# What is the pKa of CH_3COOH and H_3PO_4?

Jun 3, 2016

$\setminus m a t h b f \left(\text{CH"_3"COOH}\right)$, ACETIC ACID

Acetic acid is a monoprotic acid, so it only has one $\text{pKa}$, and it is $4.74$.

If you are more familiar with its ${K}_{a}$, which is commonly given as $1.8 \times {10}^{- 5}$, then its $\text{pKa}$ is $- \log \left({K}_{a}\right) = 4.74$.

$\setminus m a t h b f \left({\text{H"_3"PO}}_{4}\right)$, PHOSPHORIC ACID

You can see that phosphoric acid is not monoprotic.

It is polyprotic. It has three protons. So it has three $\text{pKa}$s.

• ${\text{pKa}}_{1} = 2.16$
• ${\text{pKa}}_{2} = 7.21$
• ${\text{pKa}}_{3} = 12.32$

Each of its ${K}_{a}$'s corresponds to a sequential dissociation.

${K}_{a 1} = \left(\left[{\text{H"_2"PO"_4^(-)]["H"^(+)])/(["H"_3"PO}}_{4}\right]\right)$:

${\text{H"_3"PO"_4(aq) rightleftharpoons "H"_2"PO"_4^(-)(aq) + "H}}^{+} \left(a q\right)$

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

${\text{H"_2"PO"_4^(-)(aq) rightleftharpoons "HPO"_4^(2-)(aq) + "H}}^{+} \left(a q\right)$

${K}_{a 3} = \left(\left[{\text{PO"_4^(3-)]["H"^(+)])/(["HPO}}_{4}^{2 -}\right]\right)$:

${\text{HPO"_4^(2-)(aq) rightleftharpoons "PO"_4^(3-)(aq) + "H}}^{+} \left(a q\right)$

Combining all three reactions, we get:

${\text{H"_3"PO"_4(aq) rightleftharpoons cancel("H"_2"PO"_4^(-)(aq)) + "H}}^{+} \left(a q\right)$
cancel("H"_2"PO"_4^(-)(aq)) rightleftharpoons cancel("HPO"_4^(2-)(aq)) + "H"^(+)(aq)
cancel("HPO"_4^(2-)(aq)) rightleftharpoons "PO"_4^(3-)(aq) + "H"^(+)(aq)
$\text{------------------------------------------------------}$
$\textcolor{b l u e}{{\text{H"_3"PO"_4(aq) rightleftharpoons "PO"_4^(3-)(aq) + 3"H}}^{+} \left(a q\right)}$

The full hypothetical one-step dissociation here has the ${K}_{a}$ of:

$\textcolor{b l u e}{{K}_{a , \text{one-step}} = {K}_{a 1} {K}_{a 2} {K}_{a 3}}$

$= \left(\cancel{\left[{\text{H"_2"PO"_4^(-)])["H"^(+)])/(["H"_3"PO"_4])(cancel(["HPO"_4^(2-)])["H"^(+)])/(cancel(["H"_2"PO"_4^(-)]))(["PO"_4^(3-)]["H"^(+)])/(cancel(["HPO}}_{4}^{2 -}\right]}\right)$

$= \textcolor{b l u e}{\left(\left[{\text{PO"_4^(3-)]["H"^(+)]^3)/(["H"_3"PO}}_{4}\right]\right)}$

$= {10}^{- 2.16} \cdot {10}^{- 7.21} \cdot {10}^{- 12.32} = {10}^{- 21.69}$

$= \textcolor{b l u e}{2.04 \times {10}^{- 22}}$

In other words, the one-step ${K}_{a}$ is so small that phosphoric acid heavily favors getting deprotonated one proton at a time, rather than all three in one go.

Hence, its $\text{pKa}$ is NOT given as $- \log \left({10}^{- 21.69}\right) \approx 21.69$. Instead, it is given as ${\text{pKa}}_{1}$, ${\text{pKa}}_{2}$, and ${\text{pKa}}_{3}$.