If the percent ionization of water is #3.67 xx 10^(-7)%# at a certain temperature, what is the #"pH"#?

#K_w^@ = 1.35 xx 10^(-13)#

#"pH"_"neutral" = 6.44#

#"pH" + "pOH" = 12.88#

#T^@ > 25^@ "C"#

Based on this chart, we are roughly at #50 - 60^@ "C"#.

The percent ionization is defined as the percent of substance #"HA"# that dissociated into #"A"^(-)# and #"H"^(+)#:

#%"ionization" = (["HA"]_"dissociated")/(["HA"]) xx 100%#

But the concentration of dissociated #"HA"# is the concentration of #["H"^(+)]# or of #["A"^(-)]# since their charges are the same magnitude, so

#%"ionization" = (["H"^(+)])/(["HA"]) xx 100%#

Now, your percent ionization is unreasonable. It should actually be #3.67 xx 10^(-5)%#, which is #3.67 xx 10^(-7)#, which is close to the usual #["H"^(+)] = 1.00 xx 10^(-7) "M"# at #25^@ "C"#.

Therefore:

#3.67 xx 10^(-7) = (["H"^(+)])/(["HA"])#

But with pure liquids within equilibrium constants, at a new standard reference temperature (which we do not know), we divide by the standard "concentration" #c^@# of water (which is its molar density) such that #["H"_2"O"] = 1# and the units go away.

So,

#["H"^(+)] = 3.67 xx 10^(-7) "M"#

and

#color(blue)(K_w^@) = (["H"^(+)]//c^@)(["OH"^(-)]//c^@)#

#= (3.67 xx 10^(-7))(3.67 xx 10^(-7)) = color(blue)(1.35 xx 10^(-13))#

Compared to #25^@ "C"#, #K_w# is lower, and so,

#color(blue)("pH" + "pOH" = 12.88)#

at this new temperature, and neutral pH now becomes (with #c^@ = "1 M"# for aqueous solutions):

#color(blue)("pH"_"neutral") = -log[3.67 xx 10^(-7)//c^@] = color(blue)(6.44)#

Therefore:

• Below #6.44# pH at this temperature, the solution is acidic.
• Above #6.44# pH at this temperature, the solution is basic.

Water dissociation is endothermic (bond-breaking), so at higher temperatures, this process favors the dissociation and the solution #"pH"# drops as a function of temperature.

Therefore, this temperature we are at is hotter than #25^@ "C"#.