suatu campuran terdiri dari 200 ml larutan asam asetat 0.3 M dengan 300 larutan koh 0.2 M . berapakah ph larutan yng terjadi jika diketahui ka ch3cooh=1×10^-5

200 ml CH3COOH 0,3 M + 300 ml KOH 0,2 M
Ka CH3COOH = 10^-5

Persamaan : CH3COOH + KOH => CH3COOK + H2O
Mula-mula. : 200x0,3 ~~~300x0,2
= 60mmol = 60mmol
Reaksiiiiiiii. : 60 ---------------- 60 --------60
Sisaaaaaaa : 0 ----------------- 0 ---------- 60

M = mmol/v total
= 60/600
= 0,1 M

[OH-] = akar ((kw/kb)xM)
= akar ((10^-14/10^-5)x0,1)
= akar (10^-9 x 0,1)
= akar 10^-10
= 10^-5

POH = 5
PH = 14-5 = 9