Represent the autoprotolysis of water, and explain how #pH# defines the acidity of the solution. How are #pH# and #pOH# defined?

#2H_2OrightleftharpoonsH_3O^+ + HO^-#

And #K_w=[H_3O^+][HO^-]=10^-14# at #298*K#. This equilibrium has been extensively measured.

#"6. and 7"# We can take #log_10# of both sides...........

#log_10K_w=log_10[H_3O^+]+log_10[HO^-]#

#underbrace(log_10(10^-14))_(-14)=log_10[H_3O^+]+log_10[HO^-]#

And so on rearrangement.........

#14=-log_10[H_3O^+]-log_10[HO^-]#

But by definition........

#14=underbrace(-log_10[H_3O^+])_(pH)underbrace(-log_10[HO^-])_(pOH)#

And so #14=pH+pOH#, under standard conditions of #298*K# and near #1*atm#.

If #pH=7#, then, clearly, #[H_3O^+]=[HO^-]#, and the solution is NEUTRAL.

If #pH<7#, then, clearly, #[H_3O^+]>[HO^-]#, and the solution is ACIDIC.

If #pH>7#, then, clearly, #[H_3O^+]<[HO^-]#, and the solution is ALKALINE.

#8.# We gots #[H_3O^+]=3.5xx10^-8*mol*L^-1#.

And given the former....#pH=-log_10[H_3O^+]=-log_10(3.5xx10^-8)#

#=-(-7.46)=7.46#, i.e. a basic solution where #[H_3O^+]#.

#pOH=14-7.46=6.54#, and #[HO^-]=10^(-6.54)*mol*L^-1=2.85xx10^-7*mol*L^-1#.

And if we gots #[H_3O^+]=0.0065*mol*L^-1#, #pH=-log_10(0.0065)=-(-2.19)=2.19#.

I leave it to you to calculate #pH# and #pOH# given the defining relationship.....#14=pH+pOH#