Question

Tolong jawabin no 3-5 dengan carany yaa

Answer 1

No 3
ch3cooh + NaOH -> ch3cooNa + h2o
M : 1 mmol 0.6 mmol
B : 0.6 mmol 0.6 mmol 0.6 mmol
S : 0.4 mmol - 0.6 mmol

[h+] = Ka . mol asam lemah sisa / mol garam terbentuk
= 2 x 10^-5 . 0.4 / 0.6
= 0.8/0.6 . 10^-5

pH = 5 - log 4/3
pH = 5 - log 4 + log 3
pH = 4.87
.
4.
ch3cooh + naoh -> ch3coona +h2o
M : 10 mmol 5 mmol
B : 5 mmol 5 mmol 5 mmol
S : 5 mmol - 5 mmol
[h+] = Ka . mol ch3cooh sisa / mol ch3coona
= 7 . 10^-3 . 5/5
= 7 . 10^-3
pH = 3 - log 7