If #[H_3O^+] = 1.005 * 10^-5# M, what is the pH?

Right from the start, you can say that the pH of this solution will be very close to #5#.

Here's why that is so.

As you know, the pH of pure water at room temperature is equal to #7#. This value is derived from the self-ionization of water, in which two water molecules exchange a proton, #"H"^(+)#, to form hydronium and hydroxide ions

#2"H"_2"O"_text((l]) rightleftharpoons "H"_3"O"_text((aq])^(+) + "OH"_text((aq])^(-)#

At room temperature, the ion product constant---which you may see as #K_w#---for this reaction is equal to #10^(-14)#, which is why

#10^(-14) = ["H"_3"O"^(+)] * ["OH"^(-)]#

#["H"_3"O"^(+)] = ["OH"^(-)] = sqrt(10^(-14)) = 10^(-7)"M"#

As you know, pH is defined as

#"pH" = -log( ["H"_3"O"^(+)])#

For pure water, this is equal to

#"pH" = -log(10^(-7)) = 7#

In your case, assuming that you're at room temperature as well, the concentration of hydronium ions is higher than #10^(-7)#. This tells you that

• the solution is acidic
• its pH must be lower than #7#

Since pH is defined as the negative log base #10# of the concentration of hydronium ions, a hydronium concentration that is approximately two times higher will result in a pH that is approximately two units lower.

#"pH" = -log(1.005 * 10^(-5)) = color(blue)(4.9978)#

Indeed, the resulting pH is very close to #5#, since #1.005 * 10^(-5)# is very close to #10^(-5)#.