Question #724b9

The equation for the reaction is

`"H"_3"PO"_4 + "3NaOH" → "Na"_3"PO"_4 + 3"H"_2"O"`.

1. Calculate the moles of `"NaOH"`

`"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 `"H"_3"PO"_4`

`"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 `"H"_3"PO"_4`

`"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`

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 `color(red)(3)` moles of sodium hydroxide, `"NaOH"`, to neutralize `1` mole of phosphoric acid, `"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 `color(red)(3)` times more concentrated than the phosphoric acid solution.

Now, let's assume that you have `"100 mL"` of phosphoric acid solution and `"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 `"0.1 M"`, which is approximately `8` times lower than what you'd need for `"100 mL"` of `"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 `"100 mL"`. The closest value to

`"100 mL"/8 = "12.5 mL"`

is `"13.33 mL"`, which means that this will be your answer!

`color(white)(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

`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 `"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.