Question

Question #91f35

Answer 1

`"366 mL"`

Explanation:

Before worrying about the neutralization reaction that takes place when you add the potassium hydroxide and sulfuric acid solutions, focus on finding the molarity of the diluted sulfuric acid solution.

As you know, diluting a solution essentially means keeping the number of moles of solute constant while increasing the volume of the solution.

So your strategy here is to use the molarity and volume of the initial solution to figure out how many moles of sulfuric acid you have in that sample.

`color(blue)(c = n/V implies n = c * V)`

`n_(H_2SO_4) = "3.00 M" * 12.50 * 10^(-3)"L" = "0.0375 moles H"_2"SO"_4`

After you dilute the initial solution, its volume will be equal to `"100.0 mL"`. This means that the concentration of the diluted solution will be

`c = "0.0375 moles"/(100.0 * 10^(-3)"L") = "0.375 M"`

Now focus on the balanced chemical equation for this neutralization reaction

`color(red)(2)"KOH"_text((aq]) + "H"_2"SO"_text(4(aq]) -> "K"_2"SO"_text(4(aq]) + "H"_2"O"_text((l])`

Notice that the reaction consumes `color(red)(2)` moles of potassium hydroxide for very `1` mole of sulfuric acid.

This means that, in order to have a complete neutralization, you need to provide the sulfuric acid solution will twice as many moles of potassium hydroxide as you have moles of sulfuric acid.

`0.0375color(red)(cancel(color(black)("moles H"_2"SO"_4))) * (color(red)(2)" moles KOH")/(1color(red)(cancel(color(black)("mole H"_2"SO"_4)))) = "0.0750 moles KOH"`

You know how many moles of potassium hydroxide you need, and what the molarity of the potassium hydroxide solution is, so all you have to do now is figure out what volume of this solution would contain this many moles

`color(blue)(c = n/V implies V = n/c)`

`V = (0.0750color(red)(cancel(color(black)("moles"))))/(0.205color(red)(cancel(color(black)("moles")))/"L") = "0.3659 L"`

Rounded to thre sig figs and expressed in mililiters, the answer will be

`V_(KOH) = color(green)("366 mL")`