Question #21b47

1 Answer
Truong-Son N.
Mar 19, 2017

Suppose #n# #"mol"#s of #"NaOH"# are contained in #"25 mL"#. Then,

#"n mols"/("25 mL") xx "10 mL" = 0.4n# #"mol"#s

That is, there is #40%# of the #"mol"#s of #"NaOH"# in #40%# of the volume, since #"mol"#s are an extensive quantity, just like #"grams"#.

Since #"NaOH"# is a strong base, it reacts 1:1 with the first #"H"^(+)# that oxalic acid dissociates into solution (we assume the second #K_a# is negligible).

Unless you provide actual numbers, this is as far as we can go.

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