Sodium hydroxide and hydrochloric acid react in a
#"NaOH"_ ((aq)) + "HCl"_ ((aq)) -> "NaCl"_ ((aq)) + "H"_ 2"O"_ ((l))#
This means that a complete neutralization will always consume equal numbers of moles of sodium hydroxide and of hydrochloric acid.
You know that the sodium hydroxide solution has a molarity of
In other words, if you were to mix
However, you know that it takes
This tells you that the hydrochloric acid solution is more concentrated than the sodium hydroxide solution because it contains the same number of moles of acid in a smaller volume.
In fact, the ratio that exists between the volumes of the sodium hydroxide solution and the volume of the hydrochloric acid solution
#(54.0 color(red)(cancel(color(black)("mL"))))/(23.5color(red)(cancel(color(black)("mL")))) = color(blue)(2.298)#
must also exist between the molarity of the hydrochloric acid solution and the molarity of the sodium hydroxide solution.
#c_"HCl"/c_"NaOH" = color(blue)(2.298)#
#c_"HCl" = color(blue)(2.298) * c_"NaOH"#
In your case, you will have
#color(darkgreen)(ul(color(black)(c_"HCl" = 2.298 * "0.20 M" = "0.46 M")))#
The answer is rounded to two sig figs, the number of sig figs you have for the molarity of the sodium hydroxide solution.