Question #00919

Sodium hydroxide and hydrochloric acid react in a #1:1# mole ratio to produce aqueous sodium chloride and water

#"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.

Now, a solution's molarity tells you the number of moles of solute present in exactly #"1 L"# of that solution.

You know that the sodium hydroxide solution has a molarity of #"0.20 M"#. You can say that if you were to add equal volumes of the two solutions, then the hydrochloric acid solution would also have a molarity equal to #"0.20 M"#.

In other words, if you were to mix #"54.0 mL"# of #"0.20 M"# sodium hydroxide solution with #"54.0 mL"# of hydrochloric acid solution and get a complete neutralization as a result, then the molarity of the hydrochloric acid solution would be equal to #"0.20 M"#.

However, you know that it takes #"23.5 mL"# of hydrochloric acid solution to get a complete neutralization.

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)#

Therefore,

#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.