unit 11: equilibrium / acids and bases reversible reaction: r p and r p acid dissociation is a...
Post on 13-Dec-2015
228 Views
Preview:
TRANSCRIPT
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 Eyeglasses
AgCl + 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
> 7
bitter
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
10
100100,000
Acid Nomenclature
binary acids: acids w/H and one other element
Binary Acid Nomenclature
1. Write “hydro.”
2. Write prefix of the other element, followed by “-ic acid.”
HF
HCl
HBr
hydroiodic acid
hydrosulfuric acid
hydrofluoric acid
hydrochloric acid
hydrobromic acid
HI
H2S
Oxyacids
oxyacids: acids containing H, O, and one other element
Common oxyanions (polyatomic ions that contain
oxygen) 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 acidHClO4 Perchloric acid
HClO3 Chloric acid HClO2 Chlorous acid
HClO Hypochlorous acid
phosphorus acid H3PO3
hypobromous acid HBrO
persulfuric 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+
monoproticacid
H2SO4 2 H+ + SO42–
2.3 M 4.6 M 2.3 M+
SO 42–
H+
H+
SO 42–H+
H+
+ diproticacid
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 4.5
10 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+ + SO42–
[ 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– ]
(sp
ace
)
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– ]
(sp
ace
)
= 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
(sp
ace
)
[ 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 M
pOH = –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)) consumes
extra 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–
top related