MATERI KIMIA ANALIT
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Asam dan Basa
Water1. Bagaimana struktur molekulair, kaitannya dengan sifat fisik
dan kimianya?
2. Apakah ikatan hidrogen itu?
3. Apa yang dimaksud dg asam
dan basa?
4. Apa yg dimaksud dg pH, dan
kaitannya dengan sifat sir?
5. Apa yang dimaksud dengan
kurva titrasi?
6. Apa yg dimaksud dg buffer,
dan mengapa perlu?
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Ionisasi Air
• Air netral cenderung untuk mengion
H2O H+ + OH-
• Proton bebas bergabung dengan satu
molekul air membentuk ion hydronium
H3O+
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Lompatan ProtonProton yang besar dan mobolitas hidroksida
• H3O+ : 362.4 x 10-5 cm2•V-1•s-1
• Na+: 51.9 x 10-5 cm2•V-1•s-1
• Ion Hydronium berpindah; melompat 1012 per
detik
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Bentuk Kesetimbangan
• Dinyatakan dg:
dengan K konstanta disosiasi
• Jika konsentrasi [H2O] tetap, maka:
K w
= [H+][OH-]
K =
[H+] [OH-]
[H2O]
OH H O H 2
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K w
K w
= [H+][OH-]
• dengan K w
konstanta ionisasi air
• Untuk air murni, konstanta ionisasinya
10-14 M2 pada 25ºC
• Untuk air murni
[H+] = [OH-] = (K w )1/2 = 10-7 M
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Acids and bases
• Untuk air murni (netral)
[H+] = [OH-] = (K w )1/2 = 10-7 M
• Asam, jika [H+] > 10-7 M
• Basa, jika [H+] < 10-7 M
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Acids and Bases
Definisi Asam Basa :
• Asam adalah senyawa yg dpt memberikan proton.• Basa adalah senyawa yg menerima proton.
HA + H2O H3O+
+ A-
/OH-
Acid Base Conjugate ConjugateAcid
Or
HA A- + H+
Acid Conjugate ConjugateBase Acid
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Acids and Bases
Definisi Asam Basa Lewis:
• Dengan pertimbangan bahwa yang bergerak di dalamatom adalah elektron, maka Lewis mendefinisikan
asam dan basa yang lebih modern
• “definisi asam basa Lewis” : cara modern
untukmendefinisikan asam dan basa
• Asam: electron acceptor.
• Basa: electron donor.
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Dalam reaksi kesetimbangan, perubahan konsentrasi
[H+] akan meningkatkan rasio HA dan A-.
Dengan menambah H+ , A- akan dikonsumsi
membentuk HA.
Begitu juga jika terjadi peningkatan konsentrasi [A-],
kelebihan H+ juga akan dikonsumsi dan konsentrasi
[H+] tidak akan berubah.
AHHA
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Kekuatan asam ditentukan oleh konstanta disosiasi
Pada konsentrasi molar
untuk: HA + H2O H3O+ + A-
• reactants products
HA H3O+
A- H2O
Pengukuran afinitas relatif proton untuk masing-masing
pasangan conjugate asam basa.
Rasionya:
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Bagaimana dengan air
konsentrasi H2O relatif tidak banyak berubah pada larutan asam
encer.
Bagaimana dengan konsentrasi H2O?
Ingat definisi konsentrasi: Mole per liter
1 mole H2O = 18 g = 18 ml
1000 g/liter
M
mol g
g 5.55
/18
1000
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][
]][[][ 2
HA
A H O H K K a
Sekarang kita indekskan a pada K, sehingga menjadiK a
Asam lemah (K<1)
Asam Kuat (K>1)
Asam kuat terurai sempurna: mengubah semua protonnya
H2O membentuk H3O+
HA H+ + A-
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Asam lemah
Asam lemah tidak terurai sempurna: reaksi
yang terbentuk adalah reaksi kesetimbangan:
Jika DITAMBAHKAN H+
, kesetimbangan bergeser ke kiri membentuk HA yang lebih
banyak dengan menggunakan A- yang ada.
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Disosiasi H2O
OH H O H 2
Air juga terurai atau terdisosiasi [H2O] = 55.5
][
]][[
2O H
OH H K
214 M10]][[
OH H Kw
Tetapan inonisasi air
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Since there is equal amounts of [H+] and [OH-]
M10x1][][ 7
OH H
This is neutral
At [H+] above this concentration the solution is ACIDIC
At [H+] below this concentration the solution is BASIC
9101][
x H
2101][
x H
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[H+] pH
10-7 = 7
10-3 = 3
10-2
= 2
10-10 = 10
5x10-4 = 3.3
7x10-6 = 5.15
3.3x10-8 = 7.48
pH = -Log[H+]
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Relationship between pH and [H+] / [OH-] concentration
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Henderson - Hasselbalch equation
][]][[
HA A H K
From
][
][][
A
HA K H Rearrange
Take (-)Log of each
][
][loglog
HA
A K pH
][
][log
HA
A pK pH
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Above and below this range there is insufficientamount of conjugate acid or base to combine with
the base or acid to prevent the change in pH.
[HA]
][Alog pK pH
-
110
101 fromvariesratio
]HA[][A
Buffers
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For weak acids
HA A- + H+
This equilibrium depends on concentrations of each component.
If [HA] = [A-] or 1/2 dissociated
Then
By definition the pK is the pH where [HA] = [A-]
: pH = pK
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Buffers
A buffer can resist pH changes if the pH is at or near a its pK.
Buffer range: the pH range where maximum resistance to pH
change occurs when adding acid or base. It is = +1 pH fromthe weak acid pK
If pK is 4.8 the buffering range is 3.8 to 5.8
Why?
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The buffer effect can be seen in a titration curve.
To a weak acid salt, CH3C00-, add HC1 while monitoring pH vs. the
number of equivalents of acid added.or
do the opposite with base.
Buffer capacity: the molar amount of acid which the buffer can handle
without significant changes in pH.
i.e
1 liter of a .01 M buffer can not buffer 1 liter of a 1 M solution of HCl but
1 liter of a 1 M buffer can buffer 1 liter of a .01 M solution of HCl
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Distribution curves for acetate and acetic acid
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Tit ti f h h t
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Titration curve for phosphate
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Blood Buffering System
• Bicarbonate most significant buffer
• Formed from gaseous CO2
CO2 + H2O H2CO3 H2CO3 H+ + HCO3
-
• Normal value blood pH 7.4
• Deviations from normal pH value lead toacidosis
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Henderson - Hasselbalch type problems:
You may be asked the pH, pK, the ratio of acid or base or solve for
the final concentrations of each.
[HA]
][Alog pK pH
-
Th 6 t h
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The 6 step approach
1. Write the Henderson + Hasselbalch equation.
2. Write the acid base equation
3. Make sure either an H+ or OH- is in the equation.
3. Find out what you are solving for
4. Write down all given values.
5. Set up equilibrium conditions.
6. Plug in H + H equation and solve.
What is the pH of a solution of that contains 0 1M
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What is the pH of a solution of that contains 0.1M
CH3COO- and 0.9 M CH3COOH?
1) pH = pK + Log [A-]
[HA]
2) CH3COOH CH3COO- + H+
3) Find pH
4) pK = 4.76 A- = 0.1 M HA = 0.9 M
5) Already at equilibrium
6) X = 4.76 + Log 0.1
0.9
Log 0.111 = -.95 X = 4.76 + (-.95) X = 3.81
What would the concentration of CH C00- be at pH 5 3
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What would the concentration of CH3C00 be at pH 5.3
if 0.1M CH3C00H was adjusted to that pH.
1) pH = pK + Log [A-]
[HA]
2) CH3C00H CH3C00- + H+
3) Find equilibrium value of [A-] i.e [CH3C00-]
4) pH = 5.3; pK = 4.76
5) Let X = amount of CH3C00H dissociated at equilibrium
[A-] = [X]
[HA] = [0.1 - X]
6) 5.3 = 4.76 + Log [X]
[0.1 - X]
Now solve.
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End of Chapter 2