hitunglah pH larutan (NH4)2CO3 0,1 M, jika Ka H2CO3 = 10-4 dan Kb NH4OH = 10-5 Tolong dong, nanti kan kukasih jawaban terbaik

→ Jika Ka H2CO3 = 10^-4 :

pKa = - log Ka
pKa = -log 10^-4
pKa = 4 - log10
pKa = 4 - 0
pKa = 4

pH = 1/2 (14 + pKa + log10^-1)
pH = 1/2 (14 + 4 + 1)
pH = 1/2 (19)
pH = 9,5

→ Jika Kb NH4OH = 10^-5 :

pKb = -log Kb
pKb = -log 10^-5
pKb = 5 - log10
pKb = 5

pH = 1/2 (14 - pKb - log10^-2)
pH = 1/2 (14 - 5 - 2)
pH = 1/2 (7)
pH = 3,5