" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
b. CH3COOH + NaOH ⇔ CH3COONa + H2O
(0,16x0,025) + (0,1x0,02) ⇔ 0 0
0,004 0,002 ⇔ 0 0
-0,002 -0,002 ⇔ +0,002 +0,002
0,002 0 0,002 0,002
[H+]= 0,002÷0,045= 0,4 M
= √Ka. M = √1.10^-6.4.10^-1 = 2.10^-3,5
pH = -log[H+]
= -log[2x10^-3,5]
= 3,5-log2
= 3,5-0,3
= 3,2
c. CH3COOH + NaOH ⇔ CH3COONa + H2O
(0,16x0,025) + (0,1x0,06) ⇔ 0 0
0,004 0,006 ⇔ 0 0
-0,004 -0,004 ⇔ +0,004 +0,004
0 0,002 0,004 0,004
[OH-]= 0,002÷0,085= 0,024 M
pOH = -log[OH-]
= -log[24x10^-3]
= 3-log24
= 3-1,4
= 1,6
pH = 14-pOH
= 14-1,6
= 12,4
Semoga Membantu^^