# Question 724b9

Jul 16, 2016

The answer is B 13 mL of ${\text{H"_3"PO}}_{4}$.

#### Explanation:

The equation for the reaction is

$\text{H"_3"PO"_4 + "3NaOH" → "Na"_3"PO"_4 + 3"H"_2"O}$.

1. Calculate the moles of $\text{NaOH}$

$\text{Moles of NaOH" = 0.100 color(red)(cancel(color(black)("L NaOH"))) ×( "0.1 mol NaOH")/(1 color(red)(cancel(color(black)("L NaOH")))) = "0.010 mol NaOH}$

2. Calculate the moles of ${\text{H"_3"PO}}_{4}$

${\text{Moles of H"_3"PO"_4 = 0.010 color(red)(cancel(color(black)("mol NaOH"))) × ("1 mol H"_3"PO"_4)/(3 color(red)(cancel(color(black)("mol NaOH")))) = "0.0033 mol H"_3"PO}}_{4}$

3. Calculate the volume of ${\text{H"_3"PO}}_{4}$

${\text{Volume of H"_3"PO"_4 = 0.0033 color(red)(cancel(color(black)("mol H"_3"PO"_4))) × ("1 L H"_3"PO"_4)/(0.25 color(red)(cancel(color(black)("mol H"_3"PO"_4)))) = "0.013 L H"_3"PO"_4 = "13 mL H"_3"PO}}_{4}$

Jul 16, 2016

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

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

Notice that it takes $\textcolor{red}{3}$ moles of sodium hydroxide, $\text{NaOH}$, to neutralize $1$ mole of phosphoric acid, ${\text{H"_3"PO}}_{4}$.

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 $\textcolor{red}{3}$ times more concentrated than the phosphoric acid solution.

Now, let's assume that you have $\text{100 mL}$ of phosphoric acid solution and $\text{100 mL}$ of sodium hydroxide solution. In this case, the concentration of the sodium hydroxide solution should have been

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

However, the concentration of sodium hydroxide solution is $\text{0.1 M}$, which is approximately $8$ times lower than what you'd need for $\text{100 mL}$ of $\text{0.25 M}$ phosphoric acid solution.

This means that you must have a volume of phosphoric acid solution that is approximately $8$ times smaller than $\text{100 mL}$. The closest value to

$\text{100 mL"/8 = "12.5 mL}$

is $\text{13.33 mL}$, which means that this will be your answer!

$\textcolor{w h i t e}{\frac{a}{a}}$
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

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{c = {n}_{\text{solute"/V_"solution" implies n_"solute" = c * V_"solution}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

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"))

${n}_{\text{NaOH" = "0.010 moles NaOH}}$

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 $\text{0.25 M}$ phosphoric acid solution that contains this many moles is

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.