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H 2 SO 4 2 H + + SO 4 2–. 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. . Acids and Bases. litmus paper. < 7. > 7. pH . pH . sour. bitter. - PowerPoint PPT Presentation
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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.