posted by dada78 .
how do I caluclate (H3O+) in a solution if the PH is the following 1.81
pH = -log(H3O^+)
1.81 = -log(H3O^+).
Rearrange to -1.81 = log(H3O^+). Now
punch in -1.81 on your calculator (or 1.81 and hit the change sign button to make -1.81) then hit the 10x button. That will return 0.01549 and you can round that to 0.015 to two significant figures (two are allowed with the 1.81.)