If #"0.1000 g"# of #"CO"_2# seeps into initially #"0.1026 M"# #"NaOH"# to degrade it, and the student has #"0.1142 M"# #"HCl"#, what volume of #"HCl"# is needed to neutralize #"NaOH"#? What if no #"CO"_2# seeped in? What volume then?

I got #"43.93 mL"# and #"44.92 mL"#.

All of this is using acids and bases with one proton or one #"OH"^(-)#, so molarity = normality. That said, we have #"1 L"# of #"0.1026 M NaOH"#, that absorbed #"0.1000 g CO"_2(g)#.

#a)#

The excess water in the #"NaOH"(aq)# reacts with the #"CO"_2# to form #"H"_2"CO"_3# with rate constant #"0.039 s"^(-1)# (taking about #"17.8 s"#), which is feasible.

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

The mols of #"H"_2"CO"_3# introduced was:

#"0.1000 g CO"_2 xx ("1 mol CO"_2)/("44.009 g CO"_2) = "0.00227 mols CO"_2#

#= "0.00227 mols H"_2"CO"_3#

And in #"1 L"# this is therefore #"0.00227 M"#, which does not change with the volume of the aliquot obtained. Let's see how much #"OH"^(-)# is consumed. It will push the equilibrium forward as the #"H"_2"CO"_3# is converted to #"HCO"_3^(-)#.

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

#"I"" "0.00227" "" "0.1026" "" "" "0" "" "" "" "" "-#
#"C"" "-0.00227" "-0.00227" "+0.00227" "" "-#
#"E"" "0" "" "" "" "0.1003" "" "" "0.00227" "" "" "-#

So, with this amount left, in #"50 mL"# there are

#"0.1003 mols NaOH"/"L" xx "0.050 L" = "0.005016 mols OH"^(-)#

which neutralize #"0.005016 mols H"^(+)# in #"HCl"#. The student had #"0.1142 M HCl"#, so

#"0.005016 mols HCl" xx ("1000 mL")/("0.1142 mols HCl") = color(blue)("43.93 mL HCl")#

now that some #"NaOH"# was consumed by reaction with #"CO"_2#.

#b)#

If the #"CO"_2(g)# were not to react with #"NaOH"# (that is, if the student did not let the solution be out in the open for more than a few seconds), then

#"0.1026 mols NaOH"/"L" xx "0.050 L" = "0.005130 mols OH"^(-)#

#=># #"0.005130 mols HCl"# to add

This should not be that different, so I think the answer key is not correct... This would then require

#"0.005130 mols HCl" xx ("1000 mL")/("0.1142 mols HCl") = color(blue)("44.92 mL HCl")#

not #"9.37 mL"#.

In fact, if no #"CO"_2# was absorbed, there should be more #"NaOH"# available, which means more #"HCl"# would be needed to neutralize it, not #4+# times less.