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= 0.2 × 100 × 2 = 40 mmol
mmol H + kedua = 0.3 × 100 × 2 = 60 mmol
H + campuran = mmol H + pertama + mmol H + kedua / volume total
H+ campuran = (40 + 60 ) ÷200
= 100 : 200
= 1/2
= 5 × 10^-1
ph = 1 - log 5
0,2.100+0,3.100=M3.200
20+30=M3.200
M3=0,25 M
H2SO4 => 2H+ + SO42-
Karena perbadingan koefisien setara dengan perbandingan konsentrasi maka molaritas H+ = 0,5M
pH=-log[H+]
pH=-log[0,5]
pH=1-log5