Chapters 13-14

Chapters 13-14

Acids, Bases, Buffers

Common AcidsHClHNO3H2SO4H3PO4H4SiO4HBrHIH2SO3

Patterns?

Common AcidsHClHNO3H2SO4H3PO4H4SiO4HBrHIH2SO3 Patterns? All contain H. H was written first.

Common BasesNaOHLiOHKOHCa(OH)3Ba(OH)2Mg(OH)2Al(OH)3Si(OH)4 NH4OH Patterns?

Common BasesNaOHLiOHKOHCa(OH)3Ba(OH)2Mg(OH)2Al(OH)3Si(OH)4 NH4OH Patterns? all contain OH, written second.

DefinitionsArrhenius - (common)(specific) Acid - releases contained H+ ions Base - releases contained OH- ions

Bronsted - Lowry Acid - proton (H+) donor Base - proton (H+) acceptor

Lewis (general) Acid - accepts an electron pair Base - donates an electron pair

Conjugate Acid-Base Pairs(B-L)

HB(aq) + A-(aq) HA (aq) + B -(aq)


HB(aq) + A-(aq) HA (aq) + B -(aq) D A D AWho donates, who acceptsconjugate acid base pairs

pair 1 D A Pair 2 D A



pair 1 D A Pair 2 D A



pair 1 D A Pair 2 A D

Special CaseH2O(l) + H2O(l) H3O+(aq) + OH -(aq)

autoionization of waterOR

H2O(l) + H2O(l) H+(aq) + OH -(aq)

If the thought of naked protons running around in a pool does not bother you,

This is simpler!

Special Case H2O(l) + H2O(l) H3O+(aq) + OH -(aq)


H2O(l) + H2O(l) H+(aq) + OH -(aq)


This is simpler!



H2O(l) + H2O(l) H+(aq) + OH -(aq)


This is simpler!



H2O(l) + H2O(l) H+(aq) + OH -(aq)


this is simpler!

K #3H2O(l) + H2O(l) H+(aq) + OH -(aq)

Keq(water) = Kw = H+(aq)OH-(aq)1 x 10-14

Why is the water not included???H+(aq)OH-(aq)means water is neutral. H+(aq)OH-(aq)1 x 10-14

∴H+(aq)2 1 x 10-14 Why?

∴H+(aq) 1 x 10-7 neutralWhy?

the power of Hydrogen ionsH+(aq) 1 x 10-7

neutralstronger weaker

H+(aq) 1 x 10-6 acid base? H+(aq) 1 x 10-5 acid base?H+(aq) 1 x 10-4 acid base?H+(aq) 1 x 10-3 acid base?H+(aq) 1 x 10-2 acid base?H+(aq) 1 x 10-1 acid base?H+(aq) 1 x 100 acid base?weaker stronger

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neutralstronger weaker

H+(aq) 1 x 10-6 acid base? H+(aq) 1 x 10-5 acid base?H+(aq) 1 x 10-4 acid base?H+(aq) 1 x 10-3 acid base?H+(aq) 1 x 10-2 acid base?H+(aq) 1 x 10-1 acid base?H+(aq) 1 x 100 acid base?weaker stronger

? ?


= .0000001weaker

H+(aq) 1 x 10-6 = .000001 H+(aq) 1 x 10-5 = .00001 H+(aq) 1 x 10-4 = .0001 H+(aq) 1 x 10-3 = .001 H+(aq) 1 x 10-2 = .01 H+(aq) 1 x 10-1 = .1 H+(aq) 1 x 100 =1. we aker stronger

? ?

the power of Hydrogen ionspHH+(aq) 1 x 10-7 = .00000017weaker

H+(aq) 1 x 10-6 = .000001 6H+(aq) 1 x 10-5 = .00001 5H+(aq) 1 x 10-4 = .0001 4H+(aq) 1 x 10-3 = .001 3H+(aq) 1 x 10-2 = .01 2H+(aq) 1 x 10-1 = .1 1H+(aq) 1 x 100 =1. 0 stronger

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Definitions and Other Stuff pH = - logH+pOH = - logOH-

As pH gets larger, acid content drops As pOH gets larger, base content drops

pH + pOH = 14 H+OH-= 1 x 10-14

Acid-Base Differences Strong Acids Weak acids Strong Bases Weak bases

Strong Acids (6 only)

HCl HBr HIHNO3 H2SO4 HClO4

HCl(aq) H+(aq)Cl-(aq)

So little product returns, reaction considered 100% ionized

Weak acids

a) ionizable H+ HB(aq)H2O(l) B-(aq)H3O+(aq) OR HB(aq) B-(aq)H+(aq)

So little H+ ionizes, most of the acid molecule remains intact.

b) cations B+(aq)H2O(l) BOH(aq) H+(aq)Certain anions have the ability to promote H+ ionization.


K #4 Weak acids

HB(aq)H2O(l) B-(aq)H3O+(aq)

HB(aq) B-(aq)H+(aq) weak acid ionization constant

Ka = H+B- = H3O+B- HB HB

pKa = -log Ka


Polyprotic AcidsExample: H3PO4

H3PO4(aq) H2PO4-(aq) H+(aq)

H2PO4-(aq) HPO4-2(aq) H+(aq)

HPO4 -2(aq) PO4-3(aq) H+(aq)

K1 = H2PO4-H +K2 = HPO4 -2H +K3 = PO4 -

3H +H3PO4H2PO4-

HPO4 -2

Polyprotic Acids Complex EquilibriaH3PO4(aq) H2PO4-(aq) HPO4-2(aq) PO4-3(aq) H+(aq) H+(aq) H+(aq)

Strong Bases (6 only)

LiOH NaOH KOHCa(OH)2 Sr(OH)2 Ba(OH)2

LiOH(aq) Li+(aq)OH-(aq)


Weak bases

a) molecules BH(aq)H2O(l) BH2+(aq)OH-(aq)Certain molecules have the ability to promote OH- ions

b) anions B-(aq)H2O(l) BH(aq) OH-(aq)

Certain anions have the ability to promote OH- ions


K #5 Weak bases

BH(aq)H2O(l) BH2+(aq)OH-(aq)

B-(aq)H2O(l) BH(aq) OH-(aq)

Kb = BH+OH- =BHOH- BH B-

pKb = -log Kb


weak base ionization constant

Relationships

(Ka)(Ka) = 1 x 10-14

pKa + pKa = 14

Titration Possibilities

SASBSAWBWASBWAWB

SASBStrong Acid

H+

Strong BaseOH-

Strong Acid-Strong BaseNet Ionic Equation

H+ + OH- H2O

SAWBStrong Acid

H+ + OH- H2OWeak Base

BH + H2O BH2+ + OH-

B - + H2O BH + OH-

Net Ionic EquationBH + H+ BH2+

B - + H+ BH

WASBStrong Base

H+ + OH- H2OWeak Acid

HB B - + H+

Net Ionic Equation HB + OH- B - + H2O

WAWBWeak Base

BH + H2O BH2+ + OH-

D - + H2O DH + OH-

Weak Acid HB B - + H+

Net Ionic Equation BH + HB B - + BH2+

D - + HB B - + DH

BuffersAbility to resist change

meter out change

BuffersContain:weak acids conjugate bases example: HF H+ + F-

Ka = H+F- = 3.50 x 10-4 HF

H+= Ka HFF-

Bufferssuppose: HF = .1,F-= .1 Ka = 3.50 x 10-4

H+= 3.50 x 10-4 .1.1

pH = - logH+ = 3.46

Buffer Action HF H+ + F-

F- + H2O HF + OH-

Add .01 moles OH-

HF H+ + OH- + F-

.1-.01= .09 .1+.01= .11 H+ = 3.50 x 10-4(.09) (.11) pH = 3.50

Buffer Action HF H+ + F-

F- + H2O HF + OH-

Add .01 moles H+

HF H+ + H+ + F-

.1+.01= .11 .1-.01= .09 H+ = 3.50 x 10-4(.11) (.09) pH = 3.37

Buffer ActionpH ranged from 3.37 to 3.50 in this bufferCompare to: strong acid [H+] = 1.85 x 10-5 pH = 4.74 [H+] = 1.0 x 10-14 add .01 [OH-][OH-][H+] = 1.0 x 10-14 = 1.0 x 10-12

.01 pH = 12

Summary of Buffer ActionBuffer / addition of .01 basepH ranged 3.46 to 3.50

No buffer / addition of .01 basepH range 4.74 to 12

Titration:SASB50.00 ml of 1 M HCl titrate with 1 M NaOHbeginning pH = -log[1] = 0moles 1/1000 = x/50 = .05000

Titration:SASBadd 25.00 ml OH-.05000 - .02500 = .02500 H+ in 75.00 ml [H+] = .3333 and pH = .4778

add 49.99 ml OH-.05000 - .04900 = .00001 H+ in 100.00 ml [H+] = .0001000 and pH = 4.000

add 50.00 ml OH-neutral so pH = 7.000

add 50.01 ml OH-.05000 - .05001 = .00001 OH- in 100.00 ml [OH-] = .0001000, pOH = 4.00, pH = 10.00

Titration:SASBSummary:pH changed from 4 to 10 with the addition of .02 ml of OH-.

Titration:WASBReview: acetic acid Hac H+ + ac-

Ka = H+ac-= H+2 = 1.8 x 10-5

HacH+ = 4.2 x 10-3 pH = 2.4

Titrate 50.00 ml of 1 M Hac with 1M NaOH

Titration:WASBTitrate 50.00 ml of 1 M Hac with 1M NaOH* with 25.00 ml OH-

Hac H+ + OH- + ac-

.05000 - .02500 .02500 H+ = Ka HacThis is a bufferac-Hac=ac-∴H+ = Ka, pH = pKa pH = 4.700

Titration:WASB*with 50.00 ml OH-, all Hac converted to ac- Hac H+

+ OH- + ac-.05000 - .05000 .05000equation is: ac- + H2O Hac + OH-

becomes a weak base in waterKb = [Hac][OH-] = [OH -]2

[ac-] [ac-][ac-] has same moles as original Hac, butin twice the water.Kb[ac-] = [OH-]2 5.6 x 10-10[.5000] = [OH-]2

[OH-]2 = 2.8 x 10-10 [OH-] = 1.7 x 10-5

pOH = 4.8 14 - 4.8 = 9.2 pH = 9.2

Titration:WASBSUMMARY:pH range 2.4 to 9.2 full titrationhalf titration, pH = pKa

Remember this from SASB summary?pH changed from 4 to 10 with the addition of .02 ml of OH-

Documents

Chapters 13-14