0,01 mol larutan NH4Cl ditambah air 1 liter, kb=1.10^5. Tentukan ph

M = n / v
M = 0,01 / 1
M = 0,01 M

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

pH = 1/2 (14 - pKb - log [garam])
pH = 1/2 (14 - 5 - log 1 x 10^2)
pH = 1/2 (9 + 2)
pH = 1/2 (11)
pH = 5,5