# 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)

Jun 8, 2016

You must use $p {K}_{a 2}$.

#### Explanation:

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

$\text{H"_ 3"PO"_ (4(aq)) + "H"_ 2 "O"_ ((l)) rightleftharpoons "H"_ 2"PO"_ (4(aq))^(-) + "H"_ 3 "O"_ ((aq))^(+)" } {K}_{a 1}$

$\text{H" _ 2 "PO"_ (4(aq))^(-) + "H" _ 2 "O"_ ((l)) rightleftharpoons "HPO"_ (4(aq))^(2-) + "H"_ 3 "O"_ ((aq))^(+)" } {K}_{a 2}$

$\text{HPO"_ (4(aq))^(2-) + "H"_ 2"O"_ ((l)) rightleftharpoons "PO" _ (4(aq))^(3-) + "H"_ 3"O"_ ((aq))^(+)" } {K}_{a 3}$

You can calculate the $p {K}_{a}$ values by using

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} p {K}_{a} = - \log \left({K}_{a}\right) \textcolor{w h i t e}{\frac{a}{a}} |}}}$

In this case, you will have

$\left\{\begin{matrix}p {K}_{a 1} = - \log \left({K}_{a 1}\right) = - \log \left(6.9 \cdot {10}^{- 3}\right) = 2.16 \\ p {K}_{a 2} = - \log \left({K}_{a 2}\right) = - \log \left(6.2 \cdot {10}^{- 8}\right) = 7.21 \\ p {K}_{a 3} = - \log \left({K}_{a 3}\right) = - \log \left(4.8 \cdot {10}^{- 13}\right) = 12.3\end{matrix}\right.$

Now, your buffer contains dihydrogen phosphate, ${\text{H"_2"PO}}_{4}^{-}$, and hydrogen phosphate, ${\text{HPO}}_{4}^{-}$. These two species can be found together in the second equilibrium reaction

${\text{H" _ 2 "PO"_ (4(aq))^(-) + "H" _ 2 "O"_ ((l)) rightleftharpoons "HPO"_ (4(aq))^(2-) + "H"_ 3 "O}}_{\left(a q\right)}^{+}$

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

${K}_{a 2} = \left(\left[{\text{H"_3"O"^(+)] * ["HPO"_4^(2-)])/(["H"_2"PO}}_{4}^{-}\right]\right)$

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

$\log \left({K}_{a 2}\right) = \log \left(\left(\left[{\text{H"_3"O"^(+)] * ["HPO"_4^(2-)])/(["H"_2"PO}}_{4}^{-}\right]\right)\right)$

This is equivalent to

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

Rearrange to get

$\overbrace{- \log \left(\left[{\text{H"_ 3"O"^(+)]))^(color(red)(= "pH")) = overbrace(- log(K_(a2)))^(color(red)(= pK_(a2))) + log((["HPO"_4^(2-)])/(["H"_2"PO}}_{4}^{-}\right]\right)}$

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 $p {K}_{a 2}$ here. Notice that when you have equal concentrations of dihydrogen phosphate and hydrogen phosphate, the pH of the buffer will be equal to $p {K}_{a 2}$, since

$\text{pH} = p {K}_{a 2} + \log \left(1\right)$

$\text{pH} = p {K}_{a 2}$