Question

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)`

Answer 1

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)`