# Question 8fc4f

Oct 27, 2016

Here's what I got.

#### Explanation:

The idea here is that you're trying to figure out how many milliliters of $\text{1.320 M}$ sodium hydroxide solution will contain enough moles to neutralize $\text{23.25 mL}$ of $\text{1.400 M}$ phosphoric acid solution.

The balanced chemical equation that describes this neutralization reaction looks like this

${\text{H"_ 3"PO"_ (4(aq)) + color(blue)(3)"NaOH"_ ((aq)) -> "Na"_ 3"PO"_ (4(aq)) + 3"H"_ 2"O}}_{\left(l\right)}$

Notice that the reaction consumes $\textcolor{b l u e}{3}$ moles of sodium hydroxide for every mole of phosphoric acid that takes part in the reaction.

Now, use the molarity and volume of the phosphoric acid solution to determine how many moles of acid were present in the solution

23.25 color(red)(cancel(color(black)("mL"))) * (1 color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * ("1.400 moles H"_3"PO"_4)/(1color(red)(cancel(color(black)("L"))))

$= {\text{0.03255 moles H"_3"PO}}_{4}$

Use the aforementioned mole ratio to figure out how many moles of sodium hydroxide were needed to neutralize all the moles of phosphoric acid

0.03255 color(red)(cancel(color(black)("moles H"_3"PO"_4))) * (color(blue)(3)color(white)(a)"moles NaOH")/(1color(red)(cancel(color(black)("mole H"_3"PO"_4))))

$= \text{ 0.09765 moles NaOH}$

Finally, use the molarity of the sodium hydroxide solution to calculate how many milliliters of solution would contain this many moles

0.09765color(red)(cancel(color(black)("moles NaOH"))) * (1color(red)(cancel(color(black)("L"))))/(1.320color(red)(cancel(color(black)("moles NaOH")))) * (10^3"mL")/(1color(red)(cancel(color(black)("L"))))#

$= \textcolor{g r e e n}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{73.98 mL}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answer is rounded to four sig figs.