Question #7f814

!! EXTREMELY LONG ANSWER !!

The idea here is that phosphate anion, #"PO"_4^(3-)#, will act as a base here and pick up some of the hydrogen ions (if not all of them, as you'll see soon) produced by the hydrochloric acid in the solution.

You have three possible reactions here, depending on how many hydrogen ions are present in the solution--I won't add the chloride anions here because they are not relevant.

# "PO"_ (4(aq))^(3-) + "H"_ ((aq))^(+) -> "HPO"_ (4(aq))^(2-)" " " "color(darkorange)((1))#

#"PO"_ (4(aq))^(3-) + 2"H"_ ((aq))^(+) -> "H"_ 2"PO"_ (4(aq))^(-) " " " "color(darkorange)((2))#

#"PO"_ (4(aq))^(3-) + 3"H"_ ((aq))^(+) -> "H"_ 3"PO"_ (4(aq)) " " " "color(darkorange)((3))#

Your goal here is to figure out which of these reactions will take place. In the first case, you're mixing

#50 color(red)(cancel(color(black)("mL"))) * ("0.10 moles PO"_4^(3-))/(10^3color(red)(cancel(color(black)("mL")))) = "0.0050 moles PO"_4^(3-)#

with

#50 color(red)(cancel(color(black)("mL"))) * ("0.30 moles H"^(+))/(10^3color(red)(cancel(color(black)("mL")))) = "0.0150 moles H"^(+)#

The phosphate anions and the hydrogen ions produced by the hydrochloric acid are present in a #1:3# mole ratio, so you can say that the equation #color(darkorange)((3))# will take place here.

The reaction will consume all the moles of phosphate anions and all the moles of hydrogen ions and leave you with #0.0050# moles of phosphoric acid, #"H"_3"PO"_4#, in a total volume of

#"50 mL + 50 mL = 100 mL"#

The concentration of the phosphoric acid will be

#["H"_3"PO"_4] = "0.0050 moles"/(100 * 10^(-3) quad "L") = "0.05 M"#

To find the #"pH"# of the solution, you need to use the fact that phosphoric acid is a weak acid that partially ionizes in aqueous solution to produce hydrogen ions and hydrogen phosphate anions in its first ionization.

#"H"_ 3"PO"_ (4(aq)) rightleftharpoons "H"_ 2"PO"_ (4(aq))^(-) + "H"_ ((aq))^(+)#

The acid dissociation constant for the first ionization of phosphoric acid is #-># see here.

#K_(a1) = 7.52 * 10^(-3)#

Keep in mind that phosphoric acid ionizes two more times in aqueous solution--it's a triprotic acid-- but the acid dissociation constants for these ionizations are so small that you can assume that they will have no impact on the final #"pH"# of the solution.

Now, if you take #x# #"M"# to be the concentration of hydrogen ions and of hydrogen phosphate anions produced by the first ionization, you can say that the equilibrium concentration of phosphoric acid will be

#["H"_3"PO"_4] = (0.05 -x) quad "M"#

This basically means that in order for the acid to ionize and produce #x# #"M"# of hydrogen ions and #x# #"M"# of hydrogen phosphate anions, its concentration must decrease by #x# #"M"#.

By definition, the acid dissociation constant will be

#K_(a1) = (["HPO"_4^(2-)] * ["H"^(+)])/(["H"_ 3"PO"_4])#

Plug in your values to find

#7.52 * 10^(-3) = (x * x)/(0.05 - x)#

Rearrange to quadratic equation form to get

#x^2 + 7.52 * 10^(-3) * x - 0.376 * 10^(-3) = 0#

This quadratic equation will produce two values, one positive and one negative. Since #x# represents concentration, you can discard the negative value and say that

#x = 0.015992#

This means that the concentration of hydrogen ions in this solution will be

#["H"^(+)] = "0.015992 M"#

By definition, the #"pH"# of the solution is given by

#color(blue)(ul(color(black)("pH" = - log(["H"^(+)]))))#

Plug in your value to find

#"pH" = - log(0.015992) = color(darkgreen)(ul(color(black)(1.80)))#

#color(white)(a/a)#

Now, for part (2), you once again examine the number of moles of phosphate anions and of hydrogen ions you're mixing.

This time, you're mixing #0.0050# moles of phosphate anions and only

#50 color(red)(cancel(color(black)("mL"))) * ("0.25 moles H"^(+))/(10^3color(red)(cancel(color(black)("mL")))) = "0.0125 moles H"^(+)#

So you know for a fact that you don't have enough moles of hydrogen ions for reaction #color(darkorange)((3))# to consume all the moles of phosphate anions present in the solution, but you do have enough to ensure that reaction #color(darkorange)((2))# will take place and consume all the moles of phosphate anions.

This time, the reaction will consume all the moles of phosphate anions and only

#0.0050 color(red)(cancel(color(black)("moles PO"_4^(3-)))) * "2 moles HCl"/(1color(red)(cancel(color(black)("mole PO"_4^(3-))))) = "0.0100 moles H"^(+)#

and produce #0.0050# moles of dihydrogen phosphate, #"H"_2"PO"_4^(-)#.

So after the initial reaction takes place, the solution will contain #0.0050# moles of dihydrogen phosphate and

#overbrace("0.0125 moles")^(color(blue)("what you started with")) - overbrace("0.0100 moles")^(color(blue)("what was consumed")) = "0.0025 moles H"^(+)#

Now, these extra hydrogen ions will react with the dihydrogen phosphate anions to produce phosphoric acid.

#"H"_ 2"PO"_ (4(aq))^(-) + "H"_ ((aq))^(+) -> "H"_ 3"PO"_ (4(aq))#

This reaction will consume all the moles of hydrogen ions and only

#0.0025 color(red)(cancel(color(black)("moles H"^(+)))) * ("1 mole H"_ 2"PO"_ 4^(-))/(1color(red)(cancel(color(black)("mole H"^(+))))) = "0.0025 moles H"_ 2"PO"_ 4^(-)#

and produce #0.0025# moles of phosphoric acid. So after this reaction is complete, the solution will also contain

#overbrace("0.0050 moles")^(color(blue)("what was produced initially")) - overbrace("0.0025 moles")^(color(blue)("what was consumed")) = "0.0025 moles H"_2"PO"_4^(-)#

The total volume of the solution will once again be

#"50 mL + 50 mL = 100 mL"#

which means that the concentrations of the two chemical species will be

#["H"_2"PO"_4^(-)] = ["H"_3"PO"_4] = "0.0025 moles"/(100 * 10^(-3) quad "L") = "0.025 M"#

At this point, the fact that the solution contains phosphoric acid, a weak acid, and dihydrogen phosphate, its conjugate base, should tell you that you're now dealing with a buffer solution.

As you know, the #"pH"# of a weak acid-conjugate base buffer solution can be calculated using the Henderson - Hasselbalch equation

#color(blue)(ul(color(black)("pH" = "p"K_a + log ( (["conjugate base"])/(["weak acid"])))))#

In your case, you have

#"pH" = "p"K_(a1) + log( (["H"_2"PO"_4^(-)])/(["H"_3"PO"_4]))#

with

#"p"K_(a1) = - log(K_(a1))#

#"p"K_(a1) = - log(7.52 * 10^(-3)) = 2.12#

Now, because your buffer contains equal concentrations of the weak acid and of its conjugate base, you can say that its #"pH"# will be equal to the #"p"K_a# of the weak acid.

This is the case because you have

#"pH" = 2.12 + log ((color(red)(cancel(color(black)("0.025 M"))))/(color(red)(cancel(color(black)("0.025 M")))))#

#"pH" = 2.12 + 0 = color(darkgreen)(ul(color(black)(2.12)))#

And there you have it. I'll leave the answers rounded to two decimal places, but keep in mind that you have one significant figure for almost all your values, so you should report the two #"pH"# values rounded to one decimal place.