Acids and Bases Review/Equilibrium

Acids and Bases Review/Equilibrium

reversible reaction: R P and R P

Acid dissociation is a reversible reaction and is said to be in equilibrium.

H2SO4 2 H+ + SO42–

< 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

Aliens: Acid Blood Robot Chicken: Alien Blood

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

pH = –log [H3O+]

Measuring pH• pH meter • acid-base indicators:

e.g.,

• pH paper is paper impregnated with mixtures of various indicators.

a pair of electrodeschemicals whose color

depends on pHLitmus <7 >7 Phenolphthalein <8.2 >8.2 bromthymol blue <6 >7.6 methyl red <4.4 >6.2

4 5 6 7 8 9 10

R O Y G B I V

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; extremely caustic

lemons or limes; sour candy

vitamin C

waste product of muscular exertion

LSD… “Acid”• Lysergic acid diethylamide, C20H25N3O• LSD from the German :"Lysergsäure-

diethylamid“• Ironically, not acidic, but slightly basic• Derived from ergot, a grain fungus

carbonic acid: H2CO3

• carbonated beverages

• CO2 + H2O H2CO3

dissolveslimestone(CaCO3)

rainwaterin air

H2CO3: cave formation H2CO3: natural acidity of lakes

H2CO3: beverage carbonation

Acid Attacks(primarily on women…)

• Developing Nations Around the World• UK Model Katie Piper

Katie Piper before her acid attack

…and after.

Common BasesStrong Bases

Lye: used to make soap; clogged drain cleaner

(dissociate ~100%)

sodium hydroxide: NaOH Na+ + OH– •

ammonia: NH3 + H2O NH4+ + OH–

• common household cleaning sol’n (Windex); hair dye

household bleach; pool “chlorine”sodium hypochlorite: NaClO + H2O HClO + OH– •

“Slaked lime”: limestone plus water; mortar/plastercalcium hydroxide: Ca(OH)2 Ca2+ + 2 OH– •

Weak Bases (dissociate very little)

Strong Acids and Bases • these are strong electro- lytes that exist entirely as ions in aqueous solution • memorize the names

and formulas of the seven strong acids...

...and the eight strong, hydroxide bases...

the hydroxides of...

“strong base cations”

Li, Na, K, Rb,Cs, Ca, Sr, Ba

hydrochloric, HCl hydrobromic, HBr hydroiodic, HI chloric, HClO3 perchloric, HClO4 nitric, HNO3 sulfuric, H2SO4

Polyatomic Ion Sheet

Halogens

Molarity (M) Review

molarity (M) = moles of soluteL of sol’n

• used most often in this class

Lmol M

mol

L M

Na+

How many mol solute are req’d to make1.35 L of 2.50 M sol’n?

What mass sodium hydroxide is this?

mol

L Mmol = M L = 2.50 M (1.35 L )

= 3.38 mol

mol 1g 40.03.38 mol = 135 g NaOH

OH– NaOH

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

HNO3 H+ + NO3–

NaOH Na+ + OH–

For “strongs,” we often use two arrows of differing length OR 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 right now, 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 big wheels (i.e., H3O+) doesn’t much matter.

H+

H3O+ The number of

front wheels is thesame as the number

of big wheels…

At 25oC, calculate the hydrogen ion concentration ifthe hydroxide ion concentration is 2.7 x 10–4 M. Isthis solution an acid or a base?

[H+] [OH–] = 1.0 x 10–14

[H+] (2.7 x 10–4) = 1.0 x 10–14

[H+] = 3.7 x 10–11 M

[H+] < [OH–]

base

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. Remember… 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– ]

1. If pH = 4.87,find [ H3O+ ].

pOH

pH [ H3O1+ ]

pH + pOH = 14

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[ H3O+ ] = 1.35 x 10–5 M

For the following problems, assume 100% dissociation.

2. 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– ]

3. 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

4. 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

5. 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

6. 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

7. 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

Amphoteric substances can be acids or bases,depending on the reaction conditions (e.g. H2O)

HCl(aq) + H2O(l)

NH3(aq) + H2O(l)

H3O+(aq) + Cl–(aq)

NH4+(aq) + OH–(aq)

NH3 is another example. [ ]+

[ ]–

NH3(aq) + H2O(l) NH2-(aq) + H3O+(aq)

HCO3-(aq) + H2O(l) OH–(aq) + H2CO3(aq)

• The two substances in a conjugate acid-base pair differ by a H+...

• Strong acids / bases easily / ____ H+.

• Weak acids / bases do NOT easily / ____ H+.

and the acid has the extra H+.

• In acid-base equilibria, protons are donated in forward and reverse reactions.

HNO2(aq) + H2O(l) NO2–(aq) + H3O+(aq)

ACID CONJ.BASEbase conj.

acid

donate

donate

accept

accept

• The stronger a/n acid / base,

the weaker its conj. base / acid.

__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 H+ + Cl–

KOH K+ + OH–

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

[ H3O+ ] VA = [ OH– ] VB

[ OH– ]

[ OH– ]

Acid Base

Fill up flask with acid

Then, titrate with base

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

[ H3O+ ] = 10–pH

= 10–2.87

= 1.35 x 10–3 M

[ H3O+ ] VA = [ OH– ] VB

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

[ OH– ] = 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 H+ + SO42– NaOH Na+ + OH–

0.872 M

(1.382 L)

[ H3O+ ] VA = [ OH– ] VB

0.630 M = (VB)0.872 M

0.872 M0.315 M 0.630 M

? ?

VB = 0.998 L

(H2SO4)

(NaOH)

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

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

Chemical Equilibrium

Chemical equilibrium is reached when reaction rates become equal, and the concentrations of R’s and P’s no longer change

reactants products

• system must be closed

• equilibrium is a dynamic process(although it might look static)

rate at whichR P

rate at whichP R=

amt. of R = amt. of P

law of mass action: expresses the relationship between [ ]s of R and P in any reaction

For a system at equilibrium with the balancedequation aA + bB pP + qQ the equilibrium-constant expression is:

ba

qp

c [B][A][Q][P]K i.e.,

PRODUCTSREACTANTS)(

You can only include reaction species in the gaseous or aqueous phase. Do NOT include solids/liquids.

Write the equilibrium-constant expressions for thefollowing reactions.

N2 (g) + 3 H2 (g) 2 NH3 (g)

2 SO3 (g) 2 SO2 (g) + O2 (g)

322

23

c ]][H[N][NHK

23

22

2c ][SO

]O[][SOK

Fritz Haber (1868–1934) discovereda way to generate ammonia fromhydrogen and nitrogen at highpressure. The ammonia was neededfor Germany’s munitions industry,which was cut off from the nitratesources of South America by theBritish blockade during WWI.

Write the equilibrium-constant expression forCaCO3(s) CaO(s) + CO2(g).

Write expressions for Kc.

CO2(g) + H2(g) CO(g) + H2O(l)

SnO2(s) + 2 CO(g) Sn(s) + 2 CO2(g)

][CaCO]CO[[CaO]K

3

2c ][CO'K 2c

]][H[CO[CO]K

22c

2

22

c [CO]][COK

Type of Equilibrium Constants (K)There are lots of different “K’s” they are all the same concept, just different types of equations.

N2(g) + 3H2(g) ↔ 2NH3(g)

HNO2(aq) + H2O(l) ↔ H3O+(aq) + NO2-(aq)

NH3(aq) + H2O(l) ↔ NH4+(aq) + OH-(aq)

23

32 2

[ ][ ][ ]cNHKN H

3

2 2

2

3NH

pN H

PK

P P

3 2

2

[ ][ ][ ]a

H O NOKHNO

4

3

[ ][ ][ ]b

NH OHKNH

For use with acids…

For use with bases…

Kconc ≠ Kpress

Calculating KH2(g) + I2(g) 2HI(g)

The system is allowed to reach equilibrium and the following data is collected:

1. [H2] = 4.953x10-4, [I2] =4.953x10-4, [HI] = 3.655x10-3

2. [H2] = 1.141x10-3, [I2] = 1.141x10-3, [HI] = 8.410x10-3

3. [H2] = 3.560x10-3, [I2] = 1.250x10-3, [HI] = 1.559x10-2

2 3 2

4 42 2

[ ] [3.655 10 ] 54.46[ ][ ] [4.953 10 ][4.953 10 ]HI xKH I x x

2 3 2

3 32 2

[ ] [8.410 10 ] 54.33[ ][ ] [1.141 10 ][1.141 10 ]HI xKH I x x

2 2 2

3 32 2

[ ] [1.559 10 ] 54.62[ ][ ] [3.560 10 ][1.250 10 ]HI xKH I x x

2

2 2

[ ][ ][ ]HIKH I

• they are reciprocals

The Magnitude of the Equilibrium Constant

If K >> 1...

If K << 1...

products are favored.Eq. “lies to the right.”

reactants are favored.Eq. “lies to the left.”

The K for the forward and reverse reactionsare NOT the same.

• You must write out the reaction and specify the temperature when reporting a K.

PRODUCTSREACTANTS)(

The Autoionization of Water: Kw

In ordinary water, we constantly have...

H2O(l) + H2O(l) H3O+(aq) + OH–(aq)

• reaction is very rapid in each direction

• at room temp., ~1 out of a billion m’cules are ionized, so pure water is a poor conductor

For the above equation, Kc =

If we exclude the pure liquid...

This equation is taken to be valid for purewater and for dilute aqueous solutions.

[ H+ ] > [ OH– ] [ H+ ] < [ OH– ] [ H+ ] = [ OH– ]

H2O(l) + H2O(l) H3O+(aq) + OH–(aq)

[H3O+] [OH–][H2O]2

Kw = [H3O+] [OH–] = [H+] [OH–] = 1.0 x 10–14

(@ 25oC)

BASE NEUTRALACID

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

large Ka:

Weak Acids• Most acids are weak (i.e., only partially ionized)

• For a weak acid HX... HX(aq) H+(aq) + X–(aq)

• acid-dissociation constant Ka = [H+] [X–]

[HX]

small Ka:

stronger acid

weaker acid

The % of a weak acidthat is ionized is given

by the equation:

[H+] at eq. x 100% ionization =[acid]orig.

= “small”

Ka acetic acid = 1.8 x 10–5

If you know the concentrations of only some substances at equilibrium, make a chart and use reaction stoichiometry to figure out the other concentrations at equilibrium. THEN plug and chug.

“What kind of chart?”

“Ice, ice, baby…”

I = “initial” C = “change” E = “equilibrium”

Weak Acid Equilibrium: Ka1. Write the reaction for carbonic acid (H2CO3) dissociating in water.

H2CO3(aq) + H2O(l) H3O+(aq) + HCO3-(aq)

a. Write the equilibrium expression for this reaction.

b. If [H2CO3] = 0.100M and it shows 0.212% ionization, what is the value of Ka?

H2CO3(aq) + H2O(l) H3O+(aq) + HCO3-(aq)

Initial

Change

Equilibrium

0.100M ------- 0 0

- 2.12x10-4 ------- + 2.12x10-4 + 2.12x10-4

0.099788 ------- 2.12x10-4 2.12x10-4

3 3

2 3

[ ][ ][ ]

H O HCOK

H CO

4 473 3

2 3

[ ][ ] [2.12 10 ][2.12 10 ] 4.50 10[ ] [0.099788]

H O HCO x xK xH CO

Weak Acid Equilibrium: Ka

Write the reaction for carbonic acid (H2CO3) dissociating in water.

H2CO3(aq) + H2O(l) H3O+(aq) + HCO3-

(aq)

c. If [H2CO3] = 0.0500M and it shows 0.300% ionization, what is the value of Ka? H2CO3(aq) + H2O(l) H3O+(aq) + HCO3

-(aq)ICE

0.0500M ------- 0 0

- 1.50x10-4 ------- + 1.50x10-4 + 1.50x10-4

0.04985 ------- 1.50x10-4 1.50x10-4

4 473 3

2 3

[ ][ ] [1.50 10 ][1.50 10 ] 4.51 10[ ] [0.04985]

H O HCO x xK xH CO

Weak Acid Equilibrium: KaWrite the reaction for carbonic acid (H2CO3) dissociating

in water.

H2CO3(aq) + H2O(l) H3O+(aq) + HCO3-

(aq)

c. Given the value of Ka, if [H2CO3] = 0.0100M, what will the concentrations of the products? What is the % ionization? H2CO3(aq) + H2O(l) H3O+(aq) + HCO3

-(aq)ICE

0.0100M ------- 0 0

- X ------- + X + X

0.0100-X ------- X X2

7 73 3

2 3

[ ][ ] [ ][ ] 4.50 10 4.50 10[ ] [0.0100 ] 0.01

H O HCO X X XK x xH CO X X

0X2 = 4.50x10-7(0.01)X = √(4.50x10-9)X = 6.71x10-5M

56.71 10 100 0.671%0.0100x x

< 5.00%5% Rule: at equilibrium, [H2CO3] ≈

initial [H2CO3]0. Only good if X < 5% of the initial concentration!

2 42

b b acXa

2. The reaction for benzoic acid (HC7H5O2) dissociating in water:

If [HC7H5O2] = 1.500M and it ionizes 0.651%, calculate the Ka for benzoic acid.

HC7H5O2(aq) + H2O(l) H3O+(aq) + C7H5O2 -(aq)

Calculate the pH for this solution.

Weak Acid Equilibrium: Ka

ICE

1.500M ------- 0 0

- .009765 ------- + .009765 + .009765

1.490 ------- + .009765 + .009765

253 7 5 2

7 5 2

[ ][ ] [0.009765] 6.40 10[ ] [1.490]

H O C H OK xHC H O

[H3O+] = 0.009765M

pH = -log (0.009765) = 2.01

Weak Acid Equilibrium: Ka3. Write the reaction for boric acid (H3BO3) dissociating in

water.

Given Ka = 5.4x10-10, if [H3BO3] = 0.500M, what will be the pH of the resulting solution?

H3BO3(aq) + H2O(l) H3O+(aq) + H2BO3-

(aq)ICE

0.500M ------- 0 0

- X ------- + X + X

0.500-X ------- X X2

10 103 2 3

3 3

[ ][ ] [ ][ ] 5.40 10 5.40 10[ ] [0.500 ] 0.50

H O H BO X X XK x xH BO X X

0X2 = 5.40x10-10(0.50)X = √(2.70x10-10)X = 1.64x10-5M

X = [H3O+] = 1.64x10-5M

pH = -log (1.64x10-5) = 4.78

51.64 10 100 0.00328%0.500x x

The reaction for ammonia (NH3) in water is as follows:

If Kb = 1.80x10-5 and [NH3] = 2.500M, calculate the pH of this solution.

NH3(aq) + H2O(l) NH4+(aq) + OH

-(aq)

Calculate the pH for this solution.

Weak Base Equilibrium: Kb

ICE

2.500M ------- 0 0

- X ------- + X + X

2.500 - X ------- + X + X

254

3

[ ][ ] [ ] 1.80 10[ ] [2.500 ]

NH OH XK xNH X

X= [OH-] = 0.00671MpOH = -log (0.00671M) = 2.17 11.83

0 X2 = 1.80x10-5(2.50)X = √(4.50x10-5)X = 0.00671M0.00671 100 0.268%

2.500x

Le Chatlier’s Principle

H2

NH3

N2

orig.eq.

neweq.

H2 addedat thistime

systemcounteracting

stress

N2(g) + 3 H2(g) 2 NH3(g)

As long as T stays the same, K stays the same! The changes keep K the same.

When a system at equilibrium is disturbed, it shifts to a new equilibrium that counteracts the disturbance.

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………………

Anything with a (s) or (l)no shift no shift

Change in pressure for gaseous equilibrium systems

2 NO2(g) N2O4(g)

If we decrease pressure, thesystem “wants” to...(goes to the side with more gas)

increase it

If we increase pressure, thesystem “wants” to...(goes to the side with less gas)

decrease it SHIFT

SHIFT

For H2(g) + I2(g) 2 HI(g),pressure changes result in...

NO SHIFT

Changes in temperature

These almost always result in... shifts in eq. AND changes in K.

For exothermic reactions:

R P + heat (DH is ____) • as T increases... shift , K• as T decreases... shift , K

For endothermic reactions:

R + heat P (DH is ____) • as T increases... shift , K• as T decreases... shift , K

–

+

For PCl5(s) PCl3(g) + Cl2(g), DHo = 87.9 kJ.Predict shifts for...

(a) adding Cl2

(b) increasing temperature

(c) decreasing volume

(d) adding PCl5(s)

PCl5(s) + 87.9 kJ PCl3(g) + Cl2(g)

We might want to rewrite the eq. as…

SHIFT

SHIFT

SHIFT

NO SHIFT

AgCl + energy Ago + Clo

shift to a new equilibrium:

Then go inside…

shift to a new equilibrium:

Light-Darkening Eyeglasses

“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.