Tolong dibantu ya ... dalam 100 mL larutan terlarut 3,6 gram C₆H₅COONa (Mr=144). jika Ka C₆H₅COOH = 6 X 10⁻⁵, hitunglah pH larutan tersebut !
TJulian
Ya tinggal dicari menggunakan rumus [H+] = akar Ka x M M=? M=n/v M=0.025/0.1L = 0.25 M n=? n=m/mr n= 3.6/144=0.025 [H+]=akar 6x10-5 x 0.25 = 3.87 x 10-3 pH= 3-log 3.87 :)
M=?
M=n/v M=0.025/0.1L = 0.25 M
n=?
n=m/mr
n= 3.6/144=0.025
[H+]=akar 6x10-5 x 0.25 = 3.87 x 10-3
pH= 3-log 3.87 :)