Sebanyak 0,55 gram kristal K2S dilarutkan dalam air sehingga volumenya menjadi 500 mL. Jika diketahui Ka H2S = 1 x 10^-8, Ar S = 32, pH larutan yg terbentuk seb

0,55 gram K2S
V = 500 mL
Ka H2S = 1 × 10^-8
Ar S = 32, Ar K = 39

pH ... ?

Mr K2S
Mr = 2×Ar K + Ar S
Mr = 2×39 + 32
Mr = 78 + 32
Mr = 110

Konsentrasi K2S
M = gram/Mr × 1000/mL
M = 0,55/110 × 1000/500
M = 0,005 × 2
M = 0,01

.... K2S => 2 K^{+} + S^{2-}
. 0,01 M .. 0,02 M .. 0,01 M

[OH^-] = √( Kw/Ka × M anion )
[OH^-] = √( 10^-14/10^-8 × 0,01 )
[OH^-] = √( 10^-6 × 0,01 )
[OH^-] = √( 10^-6 × 10^-2 )
[OH^-] = √( 10^-8 )
[OH^-] = 10^-4

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

pH = 14 - pOH
pH = 14 - 4
pH = 10