Question #22148

Look at the balanced chemical equation that describes your neutralization reaction

#"H"_ 2"SO"_ (4(aq)) + color(blue)(2)"KOH"_ ((aq)) -> "K"_ 2"SO"_ (4(aq)) + 2"H"_ 2"O"_ ((l))#

Notice that the reaction consumes #color(blue)(2)# moles of potassium hydroxide for every mole of sulfuric acid that takes part in the reaction.

#color(white)(a)#
QUICK ANSWER

Molarity is meant to express moles of solute present in one liter of solution. Notice that the two solutions have the same molarity.

This, of course, means that they both contain the same number of moles of solute in a given unit of volume.

Now, you know that the reaction consumes twice as many moles of potassium hydroxide, which can only mean that the volume of the potassium hydroxide must be twice as big as the volume of the sulfuric acid solution.

Why?

Because it must contain twice as many moles of solute.

Therefore, you can say that

#color(green)(bar(ul(|color(white)(a/a)color(black)("volume of KOH solution" = 2 xx "11.8 mL" = "23.6 mL")color(white)(a/a)|)))#

#color(white)(a)#
DETAILED ANSWER

Your goal when dealing with stoichiometry problems that involve solutions is to use the molarity and volume given for one reactant to figure out how many moles of said reactant took part in the reaction.

In this case, you know that the reaction consumed

#11.8 color(red)(cancel(color(black)("mL"))) * (1 color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * ("0.260 moles H"_2"SO"_4)/(1color(red)(cancel(color(black)("L solution"))))#

# = "0.003068 moles H"_2"SO"_4#

Now that you know how many moles of sulfuric acid took part in the reaction, you can use the aforementioned mole ratio to determine the number of moles of potassium hydroxide that reacted

#0.003068 color(red)(cancel(color(black)("moles H"_2"SO"_4))) * (color(blue)(2)color(white)(a)"moles KOH")/(1color(red)(cancel(color(black)("mole H"_2"SO"_4))))#

# = " 0.006136 moles KOH"#

Finally, use the molarity of the potassium hydroxide solution to calculate how many liters would contain that many moles of solute

#0.006136 color(red)(cancel(color(black)("moles KOH"))) * "1 L solution"/(0.260 color(red)(cancel(color(black)("moles KOH"))))#

# = " 0.0236 L"#

Expressed in milliliters and rounded to three sig figs, the answer will be

#color(green)(bar(ul(|color(white)(a/a)color(black)("volume of KOH solution " = " 23.6 mL")color(white)(a/a)|)))#