Question

What volume of 0.100 M `"HCl"` solution is needed to neutralize 50.0 mL of 0.350 M `"KOH"`?

Answer 1

`"175 mL"`

Explanation:

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