What volume (in milliliters of 0.120 M hclo₃ are required to neutralize 50.0 mL of 0.150 M NaOH)

Hydrochloric acid, #"HCl"#, and potassium hydroxide, #"KOH"#, react in a #1:1# mole ratio to produce aqueous potassium chloride, #"KCl"#, and water.

#"HCl"_ ((aq)) + "KOH"_ ((aq)) -> "KCl"_ ((aq)) + "H"_ 2"O"_ ((l))#

This means that a complete neutralization requires equal numbers of moles of hydrochloric acid, i.e. of hydronium cations, #"H"_3"O"^(+)#, and of potassium hydroxide, i.e. of hydroxide anions, #"OH"^(-)#.

As you know,m molarity is defined as the number of moles of solute present in #"1 L"# of solution.

Now, notice that the potassium hydroxide solution, which has a molarity of #"0.350 M"#, is

#(0.350 color(red)(cancel(color(black)("M"))))/(0.100color(red)(cancel(color(black)("M")))) = color(red)(3.5)#

times more concentrated than the hydrochloric acid solution, which has a molarity of #"0.100 M"#. In other words, for the same volume of both solutions, the potassium hydroxide solution contains #color(red)(3.5)# times more moles of solute than the hydrochloric acid solution.

This means that in order to have equal numbers of moles of both solutes, you need to have a volume of hydrochloric acid solution that is #color(red)(3.5)# times bigger than the volume of the sodium hydroixde solution.

Since the sodium hydroixde solution has a volume of #"50.0 mL"#, it follows that the volume of hydrochloric acid needed will be

#V_"HCl" = color(red)(3.5) xx "50.0 mL" = color(green)(|bar(ul(color(white)(a/a)color(black)("175 mL")color(white)(a/a)|)))-># to three sig figs

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