Question #724b9

2 Answers
Jul 16, 2016

Answer:

The answer is B 13 mL of #"H"_3"PO"_4#.

Explanation:

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#

Jul 16, 2016

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 #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.