Unit 11: Equilibrium / Acids and Bases

reversible reaction: R P and R P

Acid dissociation is a reversible reaction.

H2SO4 2 H+ + SO42–

equilibrium:

-- looks like nothing is happening, however…

--

rate at whichR P

rate at whichP R=

system is dynamic, NOT static equilibrium does NOT mean “half this & half that”

Le Chatelier’s principle:

When a system at equilibrium isdisturbed, it shifts to a new equili-

brium that counteracts the disturbance.

N2(g) + 3 H2(g) 2 NH3(g) Disturbance Equilibrium Shift

Add more N2………………Add more H2………………Add more NH3…………….

Add a catalyst…………….. Remove NH3………………

Increase pressure………… no shift

shift to a new equilibrium:

Then go inside…

shift to a new equilibrium:

Light-Darkening EyeglassesAgCl + energy Ago + Clo

“energy”

Go outside…Sunlight more intense than inside light;

GLASSES DARKEN

(clear) (dark)

“energy”

GLASSES LIGHTEN

In a chicken… CaO + CO2 CaCO3

In summer, [ CO2 ] in a chicken’sblood due to panting. --

--

--

How could we increase eggshell thickness in summer?

(eggshells)

eggshells are thinnershift ;

give chickens carbonated water

put CaO additives in chicken feed

[ CO2 ] , shift

[ CaO ] , shift

I wish I had sweat glands.

< 7

Acids and Bases

pH taste ______

react with ______

proton (H+) donor

Both are electrolytes:

turn litmus lots of H+/H3O+

react w/metals

pH taste ______

react with ______

proton (H+) acceptor

turn litmus

lots of OH– don’t react w/metals

sour

bases

red

> 7bitter

acids

blue

they conduct electricity in sol’n

litmus paper

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

ACID BASE

NEUTRAL

pH scale: measures acidity/basicity

Each step on pH scale represents a factor of ___.pH 5 vs. pH 6

(___X more acidic) pH 3 vs. pH 5 (_______X different) pH 8 vs. pH 13 (_______X different)

10

10100

100,000

Acid Nomenclaturebinary acids: acids w/H and one other element

Binary Acid Nomenclature1. Write “hydro.”2. Write prefix of the other element, followed by “-ic acid.”HFHClHBrhydroiodic acidhydrosulfuric acid

hydrofluoric acidhydrochloric acidhydrobromic acidHIH2S

Oxyacids

oxyacids: acids containing H, O, and one other element

Common oxyanions (polyatomic ions that containoxygen) that combine with H to make oxyacids:

BrO3- NO3

-

CO32- PO4

3-

ClO3- SO4

2-

IO3-

bromate

carbonate

chlorate

iodate

nitrate

phosphate

sulfate

Oxyacid Nomenclature

Write the name of the polyatmic ion and change the ending to “-ic acid.” NO HYDRO!

HBrO3

HClO3

H2CO3

H2SO4

H3PO4

bromic acid

chloric acid

carbonic acid

sulfuric acid

phosphoric acid

These are the “most common” forms of the oxyacids. Notice that each polyatomic ion ended in “-ate”

Oxyacid Nomenclature Con’tIf an oxyacid differs from the “-ate” form by the # of O

atoms, the name changes are as follows:one more Ox = per_____ic acid

“most common” # of Ox = _____ic acid one less Ox = _____ous acid two fewer Ox = hypo_____ous acid

HClO4 Perchloric acidHClO3 Chloric acid HClO2 Chlorous acidHClO Hypochlorous acidphosphorus acid H3PO3

hypobromous acid HBrOpersulfuric acid H2SO5

Becomes an acid ending with

An ion with aname ending with

-ic acid

-ous acid

-ate

-ite

Hill, Petrucci, General Chemistry An Integrated Approach 1999, page 60

Oxyacid Naming Rules

Naming Acids Practice

Formula Name

1 HCl ____________________

2 HClO ____________________

3 ________________ sulfuric acid

4 ________________ hydrofluoric acid

5 H3N ____________________

6 ________________ periodic acid

hydrochloric acid

HIO4

hypochlorous acid

H2SO4

HF

hydronitric acid

Common Oxyacid Names

The following table lists the most common families of oxy acids.one moreoxygen atom

most“common”

one lessoxygen

two lessoxygen’s

HClO4

perchloric acid

HClO3

chloric acid

HClO2

chlorous acid

HClOhypochlorous acid

H2SO4

sulfuric acid

H2SO3

sulfurous acid

H3PO4

phosphoric acid

H3PO3

phosphorous acid

H3PO2

hypophosphorous acid

HNO3

nitric acid

HNO2

nitrous acid

Common AcidsStrong Acids

stomach acid;

(dissociate ~100%)

hydrochloric acid: HCl H+ + Cl– --

pickling: cleaning metals w/conc. HCl

sulfuric acid: H2SO4 2 H+ + SO42–

-- #1 chemical; (auto) battery acid

explosives;nitric acid: HNO3 H+ + NO3

– -- fertilizer

Common Acids (cont.)Weak Acids (dissociate very little)

acetic acid: CH3COOH H+ + CH3COO– --

hydrofluoric acid: HF H+ + F– -- citric acid, H3C6H5O7 -- ascorbic acid, H2C6H6O6 -- lactic acid, CH3CHOHCOOH --

vinegar; naturally made by apples

used to etch glass

lemons or limes; sour candy

vitamin C

waste product of muscular exertion

carbonic acid, H2CO3

-- carbonated beverages

-- CO2 + H2O H2CO3

dissolveslimestone(CaCO3)

rainwaterin air

H2CO3: cave formation H2CO3: natural acidity of lakes

H2CO3: beverage carbonation

Dissociation and Ion ConcentrationStrong acids or bases dissociate ~100%.

HNO3 H+ + NO3–

HNO3 H+ + NO3–

For “strongs,” weoften use two arrows

of differing lengthOR

just a single arrow.

H+ NO3– +H+ NO3

–

1 2

100 1000/L

0.0058 M

1 2

100 1000/L

0.0058 M

1 2

100 1000/L

0.0058 M

+ + + + +

HCl H+ + Cl–

4.0 M 4.0 M 4.0 M+monoprotic

acid

H2SO4 2 H+ + SO42–

2.3 M 4.6 M 2.3 M+

SO 42–

H+

H+

SO42–H+

H++ diprotic

acid

Ca(OH)2

Ca2+

2 OH–

+ 0.025 M 0.025 M 0.050 M+

pH CalculationsRecall that the hydronium ion (H3O+) is the speciesformed when hydrogen ion (H+) attaches to water(H2O). OH– is the hydroxide ion.

For this class, in any aqueous sol’n, [ H3O+ ] [ OH– ] = 1 x 10–14

( or [ H+ ] [ OH– ] = 1 x 10–14 )

…so whether we’re counting front wheels (i.e., H+)or trikes (i.e., H3O+) doesn’t much matter.

H+ H3O+

The number offront wheels is the

same as the numberof trikes…

If hydronium ion concentration = 4.5 x 10–9 M,find hydroxide ion concentration.

[ H3O+ ] [ OH– ] = 1 x 10–14

= 2.2 x 10–6 M

0.0000022 M

2.2–6 M

10x

yx

EE1 1 4 4 . 5 EE 9 =–..– –] OH [

10 x 1 ] OH [3

-14-

M 10 x 4.510 x 1 9-

14-

OR

Given: Find: A. [ OH– ] = 5.25 x 10–6 M [ H+ ] = 1.90 x 10–9 MB. [ OH– ] = 3.8 x 10–11 M [ H3O+ ] = 2.6 x 10–4 MC. [ H3O+ ] = 1.8 x 10–3 M [ OH– ] = 5.6 x 10–12 MD. [ H+ ] = 7.3 x 10–12 M [ H3O+ ] = 7.3 x 10–12 M

Find the pH of each sol’n above. pH = –log [ H3O+ ] ( or pH = –log [ H+ ] )

A.

B. C. D. 3.59 2.74 11.13

8.72

pH = –log [ H3O+ ] = –log [1.90 x 10–9 M ]

log 1 9 EE 9 =– –.

[ H3O+ ] [ OH– ] = 1 x 10–14

A few last equations…

pOH = –log [ OH– ] pH + pOH = 14 [ OH– ] = 10–pOH

[ H3O+ ] = 10–pH

( or [ H+ ] = 10–pH )

pOH

pH

[ OH– ]

[ H3O+ ]

pH + pOH = 14 [ H3O+ ] [ OH– ] = 1 x 10–14

[ H3O+ ] = 10–pH

pH = –log [ H3O+ ]

[ OH– ] = 10–pOH

pOH = –log [ OH– ]

If pH = 4.87,find [ H3O+ ].

pOH

pH

[ OH1– ]

[ H3O1+ ]

pH + pOH = 14 [ H3O1+ ] [ OH1– ] = 1 x 10–14

[ H3O1+ ] = 10–pH

pH = –log [ H3O1+ ]

[ OH1– ] = 10–pOH

pOH = –log [ OH1– ] pOH

pH

[ OH– ]

[ H3O+ ]

pH + pOH = 14 [ H3O+ ] [ OH– ] = 1 x 10–14

[ H3O+ ] = 10–pH

pH = –log [ H3O+ ]

[ OH– ] = 10–pOH

pOH = –log [ OH– ]

[ H3O+ ] = 10–pH

= 10–4.87

On a graphing calculator…

log – . 8 =742nd

10x[ H3O1+ ] = 1.35 x 10–5 M

If [ OH– ] = 5.6 x 10–11 M,find pH.

pOH

pH

[ OH1– ]

[ H3O1+ ]

pH + pOH = 14 [ H3O1+ ] [ OH1– ] = 1 x 10–14

[ H3O1+ ] = 10–pH

pH = –log [ H3O1+ ]

[ OH1– ] = 10–pOH

pOH = –log [ OH1– ] pOH

pH

[ OH1– ]

[ H3O1+ ]

pH + pOH = 14 [ H3O1+ ] [ OH1– ] = 1 x 10–14

[ H3O1+ ] = 10–pH

pH = –log [ H3O1+ ]

[ OH1– ] = 10–pOH

pOH = –log [ OH1– ]

Find [ H3O+ ] Find pOH= 1.79 x 10–4 M

Then find pH…

pH = 3.75 pH = 3.75

= 10.25

For the following problems, assume 100% dissociation.

Find pH of a 0.00057 M nitric acid (HNO3) sol’n.

HNO3

0.00057 M 0.00057 M 0.00057 M (“Who cares?”)(GIVEN) (affects pH)

H+ + NO3–

pH = –log [ H3O+ ]

= –log (0.00057)

= 3.24 pOH

pH

[ OH1– ]

[ H3O1+ ]

pH + pOH = 14 [ H3O1+ ] [ OH1– ] = 1 x 10–14

[ H3O1+ ] = 10–pH

pH = –log [ H3O1+ ]

[ OH1– ] = 10–pOH

pOH = –log [ OH1– ] pOH

pH

[ OH1– ]

[ H3O1+ ]

pH + pOH = 14 [ H3O1+ ] [ OH1– ] = 1 x 10–14

[ H3O1+ ] = 10–pH

pH = –log [ H3O1+ ]

[ OH1– ] = 10–pOH

pOH = –log [ OH1– ]

Find pH of a 3.2 x 10–5 M barium hydroxide (Ba(OH)2)sol’n. Ba(OH)2

3.2 x 10–5 M 3.2 x 10–5 M 6.4 x 10–5 M(“Who cares?”)(GIVEN) (affects pH)

Ba2+ + 2 OH–

pOH = –log [ OH– ]= –log (6.4 x 10–5)= 4.19

pOH

pH

[ OH1– ]

[ H3O1+ ]

pH + pOH = 14 [ H3O1+ ] [ OH1– ] = 1 x 10–14

[ H3O1+ ] = 10–pH

pH = –log [ H3O1+ ]

[ OH1– ] = 10–pOH

pOH = –log [ OH1– ] pOH

pH

[ OH1– ]

[ H3O1+ ]

pH + pOH = 14 [ H3O1+ ] [ OH1– ] = 1 x 10–14

[ H3O1+ ] = 10–pH

pH = –log [ H3O1+ ]

[ OH1– ] = 10–pOH

pOH = –log [ OH1– ]

pH = 9.81

Find the concentration of an H2SO4 sol’n w/pH 3.38. H2SO4 2 H+ + SO4

2–

[ H+ ] = 10–pH = 10–3.38 = 4.2 x 10–4 M

pOH

pH

[ OH1– ]

[ H3O1+ ]

pH + pOH = 14 [ H3O1+ ] [ OH1– ] = 1 x 10–14

[ H3O1+ ] = 10–pH

pH = –log [ H3O1+ ]

[ OH1– ] = 10–pOH

pOH = –log [ OH1– ] pOH

pH

[ OH1– ]

[ H3O1+ ]

pH + pOH = 14 [ H3O1+ ] [ OH1– ] = 1 x 10–14

[ H3O1+ ] = 10–pH

pH = –log [ H3O1+ ]

[ OH1– ] = 10–pOH

pOH = –log [ OH1– ]

(spa

ce)

4.2 x 10–4 M (“Who cares?”) X M 2.1 x 10–4 M

[ H2SO4 ] =2.1 x 10–4 M

Find pH of a sol’n with 3.65 g HCl in 2.00 dm3 of sol’n.

HCl H+ + Cl–

[ HCl ]

0.05 M 0.05 M 0.05 M

pH = –log [ H+ ]

= –log (0.05)

= 1.3 pOH

pH

[ OH1– ]

[ H3O1+ ]

pH + pOH = 14 [ H3O1+ ] [ OH1– ] = 1 x 10–14

[ H3O1+ ] = 10–pH

pH = –log [ H3O1+ ]

[ OH1– ] = 10–pOH

pOH = –log [ OH1– ] pOH

pH

[ OH1– ]

[ H3O1+ ]

pH + pOH = 14 [ H3O1+ ] [ OH1– ] = 1 x 10–14

[ H3O1+ ] = 10–pH

pH = –log [ H3O1+ ]

[ OH1– ] = 10–pOH

pOH = –log [ OH1– ]

(spa

ce)

= 0.05 M HCl

= MHCl Lmol

g 36.5HCl mol 13.65 g

= 2.00 L

What mass of Al(OH)3 is req’d to make 15.6 L of asol’n with a pH of 10.72?

Al(OH)3 Al3+ + 3 OH–

pOH

pH

[ OH1– ]

[ H3O1+ ]

pH + pOH = 14 [ H3O1+ ] [ OH1– ] = 1 x 10–14

[ H3O1+ ] = 10–pH

pH = –log [ H3O1+ ]

[ OH1– ] = 10–pOH

pOH = –log [ OH1– ] pOH

pH

[ OH1– ]

[ H3O1+ ]

pH + pOH = 14 [ H3O1+ ] [ OH1– ] = 1 x 10–14

[ H3O1+ ] = 10–pH

pH = –log [ H3O1+ ]

[ OH1– ] = 10–pOH

pOH = –log [ OH1– ]

pOH = 3.28 = 10–3.28 = 5.25 x 10–4 M

(spa

ce)

[ OH– ] = 10–pOH

5.25 x 10–4 M (“w.c.?”) 1.75 x 10–4 M

mol = 1.75 x 10–4(15.6)

= 0.213 g Al(OH)3

molAl(OH) = M L 3

= 0.00273 mol Al(OH)3

Acid-Dissociation Constant, Ka

For the generic reaction in sol’n: A + B C + D

For strong acids, e.g., HCl…

HCl H+ + Cl–

= “BIG.”

Assume 100% dissociation;Ka not applicable for strong acids.

] B [ ] A [] D [ ] C [ K

] REACTANTS [] PRODUCTS [ K aa

-

a[ H ] [ Cl ]K

[ HCl ]

lots~0 lots

For weak acids, e.g., HF… HF H+ + F–

= “small”(6.8 x 10–4 for HF)

Ka’s for other weak acids: CH3COOH H+ + CH3COO–

HNO2 H+ + NO2–

HC3H5O3 H+ + C3H5O3–

Ka = 1.8 x 10–5

Ka = 1.4 x 10–4

Ka = 4.5 x 10–4

The weaker the acid, the smaller the Ka. “ stronger “ “ , “ larger “ “ .

-

a[ H ] [ F ]K

[ HF ]

~0 ~0lots

Indicators chemicals that change color,depending on the pH

Two examples, out of many:

litmus…………………

phenolphthalein……..

red in acid,blue in base

clear in acid,pink in base acid base

acid base

pinklear

bo

Measuring pH

litmus paper phenolphthalein Basically, pH < 7 or pH > 7.

pH paper -- contains a mixture of various indicators

--

--

each type of papermeasures a range of pH

pH anywhere from 0 to 14

universal indicator -- is a mixture of several indicators --

4 5 6 7 8 9 10

R O Y G B I V

pH 4 to 10

Measuring pH (cont.)

pH meter -- measures small voltages in solutions -- calibrated to convert voltages into pH -- precise measurement of pH

__HCl + __NaOH ________ + ______

__H3PO4 + __KOH ________ + ______

__H2SO4 + __NaOH ________ + ______

__HClO3 + __Al(OH)3 ________ + ______

________ + ________ __AlCl3 + ______

________ + ________ __Fe2(SO4)3 + ______

1

Neutralization Reaction ACID + BASE SALT + WATER

NaCl H2O

H2O

H2O

H2O

H2O

H2O

1

1

1

1 3

1 2

3 1

K3PO4 3

1 Na2SO4 2

Al(ClO3)3

31HCl Al(OH)3 3 1

H2SO4 Fe(OH)3 3 2 1 6

31

1

Titration If an acid and a base are mixed together in the rightamounts, the resulting solution will be perfectlyneutralized and have a pH of 7.

-- For pH = 7……………..

then mol = M L

In a titration, the above equation helps us to use…

a KNOWN conc. of acid (or base) to determinean UNKNOWN conc. of base (or acid).

, L

mol M Since

BOHAOHL M L M -11

3

BOHAOHVM VM -11

3 B

-1A

13 V] OH [ V] OH [ or

mol H3O1+ = mol OH1–

2.42 L of 0.32 M HCl are used to titrate 1.22 L of anunknown conc. of KOH. Find the molarity of the KOH.

HCl H1+ + Cl1–

KOH K1+ + OH1–

0.32 M 0.32 M

X M X M

0.32 M (2.42 L) = (1.22 L)1.22 L1.22 L

= MKOH = 0.63 M

[ H3O1+ ] VA = [ OH1– ] VB

[ OH1– ]

[ OH1– ]

458 mL of HNO3 (w/pH = 2.87) are neutralizedw/661 mL of Ba(OH)2. What is the pH of the base?

[ H3O1+ ] = 10–pH

= 10–2.87

= 1.35 x 10–3 M

[ H3O1+ ] VA = [ OH1– ] VB

(1.35 x 10–3)(458 mL) = [ OH1– ] (661 mL)

[ OH1– ] = 9.35 x 10–4 MpOH = –log (9.35 x 10–4) = 3.03

pH = 10.97

OK

If we find this,we can find the

base’s pH.

OK

How many L of 0.872 M sodium hydroxide willtitrate 1.382 L of 0.315 M sulfuric acid?

H2SO4 2 H1+ + SO42– NaOH Na1+ + OH1–

0.872 M

(1.382 L)

[ H3O1+ ] VA = [ OH1– ] VB

0.630 M = (VB)0.872 M

0.872 M0.315 M 0.630 M

? ?

VB = 0.998 L

(H2SO4)

(NaOH)

Strong Acid/Strong Base Titration

pH Equivalence Ptmol H+ = mol OH-

mL of base

HCl + NaOH H2O + NaCl

At this point, everything is water, Na+ and Cl-, so the pH = 7 EndpointOvershoot

Titrate with Bromthymol Blue

Partial Neutralization

1.55 L of0.26 M KOH

2.15 L of0.22 M HCl

pH = ?

Procedure:

1. Calc. mol of substance, then mol H+ and mol OH–.

2. Subtract smaller from larger.

3. Find [ ] of what’s left over, and calc. pH.

1.55 L of0.26 M KOH

2.15 L of0.22 M HCl

mol KOH == 0.403 mol OH–

0.26 M (1.55 L) = 0.403 mol KOH

mol HCl == 0.473 mol H+

0.22 M (2.15 L) = 0.473 mol HCl

[ H1+ ] =

= 0.070 mol H+

= 0.0189 M H+ 0.070 mol H+

1.55 L + 2.15 L

LEFT OVER

= –log (0.0189) pH = –log [ H+ ] = 1.72

mol

L M

4.25 L of 0.35 M hydrochloric acid is mixed w/3.80 L of0.39 M sodium hydroxide. Find final pH. Assume

100% dissociation.

(HCl)

(NaOH)

mol HCl == 1.4875 mol H+

0.35 M (4.25 L) = 1.4875 mol HCl

mol NaOH == 1.4820 mol OH–

0.39 M (3.80 L) = 1.4820 mol NaOH

[ H1+ ] =

= 0.0055 mol H+

= 6.83 x 10–4 M H+ 0.0055 mol H+

4.25 L + 3.80 L

LEFT OVER

= –log (6.83 x 10–4) pH = –log [ H+ ] = 3.17

5.74 L of 0.29 M sulfuric acid is mixed w/3.21 L of0.35 M aluminum hydroxide. Find final pH.

Assume 100% dissociation.

(H2SO4)

(Al(OH)3)mol H2SO4 =

= 3.3292 mol H+ 0.29 M (5.74 L) = 1.6646 mol H2SO4

mol Al(OH)3 == 3.3705 mol OH–

0.35 M (3.21 L) = 1.1235 mol Al(OH)3

= 0.0413 mol OH– LEFT OVER

[ OH1– ] = = 0.00461 M OH–0.0413 mol OH–

5.74 L + 3.21 L

pOH = –log (0.00461) = 2.34

pH = 11.66

A. 0.038 g HNO3 in 450 mL of sol’n. Find pH.

= 0.45 L

0.038 g HNO3 63 g1 mol( )= 6.03 x 10–4 mol HNO3

1000 mL1 L( )

= 6.03 x 10–4 mol H+

[ H1+ ] = = 1.34 x 10–3 M H+ 6.03 x 10–4 mol H+

0.45 L

= –log (1.34 x 10–3) pH = –log [ H+ ] = 2.87

mol

L M

B. 0.044 g Ba(OH)2 in 560 mL of sol’n. Find pH. = 0.56 L 1000 mL

1 L( )

0.044 g Ba(OH)2 171.3 g1 mol( ) = 2.57 x 10–4 mol Ba(OH)2

= 5.14 x 10–4 mol OH–

[ OH1– ] = = 9.18 x 10–4 M OH–5.14 x 10–4 mol OH–

0.56 L

pOH = –log (9.18 x 10–4) = 3.04

pH = 10.96

mol

L M

C. Mix them. Find pH of resulting sol’n.

From acid… 6.03 x 10–4 mol H+

From base… 5.14 x 10–4 mol OH–

= 8.90 x 10–5 mol H+ LEFT OVER

[ H1+ ] = = 8.81 x 10–5 M H+ 8.90 x 10–5 mol H+ 0.45 L + 0.56 L

= –log (8.81 x 10–5) pH = –log [ H+ ] = 4.05

Buffers

Example: The pH of blood is 7.4. Many buffers are present to keeppH stable.

hyperventilating: CO2 leaves blood too quickly

mixtures of chemicals that resist changes in pH

H+ + HCO3– H2CO3 H2O + CO2

[ CO2 ]

alkalosis: blood pH is too high (too basic)

shift pH

[ H+ ]

right (more basic)

Remedy: Breathe into bag.

[ CO2 ]

shift pH

[ H+ ]

left

(more acidic;closer to normal)

H+ + HCO3– H2CO3 H2O + CO2

acidosis: blood pH is too low (too acidic)

Maintain blood pH with a healthy diet and regular exercise.

More on buffers:

-- a combination of a weak acid and a salt

-- together, these substances resist changes in pH

1. 2. **Conclusion:

If you add acid… (e.g., HCl H+ + Cl–)

1. 2. **Conclusion:

If you add base…(e.g., KOH K+ + OH–)

(A) weak acid: CH3COOH CH3COO– + H+ (lots) (little) (little)

(B) salt: NaCH3COO Na+ CH3COO–

(little) (lots) (lots) +

large amt. of CH3COO– (from (B)) consumesextra H+, so (A) goes

pH remains relatively unchanged.

pH remains relatively unchanged.

extra OH– grabs H+ from the large amt. of

CH3COOH and forms CH3COO– and H2O

Equivalence Ptmol H+ = mol OH-

mL of acid

At this point, the solution

contains water, Cl-,

and NH4+, so

the pH < 7.

EndpointOvershoot

NH3 + H+ NH4+

BUFFERING REGION

Weak Base/Strong Acid Titration HCl + NH4OH H2O + NH4Cl

Titrate with Methyl Orange

pH

Equivalence Ptmol H+ = mol OH-

mL of base

At this point, the solution

contains water, Na+, and

C2H3O2 - , so

the pH > 7.

EndpointOvershoot

BUFFERING REGION

HC2H3O2 C2H3O2

- + H+

Weak Acid/Strong Base Titration HC2H3O2 + NaOH H2O + NaC2H3O2

Titrate with Phenolphthalein

pH

Amphoteric Substances

can act as acids OR bases

e.g., NH3

NH3

accepts H+

donates H1+

(BASE)

(ACID)

NH4+NH2

–

e.g., H2O

H2Oaccepts H+

donates H1+

(BASE)

(ACID)

H3O+OH–