Question

Sebanyak 0,49 gram garam NaCN dilarutkan kedalam air hingga volumenya tepat 250 Ml Jika diketahui KaHCN 1x10^-10 ,A, Na:23 C:12 N:14 Tentukan PH larutan tsb

Tolong dibantu ^^

Answer 1

Diketahui :
m NaCN = 0,49 gram
Volume larutan = 250 ml = 0,25 L
Ka HCN = 10^-10
Ar Na = 23 , C = 12 , N = 14

Ditanya : pH larutan ?

Jawab :

Mr NaCN :
= Ar Na + Ar C + Ar N
= 23 + 12 + 14
= 49 gram/mol

n NaCN :
= m / Mr
= 0,49 gram / 49 gram/mol
= 0,01 mol

Molaritas garam NaCN :
= n / V
= 0,01 mol / 0,25 L
= 0,04 mol/L

[OH-] :
= √ Kw / Ka × M garam
= √ 10^-14 / 10^-10 × 0,04
= 2 × 10^-3

pOH :
= - log [OH-]
= - log 2 × 10^-3
= - ( log 2 + log 10^-3)
= - log 2 - log 10^-3
= 3 - log 2

pH :
= 14 - pOH
= 14 - ( 3 - log 2)
= 14 - 3 + log 2
= 11 + log 2



Catatan :
Kw = Tetapan Kesetimbangan Air = 10^-14