Henderson Hasselbalch help? Phosphoric acid is a triprotic acid. To find the pH of a buffer composed of #"H"_2"PO"_(4(aq))^(-)# and #"HPO"_(4(aq))^(2-)#, which #pK_a# value would you use in the Henderson-Hasselbalch equation?

#K_(a1) = 6.9 xx 10^(–3 )#

#K_(a2) = 6.2 xx 10^(–8)#

#K_(a3) = 4.8xx 10^(–13)#

1 Answer
Jun 8, 2016

Answer:

You must use #pK_(a2)#.

Explanation:

Phosphoric acid will ionize in aqueous solution according to the following equilibrium reactions

#"H"_ 3"PO"_ (4(aq)) + "H"_ 2 "O"_ ((l)) rightleftharpoons "H"_ 2"PO"_ (4(aq))^(-) + "H"_ 3 "O"_ ((aq))^(+)" "K_(a1)#

#"H" _ 2 "PO"_ (4(aq))^(-) + "H" _ 2 "O"_ ((l)) rightleftharpoons "HPO"_ (4(aq))^(2-) + "H"_ 3 "O"_ ((aq))^(+)" "K_(a2)#

#"HPO"_ (4(aq))^(2-) + "H"_ 2"O"_ ((l)) rightleftharpoons "PO" _ (4(aq))^(3-) + "H"_ 3"O"_ ((aq))^(+)" "K_(a3)#

You can calculate the #pK_a# values by using

#color(blue)(|bar(ul(color(white)(a/a)pK_a = - log(K_a)color(white)(a/a)|)))#

In this case, you will have

#{ ( pK_(a1) = - log(K_(a1)) = - log(6.9 * 10^(-3)) = 2.16), (pK_(a2) = - log(K_(a2)) = - log(6.2 * 10^(-8)) = 7.21), (pK_(a3) = - log(K_(a3)) = - log(4.8 * 10^(-13)) = 12.3) :}#

Now, your buffer contains dihydrogen phosphate, #"H"_2"PO"_4^(-)#, and hydrogen phosphate, #"HPO"_4^(-)#. These two species can be found together in the second equilibrium reaction

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

By definition, the acid dissociation constant for this reaction will be

#K_(a2) = (["H"_3"O"^(+)] * ["HPO"_4^(2-)])/(["H"_2"PO"_4^(-)])#

Now, take the log base #10# of both sides to get

#log(K_(a2)) = log((["H"_3"O"^(+)] * ["HPO"_4^(2-)])/(["H"_2"PO"_4^(-)]))#

This is equivalent to

#log(K_(a2)) = log(["H"_3"O"^(+)]) + log((["HPO"_4^(2-)])/(["H"_2"PO"_4^(-)]))#

Rearrange to get

#overbrace(- log(["H"_ 3"O"^(+)]))^(color(red)(= "pH")) = overbrace(- log(K_(a2)))^(color(red)(= pK_(a2))) + log((["HPO"_4^(2-)])/(["H"_2"PO"_4^(-)]))#

Therefore, you will have

#color(blue)(|bar(ul(color(white)(a/a)color(black)("pH" = pK_(a2) + log((["HPO"_4^(2-)])/(["H"_2"PO"_4^(-)])))color(white)(a/a)|)))#

This is the Henderson - Hasselbalch equation that describes a buffer that contains dihydrogen phosphate, a weak acid, and hydrogen phosphate, its conjugate base.

As you can see, you must use #pK_(a2)# here. Notice that when you have equal concentrations of dihydrogen phosphate and hydrogen phosphate, the pH of the buffer will be equal to #pK_(a2)#, since

#"pH" = pK_(a2) + log(1)#

#"pH" = pK_(a2)#