MATERI KIMIA ANALIT

7/27/2019 MATERI KIMIA ANALIT

http://slidepdf.com/reader/full/materi-kimia-analit 1/33

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?



Ionisasi Air

• Air netral cenderung untuk mengion

H2O H+ + OH-

• Proton bebas bergabung dengan satu

molekul air membentuk ion hydronium

H3O+



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



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



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



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



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



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.



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



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:



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



][

]][[][ 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-



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.



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



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



[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+]



Relationship between pH and [H+] / [OH-] concentration



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



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



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



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?



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



Distribution curves for acetate and acetic acid



Tit ti f h h t



Titration curve for phosphate



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



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



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



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



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.



End of Chapter 2

MATERI KIMIA ANALIT

Documents

Transcript of MATERI KIMIA ANALIT