1) larutan 100 ml CH3COOH 0,15 Mdicampurkan dengan 50 ml larutan NaOH 0,2 M ( Ka CH3COOH=10^5) maka pH campuran tersebut adalah.....
2) Untuk membuat larutan penyangga dengan pH=6, kedalam 100 ml larutan asam asetat 0,1 M harus ditambahkan natrium asetat sebanyak( Ka=10^5; Na=23; C=12; H=1 O=16)
Tlng dngn caranya
egayoza
1. CH3COOH + NaOH ⇒ CH3COONa + H2O m 15 10 b 10 10 10 10 s 5 - 10 10
mrifqis
1) Diket : CH3COOH = 0,15 M x 100 mL = 15 mmol NaOH = 0,2 M x 50 mL = 10 mmol Ka = CH3COOH + NaOH -> CH3COONa + H2O mula 15 10 - - reaksi 10 10 10 10 ---------------------------------------------------------------------------- - hasil 5 0 10 10 [] = Ka x [] = x [] = x [] = x 0,5 [] = x [] = pH = -log [] pH = -log pH = 6-log 5
2) pH = 6 [] = Ka = Asam Asetat = 0,1 M x 100 mL = 10 mmol [] = Ka x = x (x) = (10) x = x = x = 100 mmol Jadi, harus ditambahkan natrium asetat sebanyak 100 mmol
m 15 10
b 10 10 10 10
s 5 - 10 10
H^+ = Ka × (mol asam ÷ mol garam)
= 10^-5 × ( 5 ÷ 10)
= 5 × 10^-6
pH = 6- log 5
2. pH = 6 ; H^+ = 10^-6
H^+ = Ka × (mol asam ÷ mol garam)
10^-6 = 10^-5 × (10^-2 ÷ mol garam)
mol garam = 0,1
massa garam = 0,1 × 70 = 7 gr
CH3COOH = 0,15 M x 100 mL = 15 mmol
NaOH = 0,2 M x 50 mL = 10 mmol
Ka =
CH3COOH + NaOH -> CH3COONa + H2O
mula 15 10 - -
reaksi 10 10 10 10
---------------------------------------------------------------------------- -
hasil 5 0 10 10
[
[
[
[
[
[
pH = -log [
pH = -log
pH = 6-log 5
2) pH = 6
[
Ka =
Asam Asetat = 0,1 M x 100 mL = 10 mmol
[
x =
x =
x = 100 mmol
Jadi, harus ditambahkan natrium asetat sebanyak 100 mmol