# Question #22148

##### 1 Answer

#### Explanation:

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 **moles** of potassium hydroxide **for every mole** of sulfuric acid that takes part in the reaction.

**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)|)))#

**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)|)))#