100ml larutan HCl 0,2M direaksikan dengan 100mL Be(OH)2 0,1M, jika Be(OH)2 adalah 10^-4 tentukan PH larutan

100 mL HCl 0,1 M + 100 mL Ba(OH)2 0,1 M

pH ... ?

n HCl = M × V
n HCl = 0,1 × 100
n HCl = 10 mmol

n Ba(OH)2 = M × V
n Ba(OH)2 = 0,1 × 100
n Ba(OH)2 = 10 mmol

#Reaksi
..... 2 HCl + Ba(OH)2 => BaCl2 + 2 H2O
M : ... 10 ........ 10
R : ... 10 .......... 5
______________---
S : ... --- .......... 5 ............. 5 ........... 10

yang sisa Ba(OH)2
M = n/V total
M = 5/(100 + 100)
M = 5/200
M = 0,025

[OH^-] = b × Mb
[OH^-] = 2 × 0,025
[OH^-] = 0,05
[OH^-] = 5 × 10^-2

pOH = - log [OH^-]
pOH = - log 5 × 10^-2
pOH = 2 - log 5

pH = 14 - pOH
pH = 14 - (2 - log 5)
pH = 14 - 2 + log 5
pH = 12 + log 5