If #"NaOH"(aq)# that was originally #"0.300 M"# is left exposed to the air and #"0.0200 M"# of #"CO"_2# dissolves in the solution, but right at that point the student catches his mistake and caps the #"NaOH"#, what is its new molarity?

Well, #"NaOH"# is dissolved in water, so... the #"CO"_2# absorbed would react with the excess water in the aqueous #"NaOH"# to form #"H"_2"CO"_3#.

#"CO"_2(g) + "H"_2"O"(l) rightleftharpoons "H"_2"CO"_3(aq)#

with #K_(hyd) = 1.7 xx 10^(-3)# for #"H"_2"CO"_3# getting formed like this.

The rate constant for #"H"_2"CO"_3# decomposition is #"23 s"^(-1)#, so the half-life of the first-order process is

#t_"1/2" = ln2/k = 0.693/("23 s"^(-1)) ~~ "0.03 s"#

It's such a fast decomposition, that we have to assume that #"NaOH"# reacts with the formed #"H"_2"CO"_3# right away.

By Le Chatelier's principle, when that occurs, more #"H"_2"CO"_3# gets produced, until the #"CO"_2# that made it is no longer present (good luck with that...). At that point, #"H"_2"CO"_3# is gone and we ignore any other #"CO"_2# that bothers the #"NaOH"#,

Anyhow, we form the ICE table with #"0.300 M NaOH"#, assuming the volume did not change (all units in #"M"#):

#"OH"^(-)(aq) + "H"_2"CO"_3(aq) -> "H"_2"O"(l) + "HCO"_3^(-)(aq)#

#"I"" "0.300" "" "" "0.0200" "" "" "" "-" "" "" "0#
#"C"" "-x" "" "" "" "-x" "" "" "" "-" "" "+x#
#"E"" "(0.300 - x)" "(0.0200 - x)color(white)(.)-" "" "" "x#

where #"mols CO"_2 -> "mols H"_2"CO"_3# by the reaction stoichiometry, and #"2.0 mmol CO"_2//"100 mL soln"# #=# #"0.0200 M"#.

As just mentioned, we have a strong base, so its consumption to neutralize #"H"_2"CO"_3# shall push the equilibrium with #"CO"_2# forward to make more #"H"_2"CO"_3#. This occurs until there is no more #"CO"_2# and we can move forward.

Therefore, #x = "0.0200 M"#, and so:

#color(blue)(["OH"^(-)]_"equiv pt" = "0.280 M")#.

And of course, if you wanted to titrate it with #"HCl"#, you should experimentally find #"0.280 M"# as well. What is this molarity in terms of normality?