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Earlier discussion
4 older comments kept for context.
Damilola Precious
pH = -log(2H+)
=-log(2×0.5×10
−
6)
= 6.0
Aderogba Qudus
H2SO4 => 2H+ + SO4²-
[H+] = 2(0.5×10^-6) = 10^-6M
pH = –Log[H+] = –Log(10^-6)
pH = 6.
C is correct
N
Shouldn't d 2 in d 2H+ be multiplied with it
Den d pH ll be 6
Fapuro Oluwadarasimi
pH = -log[H\(^+\)]
pH = -log[0.5×10\(^{-6}\)]
pH = 6.30