Question

100ml lar.HCl 1M +200ml lar.NH3 1M,jika kb NH3=1X10pangkat-5 maka PH campuran

Answer 1

Mmol HCl = 100 x 1 = 100 mmol
mmol NH3 = 200 x 1 = 200 mmol
Kb = 10⁻⁵

   HCl + NH4OH => NH4Cl + H2O
a : 100    200
b : 100    100           100        100
s :   -       100           100        100

[OH-] = Kb x mol basa/mol garam
[OH-] = 10⁻⁵ x 100/100
[OH-] = 10⁻⁵

pOH = - log [OH-]
pOH = - log 10⁻⁵
pOH = 5

pH = 14 - pOH = 14 - 5 = 9