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

##### 3 Answers

#### Answer:

Questionable

#### Explanation:

For an equivalence concentration of 0.10M

Using the Henderson-Hasselbalch equation and applying the

#### Answer:

For 0.01M

By Equilibrium Calculations => pH = 5.12

By H-H Calculation => pH = 5.12

#### Explanation:

Using 0.10M

Hydrolyzing

=>

=>

For Henderson-Hasselbalch Equation

Base =

Acid =

By Equilibrium Calculations => pH = 5.12

By H-H Calculation => pH = 5.12

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 **set up an ICE Table** for the association equilibrium *right after* the equivalence point is passed, i.e.

#"A"^(-)(aq) + "H"_2"O"(l) rightleftharpoons "HA"(aq) + "OH"^(-)# ,with

#K_b = (1 xx 10^(-14))/(K_a) = (["HA"]["OH"^(-)])/(["A"^(-)])# ,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

Suppose we have an ordinary, working

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

The

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

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

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

By definition,

#ul(" "" "" "" "" "" "" "" "" "" "" "" "" ")#

#|ul(" "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

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

#["CH"_3"COO"^(-)]_(eq) ~~ ["CH"_3"COOH"]_i#

And thus...

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

and so,

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.