5 gram HCN (Mr=27) di larutkan dalam air hingga volumenya 100 ml ka=1,7.10^-8 tentukan ph larutan tersebut

Diketahui :
m HCN =5 gram
Mr = 27 gram/mol
V larutan = 100 ml = 0,1 L
Ka = 1,7 × 10^-8

Ditanya : pH larutan ?

Jawab :

n (mol) HCN :
= m / Mr
= 5 gram / 27 gram/mol
= 0,2 mol

Molaritas HCN :
= n / V
= 0,2 mol / 0,1 L
= 2 M

[H+] :
= √ Ka . Ca
= √ 3,4 × 10^-8 . 2
= 2,6 × 10^-4 M

pH larutan :
= – log [H+]
= – log 2,6 × 10^-4
= – ( log 2,6 + log 10^-4)
= – log 2,6 – log 10^-4
= 4 – log 2,6

M = 5/ 27 x 1000/100
= 0.18 x 10 = 1,8
H+ = akar 1,8 x 1,7. 10^-8
= akar 3.06 x 10^-4
Ph = 4 – log 3,06
4- 0.48 = 3.5