" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
m 5 5
b 5 5 5 5
s - - 5 5
OH^- = √ (Kw ÷ Ka) × (mol ÷ V total)
= √ (10^-14 ÷ 1,8 × 1^-5) × (5 ÷ 100)
= 5,3 × 10^-6
pOH = 6 - log 5,3
pH = 8 + log 5,3
mlCH3COOH= 0,1x 50=5ml
NaOH + CH3COOH ------- CH3COONa + H2O
mula" : 5 5
beraksi: 5 5 5 5
sisa : - - 5 5
[OH^-]=√kw/ka x CH3COONa
= √10^-14/1,8x10^-5 x 5
=√2,78x10^-8
=√2,78x10^-4
POH = -log[OH^-]
= -log√2,78 x 10^-4
= 4- log √2,78
PH = 14-POH
= 14- (4- log√2,78)