1. Hitunglah pH dari larutan dengan [H+] = 6.38 x 10-6 mol/L

Diketahui : [H+] = 6.38 x 10-6 mol/L

Ditanya : pH = . . . ?

Jawab : pH = - log [H+]

= - log [6.38 x 10-6]

= 6 – log 6.38

= 6 – 0,8048

= 5,1952

pH ≈ 5,2

2. Hitunglah [H+] untuk pH larutan 8.37

Diketahui : pH = 8,37

Ditanya : [H+] = . . . ?

Jawab : pH = - log [H+]

8,37 = - log [H+]

8 + 0,37 = - log [H+]

-log - ((-log) 2,34 ) = - log [H+]

- log

. = - log [H+]

-log 0,34 . = - log [H+]

[H+] = 0,34 x

3. Hitunglah pH dari basa kuat 1.0 x 10-3 M NaOH

pH +

= 1,0 x M

( = . valensi basa

= (1,0 x M) x 1

= 1,0 x M

pOH = - log (

= - log (

= - log ( 3 )

= 3

pH = 14 - pOH

= 14 – (3) = 11

TUGAS INDIVIDU

FARMASI FISIK

NAMA : ELISA GALUH SETYORINI

NIM : 10112087

TINGKAT : I

PRODI : S1 FARMASI

Institut Ilmu Kesehatan BHAKTI WIYATA KEDIRI

2013

4. Hitunglah pH dari basa kuat 5.0 x 10-3 M Ba(OH)2

pH +

= 0,5 x M

( = . valensi basa

= ( 0,5 x M ) x 2

= 10 x

pOH = - log (

= - log (

= - log ( - 2 )

= 2

pH = 14 – pOH

= 14 – (2) = 12

5. Hitunglah pH dari asam lemah 2.0 x 10-3 M H2CO3 (Ka = 5.64 x 10-11)

pH ( asam lemah )

+

= 2,0 x M

= 5,64 x

[ ] =

[ ] =

[ ] =

= 3,35 .

pH = - log [ ]

= - log ( 3,35 . )

= 7 – log 3,35

= 7 – 0,5250

= 6,475