Transcript S-1
Slide 1
Slide 2
Slide 3
Slide 4
Slide 5
Have a sour taste. Vinegar is a solution of acetic acid. Citrus
fruits contain citric acid.
React with certain metals to produce hydrogen gas.
React with carbonates and bicarbonates to produce carbon
dioxide gas
Basar
Have a bitter taste.
Feel slippery. Many soaps contain bases.
Slide 6
þ Mynda H+ (as H3O+) jónir í vatni (the hydronium ion is a
hydrogen ion attached to a water molecule)
þ Súrar á bragðið
þ Æta málma
þ Súrar vatnslausnir leiða rafmagn
þ Hvarfast við basa og mynda salt og vatn
þ pH undir 7
Slide 7
Anion
Ending
Binary
Acid Name
-ide
hydro-(stem)-ic acid
-ate
(stem)-ic acid
-ite
(stem)-ous acid
Ternary
Slide 8
A C ID S
start with 'H '
2 elem en ts
3 elem en ts
h yd ro - p refix
-ic en d in g
n o h yd ro - p refix
-a te en d in g
b ecom es
-ic en d in g
-ite en d in g
b ecom es
-o u s en d in g
Slide 9
• HBr (aq)
• H2CO3
• H2SO3
– H2SO4
hydrobromic acid
carbonic acid
sulfurous acid
Sulfuric acid
Slide 10
HI (aq)
HydroIodic acid
HCl (aq)
Hydrochloric acid
H2SO3
Sulfurous acid
HNO3
peIodic acid
HIO4
– perIodic acid
Slide 11
Mynda OH- jónir í vatni
Beiskir á bragðið, chalky
Vatnslausnir basa leiða rafmagn
Sápukenndir, sleipir
Hvarfast við sýrur og mynda salt og vatn
pH stærra en7
Slide 12
NaOH
natrium hydroxíð
lye - vítissódi
KOH
kalíumhydroxíð
liquid soap
Ba(OH)2
barium hydroxide
stabilizer for plastics
Mg(OH)2
magnesium hydroxide
“MOM” Milk of
magnesia
Al(OH)3
aluminum hydroxide Maalox (antacid)
Slide 13
Skilgreining #1: Arrhenius (traditional)
Sýrur – mynda H+ jónir (or hydronium ions
H3O+)
Basar – mynda OH- jónir
(problem: some bases don’t have hydroxide ions!)
Slide 14
Arrhenius sýra er efni sem myndar H+ (H3O+) in water
Arrhenius basi er efni sem myndar OH- in water
Slide 15
Skilgreining#2: Brønsted –Lowry
Sýra – Róteindagjafi
Basi – Róteindaþegi
Róteind er vetnisatóm sem hefur
misst rafeindina sína!
Slide 16
A Brønsted-Lowry sýra róteindagjafi
A Brønsted-Lowry basi róteindaþegi
Basi
sýra
Tilsvarandi
sýra
Tilsvarandi
basi
Slide 17
The Brønsted definition means
NH3 is a BASE in water —
+
and
water
is itself an NH
ACID
+
NH
+ H O
3
Base
4
2
Acid
Acid
OH
Base
-
Slide 18
Slide 19
HONORS ONLY!
Finnið sýruna, basann, tilsvarandi
(konjugered) sýru og tilsvarandi basa í
efnhvörvunum hér að neðan:
HCl + OH- Cl- + H2O
H2O + H2SO4 HSO4- + H3O+
Slide 20
Skilgr. #3 – Lewis
Lewis sýra – Efni sem
þiggur rafeindapar
Lewis basi – efni sem
gefur rafeindapar
Slide 21
Myndun oxoníum jónar er ágætt dæmi.
••
• O— H
•
H
+
A C ID
H
BASE
H
••
O— H
H
•Electron pair of the new O-H bond
originates on the Lewis base.
Slide 22
Slide 23
The heme group
in hemoglobin
can interact
with O2 and CO.
The Fe ion in
hemoglobin is a
Lewis acid
O2 and CO can
act as Lewis
bases
Heme group
Slide 24
Slide 25
Acidic solution
[H+ ] > [OH- ]
Neutral solution
[H+ ]
= [OH- ]
Basic solution
[H+ ]
< [OH- ]
Timberlake, Fig 9.3
Slide 26
Slide 27
pH = - log [H+]
(Remember that the [ ] mean Molarity)
Example: If [H+] = 1 X 10-10
pH = - log 1 X 10-10
pH = - (- 10)
pH = 10
Example: If [H+] = 1.8 X 10-5
pH = - log 1.8 X 10-5
pH = - (- 4.74)
pH = 4.74
Slide 28
Find the pH of
these:
1) A 0.15 M solution
of Hydrochloric
acid
2) A 3.00 X 10-7 M
solution of Nitric
acid
Slide 29
Ef pH í Coke er 3.12, [H+] = ???
Af því pH = - log [H+] þá
- pH = log [H+]
Taka antilógaritma (10x) af báðum:
10-pH = [H+]
[H+] = 10-3.12 = 7.6 x 10-4 M
*** Til að finna antilog eða 10^x þá ýtir þú
Á shift hnappinn og svo log
Slide 30
Lausn hefur pH of 8.5. hver er
mólstyrkur vetnisjóna í lausninni?
pH = - log [H+]
8.5 = - log [H+]
-8.5 = log [H+]
Antilog -8.5 = antilog (log [H+])
10-8.5 = [H+]
3.16 X 10-9 = [H+]
Slide 31
Vatn getur virkað sem sýra og basi.
Vatn klofnar í jónir, slíkt heitir
sjálfsjónun.
Jafnvægisfasti vatns er = Kw
Kw = [H3O+] [OH-] = 1.00 x 10-14 at 25 oC
Slide 32
Autoionization
OH
-
H 3O
+
Kw = [H3O+] [OH-] = 1.00 x 10-14 at 25 oC
Í hlutlausri lausn er [H3O+] = [OH-]
Kw = [H3O+]2 = [OH-]2
[H3O+] = [OH-] = 1.00 x 10-7 M
Slide 33
Since
acids and bases are
opposites, pH and pOH are
opposites!
pOH does not really exist, but it is
useful for changing bases to pH.
pOH looks at the perspective of
a base
pOH = - log [OH-]
Since pH and pOH are on opposite
ends,
pH + pOH = 14
Slide 34
pH
[H+]
[OH-]
pOH
Slide 35
Hvað er pH í 0.0010 M NaOH lausn?
[OH-] = 0.0010 (or 1.0 X 10-3 M)
pOH = - log 0.0010
pOH = 3
pH = 14 – 3 = 11
eða Kw = [H3O+] [OH-]
[H3O+] = 1.0 x 10-11 M
pH = - log (1.0 x 10-11) = 11.00
Slide 36
The pH of rainwater collected in a certain region of the
northeastern United States on a particular day was
4.82. What is the H+ ion concentration of the
rainwater?
The OH- ion concentration of a blood sample is
2.5 x 10-7 M. What is the pH of the blood?
Slide 37
[OH-]
[H+]
pOH
pH
Slide 38
Reiknið [H3O+], pH, [OH-], and pOH
Problem 1: Efnafræðingur býr til : (a) 3.0 M og
(b) 0.0024 M.
Reiknið [H3O+], pH, [OH-], and pOH lausnanna
við 25°C.
Problem 2: Hvað er [H3O+], [OH-], og pOH
lausnar með pH = 3.67? Er lausnin súr, basisk
eða hlutlaus?
Problem 3: Problem #2 með pH = 8.05?
Slide 39
Rammar og daufar sýrur
Römm sýra (sterk) klofnar auðveldlega
í jónir (100% jónuð)
HNO3, HCl, H2SO4 and HClO4 eru rammar
sýrur.
Slide 40
Römm sýra:
HNO3 (aq)+ H2O (l)H3O+ (aq) +NO3- (aq)
HNO3 is about 100% dissociated in
water.
Slide 41
Daufar sýrur (og basar)
Klofna ekki 100% í vatni og oft mjög lítið þ.e.
innan við 1%. Sjá ediksýru
Slide 42
Rammir basar
100%
klofnun í vatni.
NaOH (aq) Na+ (aq) + OH- (aq)
Einnig KOH and Ca(OH)2.
CaO (lime) + H2O -->
Ca(OH)2 (slaked lime)
CaO
Slide 43
Daufir basar
Minna en 100% jónaðir í vatni t.d.
ammoníak
NH3
(aq)
+ H2O (l) NH4+ (aq) + OH- (aq)
Slide 44
Slide 45
Consider acetic acid, HC2H3O2 (HOAc)
HC2H3O2 + H2O H3O+ + C2H3O2 Acid
Conj. base
+
Ka
[H 3 O ][OAc
[HOAc]
-
]
1.8 x 10
(K er skrifað Ka for ACID)
K gefur hlutfall klofið/óklofið í jónir
-5
Slide 46
Sýra
Conjugate
Basi
Increase
strength
Increase
strength
Slide 47
Dauf sýra hefur Ka < 1
Leiðir til lágs [H3O+] og pH of 2 - 7
Slide 48
Daufur basi hefur Kb < 1
Leiðir til lágs [OH-] og pH of 12 - 7
Slide 49
Slide 50
Þú hefur1.00 M HOAc. Reiknið jafnvægisstyrk HOAc, H3O+, OAc- og pH.
Step 1.
[HOAc]
[H3O+]
[OAc-]
initial
1.00
0
0
change
-x
+x
+x
x
x
equilib
1.00-x
Slide 51
Step 2. Skrifið jafnvægislíkingu Ka
K a 1.8 x 10
-5
+
=
[H 3 O ][OAc
[HOAc]
-
]
x
2
1.00 - x
Þetta er annars stigs jafna. Leyst sem slík….
Eða þú getur námundað x=0 neðan við strik ef x
er mjög lítið! (Þumalfingursreglan: 10-5 or
smaller is ok eða hámark 5% skekkja)
Slide 52
Step 3. Leysið út frá Ka
K a 1.8 x 10
-5
+
=
[H 3 O ][OAc
-
]
[HOAc]
x
2
1.00 - x
Gerum ráð fyrir að x sé mjög lítið því
Ka er svo lítið.
K a 1.8 x 10
-5
Þetta er fljótlegra svona.
=
x
2
1.00
Slide 53
Step 3. Leysum með námundun
K a 1.8 x 10
-5
=
x
2
1.00
x = [H3O+] = [OAc-] = 4.2 x 10-3 M
pH = - log [H3O+] = -log (4.2 x 10-3) = 2.37
[HOAC] = 1,00 – 0,0042 ≈ 1,00
Slide 54
Til þess að mega nota námundun verður
skekkjan (alfa) að vera minna en 5%
Í þessu tilfelli 0,0042/1,00*100%= 0,42%
Slide 55
HONORS ONLY!
Reiknið pH í 0.0010 M lausn af maurasýru,
HCO2H.
HCO2H + H2O HCO2- + H3O+
Ka = 1.8 x 10-4
Með námundunaraðferð
[H3O+] = 4.2 x 10-4 M, pH = 3.37
Rétt svar með annars stigs jöfnu er
[H3O+] = [HCO2-] = 3.4 x 10-4 M
[HCO2H] = 0.0010 - 3.4 x 10-4 = 0.0007 M
pH = 3.47
Slide 56
Til þess að mega nota námundun verður
skekkjan (alfa) að vera minna en 5%
Hér er skekkjan 0,00042/0,0010*100 = 42%
sem er allt of mikið. Þá er að nota 2. stigs
jöfnuna
Eða iteration eða endurreikning.
Slide 57
[H3O+]
= 4.2 x 10-4 M með námundun
Endurreikingur gefur:
Ka = 1.8 x 10-4 = x2 / (0,0010 – 0,00042)
3,23x10-4 = x
pH = 3,49
En með 2. stigs jöfnu: 3,47
Býsna nálægt – 3,5 í báðum
Slide 58
Hvert er pH 0.010 M NH3 ?
NH3 + H2O NH4+ +
OH-
Kb = 1.8 x 10-5
Step 1. Jafnvægisuppsetning
initial
change
equilib
[NH3]
[NH4+]
[OH-]
0.010
0
0
-x
+x
+x
0.010 - x
x
x
Slide 59
NH3 + H2O NH4+ +
OH-
Kb = 1.8 x 10-5
Step 2. sett inn í jöfnu
K b 1.8 x 10
-5
+
-
[NH 4 ][OH ]
=
=
[NH 3 ]
x
0.010
2
- x
Gerum ráð fyrir að x sé mjög lítið tala
x = [OH-] = [NH4+] = 4.2 x 10-4 M
og [NH3] = 0.010 - 4.2 x 10-4 ≈ 0.010 M
Þetta er OK – 0,00042/0,01*100= 4,2% skekkja
Slide 60
Í 0.010 M NH3. er pH.
Step 3. Reikna pH í basa
[OH-] = 4.2 x 10-4 M
pOH = - log [OH-] = 3.37
Af því að pH + pOH = 14,
pH = 10.63
Eða log [OH-] +14 = 10,63
Slide 61
HONORS ONLY!
Slide 62
Margar aðferðir til að mæla pH
› Blár litmuspappír(red = acid)
› Rauður litmus pappír(blue = basic)
› pH pappír með margföldum litum
› pH penni
› pH mælir (alvöru)
› Indikatorar eins og phenolphthalein
› Rauðkál, bláber, ….
Slide 63
pH paper
Slide 64
Tests
the voltage of
the electrolyte
Converts the voltage
to pH
Very cheap,
accurate
Must be calibrated
with a buffer solution
Slide 65
Indicators are dyes that can
be added that will change
color in the presence of an
acid or base.
Some indicators only work in
a specific range of pH
Once the drops are added,
the sample is ruined
Some dyes are natural, like
radish skin or red cabbage
Slide 66
H2C2O4(aq) + 2 NaOH(aq) --->
acid
base
Na2C2O4(aq) + 2 H2O(liq)
Carry out this reaction using a TITRATION.
Oxalic acid,
H2C2O4
Slide 67
Setup for titrating an acid with a base
Slide 68
1. Add solution from the
buret.
2. Reagent (base) reacts
with compound (acid)
in solution in the flask.
3. Indicator shows when
exact stoichiometric
reaction has occurred.
(Acid = Base)
This is called
NEUTRALIZATION.
Slide 69
35.62 mL of NaOH is
neutralized with 25.2
mL of 0.0998 M HCl by
titration to an
equivalence point.
What is the
concentration of the
NaOH?
Slide 70
Add water to the 3.0 M solution to lower
its concentration to 0.50 M
Dilute the solution!
Slide 71
But how much water
do we add?
Slide 72
How much water is added?
The important point is that --->
moles of NaOH in ORIGINAL solution =
moles of NaOH in FINAL solution
Slide 73
Amount of NaOH in original solution =
M•V =
(3.0 mol/L)(0.050 L) = 0.15 mol NaOH
Amount of NaOH in final solution must
also = 0.15 mol NaOH
Volume of final solution =
(0.15 mol NaOH)(1 L/0.50 mol) = 0.30 L
or
300 mL
Slide 74
Conclusion:
add 250 mL
of water
to 50.0 mL
of 3.0 M
NaOH to
make 300
mL of 0.50
M NaOH.
Slide 75
A shortcut
M1 • V1 = M2 • V2
Slide 76
You have a stock bottle of
hydrochloric acid, which is 12.1 M.
You need 400 mL of 0.10 M HCl.
How much of the acid and how
much water will you need?
Slide 2
Slide 3
Slide 4
Slide 5
Have a sour taste. Vinegar is a solution of acetic acid. Citrus
fruits contain citric acid.
React with certain metals to produce hydrogen gas.
React with carbonates and bicarbonates to produce carbon
dioxide gas
Basar
Have a bitter taste.
Feel slippery. Many soaps contain bases.
Slide 6
þ Mynda H+ (as H3O+) jónir í vatni (the hydronium ion is a
hydrogen ion attached to a water molecule)
þ Súrar á bragðið
þ Æta málma
þ Súrar vatnslausnir leiða rafmagn
þ Hvarfast við basa og mynda salt og vatn
þ pH undir 7
Slide 7
Anion
Ending
Binary
Acid Name
-ide
hydro-(stem)-ic acid
-ate
(stem)-ic acid
-ite
(stem)-ous acid
Ternary
Slide 8
A C ID S
start with 'H '
2 elem en ts
3 elem en ts
h yd ro - p refix
-ic en d in g
n o h yd ro - p refix
-a te en d in g
b ecom es
-ic en d in g
-ite en d in g
b ecom es
-o u s en d in g
Slide 9
• HBr (aq)
• H2CO3
• H2SO3
– H2SO4
hydrobromic acid
carbonic acid
sulfurous acid
Sulfuric acid
Slide 10
HI (aq)
HydroIodic acid
HCl (aq)
Hydrochloric acid
H2SO3
Sulfurous acid
HNO3
peIodic acid
HIO4
– perIodic acid
Slide 11
Mynda OH- jónir í vatni
Beiskir á bragðið, chalky
Vatnslausnir basa leiða rafmagn
Sápukenndir, sleipir
Hvarfast við sýrur og mynda salt og vatn
pH stærra en7
Slide 12
NaOH
natrium hydroxíð
lye - vítissódi
KOH
kalíumhydroxíð
liquid soap
Ba(OH)2
barium hydroxide
stabilizer for plastics
Mg(OH)2
magnesium hydroxide
“MOM” Milk of
magnesia
Al(OH)3
aluminum hydroxide Maalox (antacid)
Slide 13
Skilgreining #1: Arrhenius (traditional)
Sýrur – mynda H+ jónir (or hydronium ions
H3O+)
Basar – mynda OH- jónir
(problem: some bases don’t have hydroxide ions!)
Slide 14
Arrhenius sýra er efni sem myndar H+ (H3O+) in water
Arrhenius basi er efni sem myndar OH- in water
Slide 15
Skilgreining#2: Brønsted –Lowry
Sýra – Róteindagjafi
Basi – Róteindaþegi
Róteind er vetnisatóm sem hefur
misst rafeindina sína!
Slide 16
A Brønsted-Lowry sýra róteindagjafi
A Brønsted-Lowry basi róteindaþegi
Basi
sýra
Tilsvarandi
sýra
Tilsvarandi
basi
Slide 17
The Brønsted definition means
NH3 is a BASE in water —
+
and
water
is itself an NH
ACID
+
NH
+ H O
3
Base
4
2
Acid
Acid
OH
Base
-
Slide 18
Slide 19
HONORS ONLY!
Finnið sýruna, basann, tilsvarandi
(konjugered) sýru og tilsvarandi basa í
efnhvörvunum hér að neðan:
HCl + OH- Cl- + H2O
H2O + H2SO4 HSO4- + H3O+
Slide 20
Skilgr. #3 – Lewis
Lewis sýra – Efni sem
þiggur rafeindapar
Lewis basi – efni sem
gefur rafeindapar
Slide 21
Myndun oxoníum jónar er ágætt dæmi.
••
• O— H
•
H
+
A C ID
H
BASE
H
••
O— H
H
•Electron pair of the new O-H bond
originates on the Lewis base.
Slide 22
Slide 23
The heme group
in hemoglobin
can interact
with O2 and CO.
The Fe ion in
hemoglobin is a
Lewis acid
O2 and CO can
act as Lewis
bases
Heme group
Slide 24
Slide 25
Acidic solution
[H+ ] > [OH- ]
Neutral solution
[H+ ]
= [OH- ]
Basic solution
[H+ ]
< [OH- ]
Timberlake, Fig 9.3
Slide 26
Slide 27
pH = - log [H+]
(Remember that the [ ] mean Molarity)
Example: If [H+] = 1 X 10-10
pH = - log 1 X 10-10
pH = - (- 10)
pH = 10
Example: If [H+] = 1.8 X 10-5
pH = - log 1.8 X 10-5
pH = - (- 4.74)
pH = 4.74
Slide 28
Find the pH of
these:
1) A 0.15 M solution
of Hydrochloric
acid
2) A 3.00 X 10-7 M
solution of Nitric
acid
Slide 29
Ef pH í Coke er 3.12, [H+] = ???
Af því pH = - log [H+] þá
- pH = log [H+]
Taka antilógaritma (10x) af báðum:
10-pH = [H+]
[H+] = 10-3.12 = 7.6 x 10-4 M
*** Til að finna antilog eða 10^x þá ýtir þú
Á shift hnappinn og svo log
Slide 30
Lausn hefur pH of 8.5. hver er
mólstyrkur vetnisjóna í lausninni?
pH = - log [H+]
8.5 = - log [H+]
-8.5 = log [H+]
Antilog -8.5 = antilog (log [H+])
10-8.5 = [H+]
3.16 X 10-9 = [H+]
Slide 31
Vatn getur virkað sem sýra og basi.
Vatn klofnar í jónir, slíkt heitir
sjálfsjónun.
Jafnvægisfasti vatns er = Kw
Kw = [H3O+] [OH-] = 1.00 x 10-14 at 25 oC
Slide 32
Autoionization
OH
-
H 3O
+
Kw = [H3O+] [OH-] = 1.00 x 10-14 at 25 oC
Í hlutlausri lausn er [H3O+] = [OH-]
Kw = [H3O+]2 = [OH-]2
[H3O+] = [OH-] = 1.00 x 10-7 M
Slide 33
Since
acids and bases are
opposites, pH and pOH are
opposites!
pOH does not really exist, but it is
useful for changing bases to pH.
pOH looks at the perspective of
a base
pOH = - log [OH-]
Since pH and pOH are on opposite
ends,
pH + pOH = 14
Slide 34
pH
[H+]
[OH-]
pOH
Slide 35
Hvað er pH í 0.0010 M NaOH lausn?
[OH-] = 0.0010 (or 1.0 X 10-3 M)
pOH = - log 0.0010
pOH = 3
pH = 14 – 3 = 11
eða Kw = [H3O+] [OH-]
[H3O+] = 1.0 x 10-11 M
pH = - log (1.0 x 10-11) = 11.00
Slide 36
The pH of rainwater collected in a certain region of the
northeastern United States on a particular day was
4.82. What is the H+ ion concentration of the
rainwater?
The OH- ion concentration of a blood sample is
2.5 x 10-7 M. What is the pH of the blood?
Slide 37
[OH-]
[H+]
pOH
pH
Slide 38
Reiknið [H3O+], pH, [OH-], and pOH
Problem 1: Efnafræðingur býr til : (a) 3.0 M og
(b) 0.0024 M.
Reiknið [H3O+], pH, [OH-], and pOH lausnanna
við 25°C.
Problem 2: Hvað er [H3O+], [OH-], og pOH
lausnar með pH = 3.67? Er lausnin súr, basisk
eða hlutlaus?
Problem 3: Problem #2 með pH = 8.05?
Slide 39
Rammar og daufar sýrur
Römm sýra (sterk) klofnar auðveldlega
í jónir (100% jónuð)
HNO3, HCl, H2SO4 and HClO4 eru rammar
sýrur.
Slide 40
Römm sýra:
HNO3 (aq)+ H2O (l)H3O+ (aq) +NO3- (aq)
HNO3 is about 100% dissociated in
water.
Slide 41
Daufar sýrur (og basar)
Klofna ekki 100% í vatni og oft mjög lítið þ.e.
innan við 1%. Sjá ediksýru
Slide 42
Rammir basar
100%
klofnun í vatni.
NaOH (aq) Na+ (aq) + OH- (aq)
Einnig KOH and Ca(OH)2.
CaO (lime) + H2O -->
Ca(OH)2 (slaked lime)
CaO
Slide 43
Daufir basar
Minna en 100% jónaðir í vatni t.d.
ammoníak
NH3
(aq)
+ H2O (l) NH4+ (aq) + OH- (aq)
Slide 44
Slide 45
Consider acetic acid, HC2H3O2 (HOAc)
HC2H3O2 + H2O H3O+ + C2H3O2 Acid
Conj. base
+
Ka
[H 3 O ][OAc
[HOAc]
-
]
1.8 x 10
(K er skrifað Ka for ACID)
K gefur hlutfall klofið/óklofið í jónir
-5
Slide 46
Sýra
Conjugate
Basi
Increase
strength
Increase
strength
Slide 47
Dauf sýra hefur Ka < 1
Leiðir til lágs [H3O+] og pH of 2 - 7
Slide 48
Daufur basi hefur Kb < 1
Leiðir til lágs [OH-] og pH of 12 - 7
Slide 49
Slide 50
Þú hefur1.00 M HOAc. Reiknið jafnvægisstyrk HOAc, H3O+, OAc- og pH.
Step 1.
[HOAc]
[H3O+]
[OAc-]
initial
1.00
0
0
change
-x
+x
+x
x
x
equilib
1.00-x
Slide 51
Step 2. Skrifið jafnvægislíkingu Ka
K a 1.8 x 10
-5
+
=
[H 3 O ][OAc
[HOAc]
-
]
x
2
1.00 - x
Þetta er annars stigs jafna. Leyst sem slík….
Eða þú getur námundað x=0 neðan við strik ef x
er mjög lítið! (Þumalfingursreglan: 10-5 or
smaller is ok eða hámark 5% skekkja)
Slide 52
Step 3. Leysið út frá Ka
K a 1.8 x 10
-5
+
=
[H 3 O ][OAc
-
]
[HOAc]
x
2
1.00 - x
Gerum ráð fyrir að x sé mjög lítið því
Ka er svo lítið.
K a 1.8 x 10
-5
Þetta er fljótlegra svona.
=
x
2
1.00
Slide 53
Step 3. Leysum með námundun
K a 1.8 x 10
-5
=
x
2
1.00
x = [H3O+] = [OAc-] = 4.2 x 10-3 M
pH = - log [H3O+] = -log (4.2 x 10-3) = 2.37
[HOAC] = 1,00 – 0,0042 ≈ 1,00
Slide 54
Til þess að mega nota námundun verður
skekkjan (alfa) að vera minna en 5%
Í þessu tilfelli 0,0042/1,00*100%= 0,42%
Slide 55
HONORS ONLY!
Reiknið pH í 0.0010 M lausn af maurasýru,
HCO2H.
HCO2H + H2O HCO2- + H3O+
Ka = 1.8 x 10-4
Með námundunaraðferð
[H3O+] = 4.2 x 10-4 M, pH = 3.37
Rétt svar með annars stigs jöfnu er
[H3O+] = [HCO2-] = 3.4 x 10-4 M
[HCO2H] = 0.0010 - 3.4 x 10-4 = 0.0007 M
pH = 3.47
Slide 56
Til þess að mega nota námundun verður
skekkjan (alfa) að vera minna en 5%
Hér er skekkjan 0,00042/0,0010*100 = 42%
sem er allt of mikið. Þá er að nota 2. stigs
jöfnuna
Eða iteration eða endurreikning.
Slide 57
[H3O+]
= 4.2 x 10-4 M með námundun
Endurreikingur gefur:
Ka = 1.8 x 10-4 = x2 / (0,0010 – 0,00042)
3,23x10-4 = x
pH = 3,49
En með 2. stigs jöfnu: 3,47
Býsna nálægt – 3,5 í báðum
Slide 58
Hvert er pH 0.010 M NH3 ?
NH3 + H2O NH4+ +
OH-
Kb = 1.8 x 10-5
Step 1. Jafnvægisuppsetning
initial
change
equilib
[NH3]
[NH4+]
[OH-]
0.010
0
0
-x
+x
+x
0.010 - x
x
x
Slide 59
NH3 + H2O NH4+ +
OH-
Kb = 1.8 x 10-5
Step 2. sett inn í jöfnu
K b 1.8 x 10
-5
+
-
[NH 4 ][OH ]
=
=
[NH 3 ]
x
0.010
2
- x
Gerum ráð fyrir að x sé mjög lítið tala
x = [OH-] = [NH4+] = 4.2 x 10-4 M
og [NH3] = 0.010 - 4.2 x 10-4 ≈ 0.010 M
Þetta er OK – 0,00042/0,01*100= 4,2% skekkja
Slide 60
Í 0.010 M NH3. er pH.
Step 3. Reikna pH í basa
[OH-] = 4.2 x 10-4 M
pOH = - log [OH-] = 3.37
Af því að pH + pOH = 14,
pH = 10.63
Eða log [OH-] +14 = 10,63
Slide 61
HONORS ONLY!
Slide 62
Margar aðferðir til að mæla pH
› Blár litmuspappír(red = acid)
› Rauður litmus pappír(blue = basic)
› pH pappír með margföldum litum
› pH penni
› pH mælir (alvöru)
› Indikatorar eins og phenolphthalein
› Rauðkál, bláber, ….
Slide 63
pH paper
Slide 64
Tests
the voltage of
the electrolyte
Converts the voltage
to pH
Very cheap,
accurate
Must be calibrated
with a buffer solution
Slide 65
Indicators are dyes that can
be added that will change
color in the presence of an
acid or base.
Some indicators only work in
a specific range of pH
Once the drops are added,
the sample is ruined
Some dyes are natural, like
radish skin or red cabbage
Slide 66
H2C2O4(aq) + 2 NaOH(aq) --->
acid
base
Na2C2O4(aq) + 2 H2O(liq)
Carry out this reaction using a TITRATION.
Oxalic acid,
H2C2O4
Slide 67
Setup for titrating an acid with a base
Slide 68
1. Add solution from the
buret.
2. Reagent (base) reacts
with compound (acid)
in solution in the flask.
3. Indicator shows when
exact stoichiometric
reaction has occurred.
(Acid = Base)
This is called
NEUTRALIZATION.
Slide 69
35.62 mL of NaOH is
neutralized with 25.2
mL of 0.0998 M HCl by
titration to an
equivalence point.
What is the
concentration of the
NaOH?
Slide 70
Add water to the 3.0 M solution to lower
its concentration to 0.50 M
Dilute the solution!
Slide 71
But how much water
do we add?
Slide 72
How much water is added?
The important point is that --->
moles of NaOH in ORIGINAL solution =
moles of NaOH in FINAL solution
Slide 73
Amount of NaOH in original solution =
M•V =
(3.0 mol/L)(0.050 L) = 0.15 mol NaOH
Amount of NaOH in final solution must
also = 0.15 mol NaOH
Volume of final solution =
(0.15 mol NaOH)(1 L/0.50 mol) = 0.30 L
or
300 mL
Slide 74
Conclusion:
add 250 mL
of water
to 50.0 mL
of 3.0 M
NaOH to
make 300
mL of 0.50
M NaOH.
Slide 75
A shortcut
M1 • V1 = M2 • V2
Slide 76
You have a stock bottle of
hydrochloric acid, which is 12.1 M.
You need 400 mL of 0.10 M HCl.
How much of the acid and how
much water will you need?