Page 47 - Modul Titrasi Asam Basa Kelompok 7A _Float

-2
pH = -log (6,67 x 10 ) = 1,18
c) pH pada penambahan 50 mL NaOH
+ –
(H O ) + (OH )  2 H O
3 2

awal : 5 mmol
5 mmol V = 50 mL

bereaksi : -5 mmol -5 mmol V = 50 mL

akhir reaksi : 0 mmol 0 mmol V = 100 mL

Berada pada keseimbangan (stoikiometrik):

–
+
2 H O H O + OH
2 3
+
–14
+
–
[H O ][OH ] = K = 1,0 x 10 [H O ]
3 w 3
–
–7
= [OH ] = 1,0 x 10
pH = 7
d) pH pada penambahan 60 mL NaOH
+ –
(H O ) + (OH )  2 H O
3 2

awal : 6 mmol
5 mmol V = 50 mL

bereaksi : -5 mmol -5 mmol V = 60 mL

akhir reaksi : 0 mmol 0 mmol V = 110 mL

–3
–
[OH ] = 1,0 mmol/110 mmol = 9,1 x 10 M
–3
pOH = -log (9,1 x 10 )= 2,04
pH = 14 – 2,04 =11,96 (skor 30)

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