0
$\begingroup$

Context

The Henderson-Hasselbalch equation is as follows. $$pH=-\log\big(K_a\big)+\log\bigg(\frac{[A^-]}{[HA]}\bigg)$$ One may follow its derivation in order to understand how it came to be, yet there are still facets of it that the proof fails to explain. For instance, the equation provides mathematical evidence for why the pH at the midpoint of a monoprotic acid-base titration is equal to the $\ce{pK_a}$. Since $[A^-]=[HA]$ at the midpoint, the fraction reduces to 1. Because $\log(1)=0$, $pH=-\log\big(K_a\big)+0=-\log\big(K_a\big)$, which, of course, gives the pH at the midpoint.

My question is about the same sort of 'insight.'

My Question

Given A mL of B M HA with $\ce{K_a}=C$ titrated by D M xOH at 25oC, where HA is a monoprotic weak acid and xOH is a strong base akin to NaOH, the following expression equates to the pH at the equilibrium point (try it out on Example #1-e here!).

$$pH=14+\log\Bigg(\sqrt{\frac{B*D*10^{-14}}{C(D+B)}}\Bigg)$$

While deriving this, it became necessary for me to determine the $[A^-]$ (it is the major pH determining species at the EQPt). I reasoned as follows.

To find the $[A^-]$, find the mmol of conjugate base and divide it by the total volume.

Since the mmol of conjugate base is equal to the mmol of weak acid used, and the latter quantity is given by the initial volume of acid times its initial molarity ($A*B$), the former quantity is equal to $A*B$.

The total volume, on the other hand, is equal to the initial volume plus what is needed to reach the equivalence point. The initial volume is given by A, so the expression will be $A+V_2$, where $\ce{V2}$ is what’s needed to reach the equivalence point. Take $\ce{M1V1}=\ce{M2V2}$ and plug in the given variables to get $B*A=D*V_2$. Solve for $\ce{V2}$ to get $V_2=\frac{B}{D}*A$. Substitute the value of $\ce{V2}$ into $A+V_2$ to find the total volume, or $A+\frac{B}{D}*A$.

As mentioned before, divide the two expressions to give the final equation, listed below.

$$[A^-]=\frac{A*B}{A+\frac{B}{D}*A}$$

However, the expression simplifies as follows.

\begin{split} [A^-]&=\frac{A*B}{A+\frac{B}{D}*A}\\ &=\frac{A(B)}{A(1+\frac{B}{D})}\\ &=\frac{B}{1+\frac{B}{D}}\\ &=\frac{B}{\frac{D}{D}+\frac{B}{D}}\\ &=\frac{B}{\frac{D+B}{D}}\\ &=\frac{B*D}{D+B}\\ &=\frac{B*D}{B+D}\\ \end{split}

I find it interesting that such a complicated concept (at least to me) simplifies down to the product of two molarities over their sum. Similar to the way that I justified my initial expression and a facet of the Henderson-Hasselbalch equation, can anyone straight-up explain why this final expression yields the EQPt concentration of conjugate base in the given scenario? If it is possible to keep the answer to a relatively basic, non-organic point of view, that would be much appreciated!

Please note that I have done the best to explain what I am asking -- I think it is a legal but unordinary question for this site. If I can clarify, do not hesitate to comment!

$\endgroup$
  • $\begingroup$ Your insight is a Gordian knot. The Henderson-Hasselbalch equation is thrown around on this site like it is a magic Swiss army knife. Do you understand the implications of using it with a "weak" acid? $\endgroup$ – MaxW Jul 17 '18 at 1:10
  • $\begingroup$ @MaxW Probably not, but I’d like to try! Any further clarification? $\endgroup$ – Shady Puck Jul 17 '18 at 3:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.