# Question #724b9

##### 2 Answers

#### Answer:

The answer is B 13 mL of

#### Explanation:

The equation for the reaction is

**1.** Calculate the moles of

**2.** Calculate the moles of

**3.** Calculate the volume of

#### Answer:

Here's what I got.

#### Explanation:

Here's how you can solve this problem by using a *more intuitive approach*.

Look at the chemical equation given to you

#"H"_ 3"PO"_ (4(aq)) + color(red)(3)"NaOH"_ ((aq)) -> "Na"_ 3"PO"_ (4(aq)) + 3"H"_ 2"O"_ ((l))#

Notice that it takes **moles** of sodium hydroxide, **mole** of phosphoric acid,

This tells you that when you're working with solutions of sodium hydroxide and phosphoric acid of **equal volumes**, the sodium hydroxide solution must be **times more concentrated** than the phosphoric acid solution.

Now, let's * assume* that you have

**should have been**

#["NaOH"] = color(red)(3) xx "0.25 M" = "0.75 M"#

However, the concentration of sodium hydroxide solution is **approximately** **times lower** than what you'd need for

This means that you must have a volume of phosphoric acid solution that is **approximately** **times smaller** than

#"100 mL"/8 = "12.5 mL"#

is

**ALTERNATIVELY**

You can prove that this is the answer by using the *molarity* and *volume* of the sodium hydroxide solution to figure out how many **moles** it contains

#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution" implies n_"solute" = c * V_"solution")color(white)(a/a)|)))#

You will have

#n_"NaOH" = "0.1 mol" color(red)(cancel(color(black)("L"^(-1)))) * overbrace(100 * 10^(-3)color(red)(cancel(color(black)("L"))))^(color(blue)("volume in liters"))#

#n_"NaOH" = "0.010 moles NaOH"#

This many moles of sodium hydroxide would require

#0.010 color(red)(cancel(color(black)("moles NaOH"))) * ("1 mole H"_3"PO"_4)/(color(red)(3)color(red)(cancel(color(black)("moles NaOH")))) = "0.00333 moles H"_3"PO"_4#

The volume of

#V_("H"_3"O"_4) = (0.00333 color(red)(cancel(color(black)("moles"))))/(0.25 color(red)(cancel(color(black)("mol"))) "L"^(-1)) = "0.01333 L"#

Expressed in *milliliters*, the answer will once again be

#V_("H"_3"PO"_4) = color(green)(|bar(ul(color(white)(a/a)color(black)("13.33 mL")color(white)(a/a)|)))#

I won't bother with **sig figs**, but keep in mind that you don't have four sig figs for the values given to you by the problem.