# Can the Henderson-Hasselbalch equation be used to estimate the pH values at the equivalence point?

Jul 7, 2017

Questionable

#### Explanation:

For an equivalence concentration of 0.10M $O A {c}^{-}$ and applying equilibrium principles, the equivalence point pH = 8.87.

Using the Henderson-Hasselbalch equation and applying the $H O A c$ and $O A {c}^{-}$ concentrations of $7.45 \times {10}^{-} 6 M$ for the acid and 0.10M for the base (typical for the hydrolysis equilibrium of 0.10M $O A {c}^{-}$), the pH = 8.88. There is a concentration of $O {H}^{-}$ ions that is not considered in the H-H Equation, so I'd say for exact pH values no, but for relative pH as being basic or acidic ... maybe, but this is the only comparative calculation that I've done.

Jul 7, 2017

For 0.01M $N {H}_{4} C l$ at equivalence point
By Equilibrium Calculations => pH = 5.12
By H-H Calculation => pH = 5.12

#### Explanation:

Using 0.10M $N {H}_{4} C l$=> $0.10 M N {H}_{4}^{+}$ + $0.10 M C {l}^{-}$

Hydrolyzing $0.10 M N {H}_{4}^{+}$
=> $7.45 \times {10}^{- 6} M \left(N {H}_{4} O H\right) + 7.4 \times {10}^{-} 6 M \left({H}^{+}\right)$
=>$\left[{H}^{+}\right] = 5.47 \times {10}^{-} 6$ => pH = 5.12

For Henderson-Hasselbalch Equation
${K}_{a} = {K}_{w} / {K}_{b} = \frac{1 \times {10}^{-} 14}{1.8 \times {10}^{-} 5} = 5.56 \times {10}^{-} 10$

$p {K}_{a} = - \log \left({K}_{a}\right) = - \log \left(5.56 \times {10}^{-} 10\right) = - \left(- 9.25\right) = 9.25$
Base = $7.45 \times {10}^{-} 6 M N {H}_{4} O H$
Acid = $0.10 M N {H}_{4}^{+}$

$p H = p {K}_{a} + \log \left(\frac{\left[b a s e\right]}{\left[a c i d\right]}\right) = 9.25 + \log \left(\frac{7.45 \times {10}^{-} 6}{0.10}\right) = 9.25 + \left(- 4.13\right) = 9.25 + 4.13 = 5.12$

By Equilibrium Calculations => pH = 5.12
By H-H Calculation => pH = 5.12

Jul 7, 2017

It shouldn't be used exactly at the equivalence point: in principle, only one species exists at the equivalence point, by definition... so the logarithm blows up.

What you instead have to do, to estimate the $\text{pH}$ near the equivalence point, is realize that this singularity will occur, and set up an ICE Table for the association equilibrium right after the equivalence point is passed, i.e.

${\text{A"^(-)(aq) + "H"_2"O"(l) rightleftharpoons "HA"(aq) + "OH}}^{-}$,

with ${K}_{b} = \frac{1 \times {10}^{- 14}}{{K}_{a}} = \left(\left[{\text{HA"]["OH"^(-)])/(["A}}^{-}\right]\right)$,

in which, remember, we've moved OFF of the equivalence point.

SO WE CAN'T USE IT RIGHT AT THE EQUIV PT?

This rapid variation of concentration at the equivalence point is not accounted for in the Henderson-Hasselbalch equation, which is founded on already establishing a fixed equilibrium in the first place.

In fact, it's in the derivation, simply because we must use the ${K}_{a}$, not, e.g. some sort of time-dependent reaction quotient, ${Q}_{a}$. If one species doesn't exist, the logarithm makes the whole solution asymptotic.

Suppose we have an ordinary, working ${K}_{a}$ (i.e. equilibrium is established and time-independent!).

${K}_{a} = \left(\left[\text{A"^(-)]["H"_3"O"^(+)])/(["HA}\right]\right)$

The $- \log$ of both sides gives:

"pKa" = -log((["A"^(-)]["H"_3"O"^(+)])/(["HA"]))

$= - \log \left(\left[\text{A"^(-)]["H"_3"O"^(+)]) - (-log["HA}\right]\right)$

= -log((["A"^(-)])/(["HA"])) - log(["H"_3"O"^(+)])

By definition, "pH" = -log["H"_3"O"^(+)], so add the lefthand logarithm to the other side to get:

$\underline{\text{ "" "" "" "" "" "" "" "" "" "" "" "" }}$
$| \underline{\text{ "color(blue)("pH" = "pKa" + log((["A"^(-)])/(["HA"])))" }} |$

Now, exactly at the equivalence point, by definition, ONE species is already completely neutralized. So if, say, we put $\text{NaOH}$ in to completely neutralize some $\text{CH"_3"COOH}$, we would have

["CH"_3"COOH"]_(eq) ~~ "0 M"

${\left[\text{CH"_3"COO"^(-)]_(eq) ~~ ["CH"_3"COOH}\right]}_{i}$

And thus...

"pH" = 4.76 + log((["CH"_3"COO"^(-)]_(eq))/("0 M"))

and so, $\text{pH} \to \infty$.

That is reflected in many titration curves, where the steep rise occurs at the equivalence point, and that is when one species has vanished and what's left is what the strong base has not destroyed, i.e. other base.

That's why in practice, you have to slow down in adding your titrant when you are close to the equivalence point --- so you can catch as many of those data points as you can.