Asam Basa kimia

(1)

PERHITUNGAN ASAM

BASA

OLEH :

ILHAM FEBRIANSYAH

: D1A013077

NOVRYANDI HUTAGALUNG : D1A013072

ABDUL GHOFAR

: D1A013099

(2)

K O N S E P P E R H I T U N G A N A S A M B A S A

pH + pOH = 14

(3)

Menghitung pH Asam Kuat Dan

Basa Kuat



Asam Kuat

pH = - Log [H

+

]

contoh : pH HCl 0,01 M ?

jawab : [H

+

] = valensi Asam x M

= 1 x 10

-2

= 10

-2

pH =

- Log [H

+

]

= - Log [10

-2

]

= 2

[H+] = valensi Asam x

M

[H+_{] = valensi Asam x}

(4)



Basa Kuat

pOH = - Log [OH

-

]

contoh : pH Mg(OH)

₂

0,001 M ?

jaawab : OH

-

= valensi Basa x M

= 2 x 10

-3

pOH = - Log [OH

-

]

= - Log [2 x 10

-3

]

= 3 – Log 2

pH = 14 – pOH

= 14 – (3 – Log 2) = 11 + Log 2

[OH-] = valensi Basa x

M

[OH-] = valensi Basa x

(5)



Asam Lemah

H

+

=

Contoh : pH CH

3

COOH 0,1 M (Ka = 10

5

)

H

+

= =

H

+

=

pH = Log [H

+

] = Log [10

3

]

= 3

A

Ka

.

10

5 1

. 



10

6

10

3

(6)



Basa Lemah

OH

-

=

Contoh : pH Al(OH)

3

0,01 M (Kb = 10

-8

)

OH

-

=

pOH = - Log [OH

-

] = - Log [10

-5

]

= 5

pH = 14 - pOH

= 9

B

Kb

.

B

Kb

.

10 810 2

. 



1010

(7)

H

2

O(l)

H

+

(aq) + OH

-

(aq)

Harga tetapan air adalah :

K [H

₂

O] = [H

+

] [OH

-

]

Kw = [H

+

] [OH

-

]

Pada suhu 25 °C, Kw yang didapat dari percobaan adalah 1,0 × 10–14.

Tetapan Kesetimbangan Air (K

w

)

O]

H

[

]

OH

][

H

[

K

2

 

(8)

LARUTAN BUFFER

 Asam

pH = pKa – log

 Basa

pOH = pKb – Log

Ket :

Ka = tetapan ionisasi asam lemah a = jumlah mol asam lemah g = jumlah mol basa konjugasi

Ket :

Kb = tetapan ionisasi basa lemah b = jumlah mol basa lemah g = jumlah mol asam konjugasi

g

a

(9)

Contoh buffer asam :

Tentukan pH larutan penyangga yang dibuat dengan

mencampurkan 50 mL larutan CH3COOH 0,1 M dengan 50 mL

larutan CH3COONa 0,1 M.

(KaCH3COOH =10–5)

Jawab:

50 mL CH3COOH 0,1 M + 50 mL CH3COONa 0,1 M

mol CH3COOH = 50 mL × 0,1 mmol/mL= 5 mmol

mol CH3COONa = 50 mL × 0,1 mmol/mL= 5 mmol

PKa = - Log [Ka] = 5

pH = pKa – log

pH = 5 – log pH = 5

g a

(10)

Contoh buffer basa :

Sebanyak 50 mL larutan NH3 0,1 M (Kb = 10–5) dicampur

dengan 100 mL larutan NH4Cl 0,5 M. Hitunglah pH larutan

tersebut!

Jawab:

50 mL NH3 0,1 M + 100 mL NH4Cl 0,5 M

mol NH3 = 50 mL × 0,1 mmol/mL = 5 mmol

mol NH4Cl = 100 mL × 0,5 mmol/mL = 50 mmol

pOH = pKb – log pOH = 5 – log

pOH = 5 – log 0,1 pOH = 5 + 1 = 6 pH = 14 – pOH = 14 – 6 = 8

g b

(11)

TERIMA KASIH

Ingat :