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e-modul KIMIA 2022

pOH = - log [OH ]
-
= - log 2 x 10
-5
= 5 – log 2
pH = 14 – pOH
= 14 – (5 –log 2)
= 9 + log 2
= 9 + (1 – log 5)
= 10 – log 5

11. A. 4 – log 2
Pembahasan:
pH HX = 2,5 log 2
[H ] = 2 x 10 -2,5
+
[H ] = Ka x M
+
(2 x 10 -2,5 2 2
) = Ka x 0,2
4 x 10 = Ka x 0,2
-5
4 10
Ka =
2 10
= 2 x 10
-4
Reaksinya:
HX + NaOH  NaX
m = 0,2 mol 0,1 mol -
r = -0,1 mol -0,1 mol +0,1 mol
s = 0,1 mol – 0,1 mol

[H ] = Ka x
+

0,1
= 2 x 10 x
-4
0,1
= 2 x 10
-4
pH = - log [H ]
+
= - log 2 x 10
-4
= 4 – log 2

12. C. 5 – log 1,25
Pembahasan:
pH NaOH = 13
pOH = 14 – 13
= 1
[OH ] = 10 -1
-
10 = 1 x M
-1
M = 10
-1
Reaksinya:
CH3COOH + NaOH  CH3COONa
m = 10 mmol 5 mmol -
r = -5 mmol -5 mmol +5 mmol
s = 5 mmol – 5 mmol

e-modul Kesetimbangan Ion dan pH Larutan Penyangga Halaman 62