Question

500mL NH3 0,001mol derajat ionisasi/alfa=10%
pH?

Answer 1

500 mL NH3 0,001 mol
(alfa) = 10% = 0,1

M = mol/liter
M = 0,001/(500/1000)
M = 0,001 / 0,5
M = 0,002
M = 2 × 10^-3

[OH^-] = (alfa) × M
[OH^-] = 0,1 × 2 × 10^-3
[OH^-] = 2 × 10^-4

pOH = - log [OH^-]
pOH = - log 2 × 10^-4
pOH = 4 - log 2

pH = 14 - pOH
pH = 14 - (4 - log 2)
pH = 14 - 4 + log2
pH = 10 + log 2