chem
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 10^{x} 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.)