +xC -x +x +xE 1.00-x x xE 1.00-x x x

7.2 x 107.2 x 10-4-4 = (x)(x) = (x)(x) (1.00-x)(1.00-x)So: 7.2 x 10So: 7.2 x 10-4-4 = x = x22 Solve for x: x = 0.027 Solve for x: x = 0.027 1.001.00

Simplification: since Ka is small, can assume the initial [HF] won’t change noticeably, so at equilibrium [HF] ≈ 1.00.

Example #2Example #2(5) (5) Verify assumption was OK- x divided by Verify assumption was OK- x divided by

the initial concentration of HF must be the initial concentration of HF must be less than or equal to 5%:less than or equal to 5%:

0.027 x 100% = 2.7%0.027 x 100% = 2.7% 1.001.00 Since 2.7% < 5%, the assumption that x Since 2.7% < 5%, the assumption that x

was small enough to be neglected was was small enough to be neglected was valid.valid.

(6) Calculate [H(6) Calculate [H++] and then the pH:] and then the pH:[H[H++] = x = 0.027M] = x = 0.027MpH = -log(0.027) = pH = -log(0.027) = 1.571.57

Practice Problem #1Practice Problem #1 Calculate the pH of a 0.100M Calculate the pH of a 0.100M

solution of hypochlorous acid solution of hypochlorous acid (HOCl). The K(HOCl). The Kaa value = 3.5 x 10 value = 3.5 x 10-8-8..

[H[H++] = 5.9 x 10] = 5.9 x 10-5-5M, so pH = 4.23M, so pH = 4.23

A mixture of Weak AcidsA mixture of Weak Acids The process is the same.The process is the same. Determine the major species.Determine the major species. The stronger will predominate The stronger will predominate

(whichever has the largest K(whichever has the largest Kaa value).value).

Doubt you’ll see this on the AP Doubt you’ll see this on the AP exam, but just in case!exam, but just in case!

ExampleExample Calculate the pH of a solution that Calculate the pH of a solution that

contains 1.00MHCN whose Kcontains 1.00MHCN whose Kaa = 6.2x10 = 6.2x10--

1010 and 5.00MHNO and 5.00MHNO22 whose K whose Kaa = 4.0x10 = 4.0x10-4-4. . Approach: same as before, now just Approach: same as before, now just

need to consider THREE K values (the need to consider THREE K values (the two above and water).two above and water).

Since KSince Kaa for HNO for HNO22 is much larger than K is much larger than Kww and the Kand the Kaa for HCN, this is the only one for HCN, this is the only one that needs to be used for finding [Hthat needs to be used for finding [H++].].

[H[H++] = 4.5 x 10] = 4.5 x 10-2-2, so pH = 1.35, so pH = 1.35

AP Practice QuestionAP Practice QuestionA 0.1M solution of acetic acid A 0.1M solution of acetic acid (CH(CH33COOH) has a pH of about:COOH) has a pH of about:

a)a)11b)b)33c)c)77d)d)1010You can answer this question without You can answer this question without doing any math!doing any math!

AP Practice QuestionAP Practice QuestionWhat is the ionization constant, KWhat is the ionization constant, Kaa, for , for a weak monoprotic acid if a 0.30M a weak monoprotic acid if a 0.30M solution has a pH of 4.0?solution has a pH of 4.0?a) 3.3 x 10a) 3.3 x 10-8-8 c) 1.7 x 10c) 1.7 x 10-6-6

b) 4.7 x 10b) 4.7 x 10-2-2 d) 3.0 x 10d) 3.0 x 10-4-4

Solve without using a calculator:Solve without using a calculator:-Estimate [H-Estimate [H++]: pH = 4.0, so [H]: pH = 4.0, so [H++] = 1 x 10] = 1 x 10--44

-K-Kaa = (1 x 10 = (1 x 10-4-4))22/0.30 /0.30 ÷1 = 1 x 10 ÷1 = 1 x 10-8-8 so K so Kaa should be a little biggershould be a little bigger

When KWhen Kww Matters… Matters…

Percent DissociationPercent Dissociation = = amount dissociated (M) amount dissociated (M) x 100 x 100

initial concentration (M) initial concentration (M) ExampleExample: Calculate the % dissociation : Calculate the % dissociation

of of 1.00 M acetic acid,1.00 M acetic acid, Ka = 1.8 x 10 Ka = 1.8 x 10-5-5.. Approach is the same as weak acid Approach is the same as weak acid

problems! Solve like an equilibrium problems! Solve like an equilibrium problem for the necessary problem for the necessary concentrations, then calculate % concentrations, then calculate % dissociation.dissociation.

% dissociation = 0.42%% dissociation = 0.42%

[H+] at equilibrium

Sections 2-3 Homework Problems

Pg. 673 #17, 22, 40, 43, 47, 51, 57

Section 14.7

Polyprotic Acids

Polyprotic acidsPolyprotic acids Always dissociate stepwise.Always dissociate stepwise. The first HThe first H++ comes off much easier comes off much easier

than the second.than the second. Ka for the first step is much bigger Ka for the first step is much bigger

than Ka for the second, the second than Ka for the second, the second is bigger than the third, etc.is bigger than the third, etc.– More difficult to lose the next H+ because

the negative charge increases. Denoted KaDenoted Ka11, Ka, Ka22, Ka, Ka33..

Polyprotic acidPolyprotic acid HH22COCO33 HH++ + HCO + HCO33

-- Ka Ka11= 4.3 x 10= 4.3 x 10-7-7

HCOHCO33-- HH++ + CO + CO33

-2-2 Ka Ka22= 4.3 x = 4.3 x 1010-10-10

Conjugate base in first step is the acid Conjugate base in first step is the acid in second.in second.

In calculations we can normally ignore In calculations we can normally ignore the second, third, etc. dissociation.the second, third, etc. dissociation.

Sulfuric Acid is SpecialSulfuric Acid is Special In the first step it is a strong acid.In the first step it is a strong acid.

– No Ka value given- complete dissociation. Second step is a weak acid.Second step is a weak acid.

– Ka2 = 1.2 x 10-2 – Small, but not always small enough to

ignore.– If the initial concentration of H2SO4 is low

enough, the second H+ impacts pH!

Sulfuric Acid ExampleSulfuric Acid Example Calculate the pH of a 0.0100M HCalculate the pH of a 0.0100M H22SOSO44

solution. Kasolution. Ka22 = 0.012. = 0.012. The first acidic proton (HThe first acidic proton (H++) dissociates ) dissociates

completely (Hcompletely (H22SOSO44 is strong at first): is strong at first): HH22SOSO44 H H++ + HSO + HSO44

-- Thus: [HThus: [H22SOSO44] = [H] = [H++] = [HSO] = [HSO44

--] = ] = 0.0100M0.0100M

Now, we need to consider the second Now, we need to consider the second acidic proton.acidic proton.

Sulfuric Acid ExampleSulfuric Acid Example The second acidic proton comes from The second acidic proton comes from

HSOHSO44--, which is a weak acid. So this is , which is a weak acid. So this is

treated like an equilibrium problem:treated like an equilibrium problem: HSOHSO44

-- H H++ + SO + SO44-2-2

I 0.0100 0.0100 0I 0.0100 0.0100 0C -x +x +xC -x +x +xE 0.0100-x 0.0100+x xE 0.0100-x 0.0100+x x Plug into KaPlug into Ka22 expression: expression:0.012 = (x)(0.0100+x)/(0.0100-x)0.012 = (x)(0.0100+x)/(0.0100-x)

Sulfuric Acid ExampleSulfuric Acid Example We can try to use the simplification that We can try to use the simplification that

x is negligible with respect to HSOx is negligible with respect to HSO44-- and and

HH++: 0.012 = (x)(0.0100)/(0.0100),: 0.012 = (x)(0.0100)/(0.0100), however the value for x = 0.012, which however the value for x = 0.012, which

does not make sense. Thus the does not make sense. Thus the simplification is not valid!simplification is not valid!

Use quadratic formula to solve!Use quadratic formula to solve! x = 0.0045x = 0.0045 [H[H++] = 0.0100+0.0045 = 0.0145M] = 0.0100+0.0045 = 0.0145M pH = 1.84 pH = 1.84

Notice that most Ka1 values are significantly larger than the Ka2 values, and thus typically do not impact the [H+] or the pH. Typically, exceptions are sulfuric acid and oxalic acid.

Section 7 HomeworkPg. 676 #93, 96, 97, 98

Lesson Essential Question:

How do calculations with acids differ from calculations with bases?

Section 14.6

Bases

BasesBases The OHThe OH--

is a strong base. is a strong base. Hydroxides of the alkali metals are Hydroxides of the alkali metals are

strong bases because they dissociate strong bases because they dissociate completely when dissolved.completely when dissolved.

The hydroxides of alkaline earth The hydroxides of alkaline earth metals (Ca(OH)metals (Ca(OH)2,2, etc.) are also strong, etc.) are also strong, but they donbut they don’’t dissolve well in water.t dissolve well in water.

Used as antacids because [OHUsed as antacids because [OH-- ] can] can’’t t

build up.build up.

Bases without OHBases without OH--

Bases are proton acceptors:Bases are proton acceptors: NHNH33 + H + H22O O NHNH44

++ + OH + OH--

It is the lone pair on N that accepts the It is the lone pair on N that accepts the proton.proton.• Many weak bases contain N

Use KUse Kbb instead: instead: BB(aq) (aq) + H+ H22O(O(ll)) HB HB++(aq) (aq) + OH+ OH- - (aq)(aq) KKbb = [ = [HBHB++][][OHOH- - ]] [ [BB] ]

Strength of BasesStrength of Bases

Hydroxides are strong.Hydroxides are strong. Others are weak.Others are weak. Smaller KSmaller Kbb = weaker base. = weaker base.

Strong Base ExampleStrong Base Example

Calculate the pH of a 0.050M solution NaOH.Treat it like a strong acid problem, just use pOH.

pOH = 1.3; pH = 14 – 1.3 = 12.7

Weak Base ExampleWeak Base Example

Calculate the pH for a 1.0M solution of CH3NH2, whose Kb=1.8x10-5. Treat it like a weak acid problem, just with OH- instead of H+.

[OH-] = 0.016M; pOH = 1.8; pH = 12.2

Another Important RelationshipAnother Important RelationshipFor conjugate acid-base pairs, Ka x Kb = Kw.This can be particularly useful if you’re dealing with a weak base and only the Ka value of the conjugate acid is known.The Kb for the base can be calculated and the equilibrium concentrations can then be found.

ExampleExample

0.400M of the acetate ion is dissolved in solution. What is the pH of the solution? Ka = 1.8x10-5 for acetic acid. Write the equilibrium equation. Then find Kb for acetate and solve.[OH-] = 1.5x10-5 M; pH = 9.18

Section 14.8

Acid-Base Properties of Salts

Salts Can Affect pHSalts Can Affect pH Salts are ionic compounds.Salts are ionic compounds. Most salts are soluble in water, and Most salts are soluble in water, and

completely dissociate.completely dissociate. Depending upon the ions making Depending upon the ions making

up the salt, they COULD affect the up the salt, they COULD affect the pH once dissociated.pH once dissociated.

Neutral SaltsNeutral Salts Salts whose ions could react with Salts whose ions could react with

water to produce water to produce a strong acid or basea strong acid or base DO NOTDO NOT effect the pH. effect the pH.– Ex: Add NaCl to water. Two possible

reactions: Na+ + H2O NaOH + H+

Cl- + H2O HCl + OH- Both possible reactions favor the Both possible reactions favor the

reactants, meaning no Hreactants, meaning no H++ or OH or OH-- is is produced, so pH remains neutral.produced, so pH remains neutral.

Strong acids Strong acids and bases and bases dissociate dissociate completelycompletely

Acidic SaltsAcidic Salts If the cation of a salt reacts with If the cation of a salt reacts with

water to water to produce a weak base and produce a weak base and HH++, the solution is acidic., the solution is acidic.

Ex: NHEx: NH44Cl has two possible reactions:Cl has two possible reactions: We already know ClWe already know Cl-- has no effect. has no effect. NHNH44

++ NH NH33 + H + H++ NHNH33 is a weak base, so equilibrium is a weak base, so equilibrium

does NOT favor the reactants; thus does NOT favor the reactants; thus the formation of Hthe formation of H++ is significant, and is significant, and pH is effected.pH is effected.

Basic SaltsBasic Salts If the anion of a salt reacts with If the anion of a salt reacts with

water to water to produce a weak acid and produce a weak acid and OHOH--, the solution is basic., the solution is basic.

Ex: NaF has two possible reactions:Ex: NaF has two possible reactions: We already know NaWe already know Na++ has no effect. has no effect. FF-- + H + H22O HF + OHO HF + OH-- HF is a weak acid, so equilibrium HF is a weak acid, so equilibrium

does NOT favor the reactants; thus does NOT favor the reactants; thus the formation of OHthe formation of OH-- is significant, is significant, and pH is effected.and pH is effected.

Acidic & Basic SaltsAcidic & Basic Salts It is possible that a salt could contain a It is possible that a salt could contain a

cation that forms a weak base and an cation that forms a weak base and an anion that forms a weak acid.anion that forms a weak acid.

If this is the case, KIf this is the case, Kaa and K and Kbb must be must be compared.compared.

The larger value dictates the pH. If KThe larger value dictates the pH. If Kaa is is larger, the solution is acidic. If Klarger, the solution is acidic. If Kbb is larger, is larger, the solution is basic. If both are equal, the the solution is basic. If both are equal, the pH is neutral.pH is neutral.– Ex: NH4CN in water: Ka & Kb = 5.6 x 10-10

PracticePracticeGiven the following salts, predict what Given the following salts, predict what the pH would be if they were added the pH would be if they were added to water (acidic, basic, or neutral).to water (acidic, basic, or neutral).

1)1)KNOKNO33

2)2)NaNa22COCO33

3)3)NHNH44Br (note: HBr = strong)Br (note: HBr = strong)4)4)NHNH44F (note: HF = weak)F (note: HF = weak)

neutralbasic

acidic

Comparison of Ka and Kb needed.

Sample Problem:Sample Problem: Calculate the pH of a 0.10M NHCalculate the pH of a 0.10M NH44Cl Cl

solution. Ksolution. Kbb = 1.8 x 10 = 1.8 x 10-5-5 for NH for NH33.. Only NHOnly NH44

++ contributes to pH change: contributes to pH change: NHNH44

++ NH NH33 + H + H++ Need KNeed Kaa because NH because NH44

++ is an acid! is an acid! KKww = K = Kaa x K x Kbb for a conjugate acid/base for a conjugate acid/base

pair.pair. So: KSo: Kaa = (1.0x10 = (1.0x10-14-14)/(1.8x10)/(1.8x10-5-5)) KKaa = 5.6x10 = 5.6x10-10-10

Sample Problem:Sample Problem: Then, treat calculations like an Then, treat calculations like an

equilibrium problem:equilibrium problem: NHNH44

++ NH NH33 + H + H++ I 0.10 0 0I 0.10 0 0C -x +x +xC -x +x +xE 0.10-x x xE 0.10-x x x 5.6 x 105.6 x 10-10-10 = (x)(x) = (x)(x) 0.10-x0.10-x5.6 x 105.6 x 10-10-10 = x = x22/0.10 => x = 7.5x10/0.10 => x = 7.5x10-6-6

Sample Problem:Sample Problem: Simplification: 0.10 – x ≈ 0.10Simplification: 0.10 – x ≈ 0.10 5.6 x 105.6 x 10-10-10 = x = x22/0.10 => x = 7.5x10/0.10 => x = 7.5x10--

66 Verify:Verify: 7.5x107.5x10-6-6 x 100 = 0.0075% x 100 = 0.0075% 0.100.10 Verification is acceptable.Verification is acceptable. [H[H++] = 7.5x10] = 7.5x10-6-6 M M pH = 5.12pH = 5.12

Section 8 HomeworkPg. 676 #99, 101, 104, 111

Section 15.2-15.3

Buffers

BuffersBuffers Solutions that maintain a constant Solutions that maintain a constant

pH when an acid or base is added.pH when an acid or base is added.– In other words, change in pH is resisted.

One species present can One species present can consume/react with Hconsume/react with H++ and another and another species present can consume OHspecies present can consume OH--. .

Very useful and important in many Very useful and important in many ways, especially for life. ways, especially for life. – Blood contains buffers to maintain a constant

pH!

BuffersBuffers Typically achieved by mixing a Typically achieved by mixing a

weak acid with its conjugate base, weak acid with its conjugate base, or a weak base with its conjugate or a weak base with its conjugate acid. acid. – The conjugate acid or base is usually added as a

salt. Examples:Examples:

– Acetic acid mixed with sodium acetate.– Ammonia mixed with ammonium chloride.

• Notice these species will not effect the pH.

Acetic Acid & Sodium AcetateAcetic Acid & Sodium Acetate HCHC22HH33OO22 H H++ + C + C22HH33OO22

--

Sodium acetate completely dissolves.Sodium acetate completely dissolves. Addition of acetate ion shifts Addition of acetate ion shifts

equilibrium to the left.equilibrium to the left. At this point there is a lot of acetic At this point there is a lot of acetic

acid and the acetate ion.acid and the acetate ion. Acetic acid (and HAcetic acid (and H++) can neutralize ) can neutralize

any base added, and the acetate ion any base added, and the acetate ion can neutralize any acid added.can neutralize any acid added.

Ammonia & Ammonium Ammonia & Ammonium ChlorideChloride NHNH33 + H + H22O O NH NH44

++ + OH + OH--

Ammonium chloride completely dissolves.Ammonium chloride completely dissolves. Addition of ammonium ion shifts Addition of ammonium ion shifts

equilibrium to the left.equilibrium to the left. At this point there is a lot of ammonia At this point there is a lot of ammonia

and the ammonium ion.and the ammonium ion. Ammonia (and OHAmmonia (and OH--) can neutralize any ) can neutralize any

acid added, and the ammonium ion acid added, and the ammonium ion can neutralize any base added.can neutralize any base added.

Why Weak Acids & Bases?Why Weak Acids & Bases? You probably noticed that only weak acids You probably noticed that only weak acids

and bases are used for buffer solutions. and bases are used for buffer solutions. Why is this so?Why is this so?

Consider the strong acid HCl and adding Consider the strong acid HCl and adding NaCl to it.NaCl to it.

HCl completely dissociates; no equilibrium:HCl completely dissociates; no equilibrium: HCl HCl H H++ + Cl + Cl--

Addition of NaCl Addition of NaCl doesn’tdoesn’t cause any shift in cause any shift in equilibrium, and only acid (Hequilibrium, and only acid (H++) remains. ) remains. This could only neutralize the addition of a This could only neutralize the addition of a base, not the addition of an acid.base, not the addition of an acid.

Why Weak Acids & Bases?Why Weak Acids & Bases?

So… So… EQUILIBRIUMEQUILIBRIUM THAT THAT EXISTS FOR WEAK ACIDS EXISTS FOR WEAK ACIDS AND BASES IS AND BASES IS KEY KEY FOR FOR DEVELOPING A DEVELOPING A SUCCESSFUL BUFFER!SUCCESSFUL BUFFER!

It’s the only way for both It’s the only way for both acid & base to exist acid & base to exist together!together!

pH & pOH of BufferspH & pOH of Buffers Henderson-Hasselbalch equation Henderson-Hasselbalch equation

can be used to find pH or pOH of can be used to find pH or pOH of buffers.buffers.– Must know Ka or Kb and initial concentrations of

the acid/conjugate base or base/conjugate acid. pH = pKpH = pKaa + log([A + log([A--]/[HA])]/[HA]) pOH = pKpOH = pKbb + log([HB + log([HB++]/[B])]/[B])

– Note: pKa or pKb means you take –log of the K value.

pH & pOH of BufferspH & pOH of Buffers pH = pKpH = pKaa + log([A + log([A--]/[HA])]/[HA]) pOH = pKpOH = pKbb + log([HB + log([HB++]/[B])]/[B])

– Also notice above that the desired pH or pOH can be ‘fine-tuned’ by adjusting the ratio of weak acid/base to its conjugate base/acid.

– Ex: the more A- you add, and the less HA you add, the higher the buffer pH will be.

Can also find pH as we’ve been Can also find pH as we’ve been doing.doing.– Set up as an equilibrium problem and solve.

AP Practice QuestionAP Practice QuestionThe following questions refer to aqueous The following questions refer to aqueous solutions containing 1:1 mole ratios of solutions containing 1:1 mole ratios of the following pairs of substances. the following pairs of substances. Assume all concentrations are 1M. Assume all concentrations are 1M. a) NHa) NH33 and H and H33CCOOH (acetic acid)CCOOH (acetic acid)b) KOH and NHb) KOH and NH33

c) HCl and KClc) HCl and KCld) Hd) H33POPO44 and KH and KH22POPO44

e) NHe) NH33 and NH and NH44ClCl

AP Practice Question Cont.AP Practice Question Cont.The solution with the highest pH.The solution with the highest pH.

a) NHa) NH33 and H and H33CCOOH (acetic acid)CCOOH (acetic acid)b) KOH and NHb) KOH and NH33

c) HCl and KClc) HCl and KCld) Hd) H33POPO44 and KH and KH22POPO44

e) NHe) NH33 and NH and NH44ClCl

AP Practice Question Cont.AP Practice Question Cont.The solution with the lowest pH.The solution with the lowest pH.

a) NHa) NH33 and H and H33CCOOH (acetic acid)CCOOH (acetic acid)b) KOH and NHb) KOH and NH33

c) HCl and KClc) HCl and KCld) Hd) H33POPO44 and KH and KH22POPO44

e) NHe) NH33 and NH and NH44ClCl

AP Practice Question Cont.AP Practice Question Cont.The solution with the pH closest to The solution with the pH closest to neutral.neutral.

a) NHa) NH33 and H and H33CCOOH (acetic acid)CCOOH (acetic acid)b) KOH and NHb) KOH and NH33

c) HCl and KClc) HCl and KCld) Hd) H33POPO44 and KH and KH22POPO44

e) NHe) NH33 and NH and NH44ClCl

AP Practice Question Cont.AP Practice Question Cont.A buffer at an alkaline pH.A buffer at an alkaline pH.

a) NHa) NH33 and H and H33CCOOH (acetic acid)CCOOH (acetic acid)b) KOH and NHb) KOH and NH33

c) HCl and KClc) HCl and KCld) Hd) H33POPO44 and KH and KH22POPO44

e) NHe) NH33 and NH and NH44ClCl

AP Practice Question Cont.AP Practice Question Cont.A buffer at an acidic pH.A buffer at an acidic pH.

a) NHa) NH33 and H and H33CCOOH (acetic acid)CCOOH (acetic acid)b) KOH and NHb) KOH and NH33

c) HCl and KClc) HCl and KCld) Hd) H33POPO44 and KH and KH22POPO44

e) NHe) NH33 and NH and NH44ClCl

Buffer Calculation ExampleBuffer Calculation ExampleWhat is the pH of a solution that has What is the pH of a solution that has 2.00mol NH2.00mol NH33 and 3.00mol NH and 3.00mol NH44Cl in 1.00L Cl in 1.00L of solution? The Kof solution? The Kbb for NH for NH33 = 1.81 x 10 = 1.81 x 10-5-5..Henderson-Hasselbalch:Henderson-Hasselbalch:pOH = -log(1.81 x 10pOH = -log(1.81 x 10-5-5) + log(3.00/2.00)) + log(3.00/2.00)pOH = 4.918, so pH = 14 – 4.918 = pOH = 4.918, so pH = 14 – 4.918 = 9.0829.082* * Can verify answer by solving as an Can verify answer by solving as an equilibrium problem. Write equilibrium for equilibrium problem. Write equilibrium for NHNH33 producing NH producing NH44

++ since you’re given K since you’re given Kbb..

Buffer Calculation PracticeBuffer Calculation PracticeWhat is the pH of a solution of 0.75M What is the pH of a solution of 0.75M lactic acid (Klactic acid (Kaa = 1.4 x 10 = 1.4 x 10-4-4) and ) and 0.25M sodium lactate? Lactic acid 0.25M sodium lactate? Lactic acid (HC(HC33HH55OO33) is often a component of of ) is often a component of of biologic systems (such as milk and biologic systems (such as milk and muscle tissue during exercise).muscle tissue during exercise).

Use Henderson-Hasselbalch; answer: Use Henderson-Hasselbalch; answer: pH = 3.38pH = 3.38

HomeworkHomework Pg. 740 # 21 & 23Pg. 740 # 21 & 23

Buffer CapacityBuffer Capacity More concentrated acid-base pairs More concentrated acid-base pairs

= more acid or base can be = more acid or base can be neutralized = less change in pH.neutralized = less change in pH.

This is the This is the buffer capacitybuffer capacity: how : how well a buffer solution can maintain well a buffer solution can maintain a constant pH.a constant pH.

Making BuffersMaking Buffers To make a buffer with an acidic pH: To make a buffer with an acidic pH:

weak acid + salt containing weak acid + salt containing conjugate base.conjugate base.

To make a buffer with a basic pH: To make a buffer with a basic pH: weak base + salt containing weak base + salt containing conjugate acid.conjugate acid.

When preparing a buffer solution you When preparing a buffer solution you must choose an acid whose pKmust choose an acid whose pKaa value value is close to the desired pH.is close to the desired pH.– Look at exponent for easy estimation!

AP Practice QuestionAP Practice QuestionWhat is a solution with an initial HWhat is a solution with an initial H33POPO22 concentration of 1M and an initial KHconcentration of 1M and an initial KH22POPO22 concentration of 1M?concentration of 1M?

a) solution with pH > 7, which is a buffera) solution with pH > 7, which is a bufferb) solution with pH < 7, which isn’t a bufferb) solution with pH < 7, which isn’t a bufferc) solution with pH < 7, which is a bufferc) solution with pH < 7, which is a bufferd) solution with pH > 7, which isn’t a bufferd) solution with pH > 7, which isn’t a buffer

AP Practice QuestionAP Practice QuestionWhat is a solution with an initial KCOOH What is a solution with an initial KCOOH concentration of 1M and an initial KHconcentration of 1M and an initial KH22POPO22 concentration of 1M?concentration of 1M?

a) solution with pH > 7, which is a buffera) solution with pH > 7, which is a bufferb) solution with pH < 7, which isn’t a bufferb) solution with pH < 7, which isn’t a bufferc) solution with pH < 7, which is a bufferc) solution with pH < 7, which is a bufferd) solution with pH > 7, which isn’t a bufferd) solution with pH > 7, which isn’t a buffer

AP Practice QuestionAP Practice QuestionUsing the following information, choose the Using the following information, choose the best answer for preparing a pH = 8 buffer.best answer for preparing a pH = 8 buffer.HH33POPO44: K: Kaa = 7.2 x 10 = 7.2 x 10-3-3

HH22POPO44--: K: Kaa = 6.3 x 10 = 6.3 x 10-8-8

HPOHPO44-2-2: K: Kaa = 4.2 x 10 = 4.2 x 10-13-13

a) Ka) K22HPOHPO44 + KH + KH22POPO44

b) Hb) H33POPO44

c) Kc) K22HPOHPO44 + K + K33POPO44

d) Kd) K33POPO44

Adding Acid/Base to a BufferAdding Acid/Base to a BufferCalculate the pH of a solution when 0.10mol Calculate the pH of a solution when 0.10mol of gaseous HCl is added to 1.0L of a solution of gaseous HCl is added to 1.0L of a solution containing 0.25M NHcontaining 0.25M NH33 and 0.40M NH and 0.40M NH44Cl.Cl.Acid is being added, so the base NHAcid is being added, so the base NH33 will will react with it.react with it.When an acid reacts with a base, the When an acid reacts with a base, the reaction goes to completion: all dissociated reaction goes to completion: all dissociated HH++ reacts with NH reacts with NH33..Use stoichiometry to calculate moles of Use stoichiometry to calculate moles of species left species left afterafter the reaction. the reaction.

Adding Acid/Base to a BufferAdding Acid/Base to a BufferCalculate the pH of a solution when 0.10mol Calculate the pH of a solution when 0.10mol of gaseous HCl is added to 1.0L of a solution of gaseous HCl is added to 1.0L of a solution containing 0.25M NHcontaining 0.25M NH33 and 0.40M NH and 0.40M NH44Cl.Cl.Set up an ‘IF’ table (for a reaction that goes Set up an ‘IF’ table (for a reaction that goes to completion- initial and final values). to completion- initial and final values).

– Don’t have to do this; IF table just helps set up your given information to solve the problem.

NHNH33 + H + H++ NH NH44++

I 0.25mol 0.10mol 0.40mol I 0.25mol 0.10mol 0.40mol FF

Must be in moles for stoichiometry!

Adding Acid/Base to a BufferAdding Acid/Base to a BufferCalculate the pH of a solution when 0.10mol Calculate the pH of a solution when 0.10mol of gaseous HCl is added to 1.0L of a solution of gaseous HCl is added to 1.0L of a solution containing 0.25M NHcontaining 0.25M NH33 and 0.40M NH and 0.40M NH44Cl.Cl.Once you have the initial moles, determine Once you have the initial moles, determine which reactant is limiting (usually the added which reactant is limiting (usually the added strong acid or base). Then find moles of all strong acid or base). Then find moles of all species remaining using stoichiometry.species remaining using stoichiometry. NHNH33 + H + H++ NH NH44

++

I 0.25mol 0.10mol 0.40mol I 0.25mol 0.10mol 0.40mol F 0.15mol 0mol 0.50molF 0.15mol 0mol 0.50mol

LR

Adding Acid/Base to a BufferAdding Acid/Base to a BufferCalculate the pH of a solution when 0.10mol of Calculate the pH of a solution when 0.10mol of gaseous HCl is added to 1.0L of a solution gaseous HCl is added to 1.0L of a solution containing 0.25M NHcontaining 0.25M NH33 and 0.40M NH and 0.40M NH44Cl.Cl.Use final mole values to find final concentrations: Use final mole values to find final concentrations: [NH[NH33] = 0.15M; [NH] = 0.15M; [NH44

++] = 0.50M] = 0.50MThen use final concentrations of acid and base left Then use final concentrations of acid and base left in the Henderson-Hasselbalch equation.in the Henderson-Hasselbalch equation.Note: KNote: Kbb for NH for NH33 = 1.8 x 10 = 1.8 x 10-5-5

pOH = -log(1.8 x 10pOH = -log(1.8 x 10-5-5) + log(0.50/0.15) = 5.27) + log(0.50/0.15) = 5.27pH = 14 – 5.27 = pH = 14 – 5.27 = 8.738.73pH of original buffer solution = 9.05. Not much pH of original buffer solution = 9.05. Not much change! change!

Adding Acid/Base to a BufferAdding Acid/Base to a BufferCalculate the pH of a solution when 0.10mol Calculate the pH of a solution when 0.10mol of gaseous HCl is added to 1.0L of a solution of gaseous HCl is added to 1.0L of a solution containing 0.25M NHcontaining 0.25M NH33 and 0.40M NH and 0.40M NH44Cl.Cl.Alternative calculation: instead of using KAlternative calculation: instead of using Kbb, you , you could also have used Kcould also have used Kaa..KKaa = (1 x 10 = (1 x 10-14-14)/(1.8 x 10)/(1.8 x 10-5-5) = 5.6 x 10) = 5.6 x 10-10-10

pH = -log(5.6 x 10pH = -log(5.6 x 10-10-10) + log(0.15/0.50) = ) + log(0.15/0.50) = 8.738.73Note the change in the log taken of the ratio!Note the change in the log taken of the ratio!Note also that the pH is the same!Note also that the pH is the same!

PracticePracticeWhat is the pH of a solution containing What is the pH of a solution containing 5.00M HC5.00M HC22HH33OO22 and 5.00M NaC and 5.00M NaC22HH33OO22 when when 0.010mol of HCl gas is added to 1.0L of this 0.010mol of HCl gas is added to 1.0L of this solution? Ksolution? Kaa = 1.8 x 10 = 1.8 x 10-5-5 for acetic acid. for acetic acid.*Reaction: C*Reaction: C22HH33OO22

-- + H + H++ HC HC22HH33OO22

*Set up IF table! Find final concentrations.*Set up IF table! Find final concentrations.LR = HLR = H++ added addedFinal concentrations: [CFinal concentrations: [C22HH33OO22

--] = 4.99M and ] = 4.99M and [HC[HC22HH33OO22] = 5.01M] = 5.01MFinal pH = 4.74Final pH = 4.74

HomeworkHomework Pg. 741 #27 & 29 (d) onlyPg. 741 #27 & 29 (d) only