The Problem

Calculate the pH of a buffer that is $\pu{0.200M}$ $\ce{H3BO3}$ and $\pu{0.122M}$ $\ce{KH2BO3}$. The $K_\mathrm{a}$ for $\ce{H3BO3}$ is $7.3\times10^{-10}$

What I Tried

For each of these methods, I used $\pu{0.200M}$ as the concentration for the acid, $\ce{H3BO3}$, and $\pu{0.122M}$ as the concentration of its conjugate base, $\ce{H2BO3-}$.

  1. Method one – Henderson–Hasselbalch equation

    $\mathrm{p}K_\mathrm{a} = -\log(K_\mathrm{a}) = 9.13668$

    ${\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \left(\frac{0.122}{0.200}\right) = 9.13668 + (-0.21467) = 8.92201}$

  2. Method two – Rice Table $$ \begin{array}{c|lcr} & \text{$\ce{H3BO3}$} & \text{$\ce{H3O+}$} & \text{$\ce{H2BO3-}$} \\ \hline \text{I} & 0.200 & 0 & 0.122 \\ \text{C} & -x & +x & +x \\ \text{E} & 0.200-x & x & 0.122+x \end{array} $$ use x is small approximation $K_\mathrm{a} = \ce{\frac{[\ce{H3O}][\ce{A-}]}{[\ce{HA}]}}$
    $\ce{K_\mathrm{a} = \frac{0.122x}{0.200}}$
    $x = 0.200 \times \frac{K_\mathrm{a}}{0.122}$
    $x = 1.19672*10^{-9}$ – This equals the number of moles $\ce{H3O+}$ $\mathrm{pH} = -\log([\ce{H3O+}]) = 8.92201$

Check $x$ is small approximation: $\frac{x}{0.122} = 9.081\times10^{-9}$ which is less than $5\ \%$, so the approximation is valid.

My question

The answer I got using both of these methods, $\mathrm{pH} = 8.92$, is not correct. The only conclusion I can come to is that I have missed some nuance of this question or I am making a careless mistake. Can somebody please tell me what I'm overlooking here?

  • $\begingroup$ How do you know that the answer of 8.92 is wrong? What is the answer supposed to be? $\endgroup$ Commented Mar 29, 2016 at 2:13
  • $\begingroup$ @pH13 boric acid does not give a proton but rather accepts an $\ce{OH-}$ to form $\ce{[B(OH)4]}$. So isn't it $\ce{K[B(OH)4]}$ instead of $\ce{KH2BO3}$? $\endgroup$
    – Aditya Dev
    Commented Mar 29, 2016 at 2:45
  • $\begingroup$ Boric acid is quite complex, so I don't really start without knowing where to go. $\endgroup$ Commented Mar 29, 2016 at 2:47
  • $\begingroup$ I don't know what the answer is supposed to be; it is a multiple choice question on Mastering Chemistry, and when I picked 8.92 it said I was wrong. I didn't list the other choices because I didn't want anyone to think I was trying to get someone to answer the problem. Really I'm just looking for some insight as to what I could be missing. $\endgroup$
    – Jenny Ann
    Commented Mar 29, 2016 at 3:11
  • 3
    $\begingroup$ @JennyAnn There's nothing wrong with what you did, you just need more confidence in knowing that you are right. Check your work in a problem like this intuitively saying to yourself, "there is a little more acid than conjugate base, so the pH will be a little below the pK". $\endgroup$
    – DavePhD
    Commented Mar 29, 2016 at 14:59

2 Answers 2


Update: Per the advice of @DavePhD , I decided to trust my work and I approached my teacher with the problem. He just emailed me to say there must have been some error in the online system; 8.92 is the correct answer.

Thanks so much to everyone who took the time to help me with this!

  • 1
    $\begingroup$ Thanks for letting us know. My perception was that when you really beat on a problem like this you learn something. // If the problem was supposed to be analyzed at a more complex level then you would have needed a whole lot more equilibrium constants. $\endgroup$
    – MaxW
    Commented Mar 30, 2016 at 3:31
  • 1
    $\begingroup$ Hi Jenny Ann. I agree with MaxW about how much you learn from really going after a problem! It throws all of us into a tizzy when there is an error in a test question, which is what I suspected from the beginning. @DavePhD was right when he suggested that you should trust your work. Still, the chemistry adventure we embarked on was cool. I see the possibility of a chem major in your future ... $\endgroup$
    – user28106
    Commented Mar 30, 2016 at 13:35

Depending on the level of rigor required, the solution to this problem could range from honors high school level chemistry to upper level graduate school inorganic chemistry. The complexity of the chemistry of aqueous boric acid is well described in this reference:


The author cites one source in which it is stated that there are 10 different equilibrium systems in boric acid solutions!

If you are in honors or AP high school chemistry or even in freshman inorganic chemistry, I suspect that you would be expected to solve the problem exactly as you did. Perhaps the test writers wanted an interesting weak acid and conjugate base for the problem and either did not worry about the true complexity of the boric acid chemistry or were not aware of it. In addition, the author(s) of the questions may not be the ones who provided the solutions to the problems. Now here are the twists.

There is some debate about whether or not boric acid behaves as a triprotic acid with three successive hydrogen ion transfers to finally produce BO3- + 3 H3O+. (https://en.wikipedia.org/wiki/Boric_acid#Properties) In accordance with the triprotic model three separate university websites (North Carolina State University, University of California Santa Barbara, and the University of Washington) cited the Ka1 of boric acid to be 5.8 x 10-10.

The competing boric acid dissociation model is well described in the crscientific source above and, in summary, begins with B(OH)3 (another way to write boric acid) acting as a Lewis acid:

    B(OH)3 (aq)  +  H2O  ⇄   B(OH)4- (aq) +  H+ (aq) 

Further reactions involving B(OH)4- (aq) introduce species such as H2B4O7, HB4O7- and B4O72-. What is very curious is that the K of the reaction of B(OH)3 (aq) and H2O is said to equal 7.3 x 10-10 in a Wikipedia article (https://en.wikipedia.org/wiki/Boric_acid#Properties). This is the value of Ka1 you were given in the problem, but the problem is based on the triprotic boric acid model! I encourage you to rework the problem with a value of 5.8 x 10-10 for the Ka1 just to see if the answer for this calculation equals one of the other choices in the problem.

  • $\begingroup$ LOL - I got hooked on this. Using the Henderson-Hasselback Equation as @JennyAnn did you get $${pKa = -log(K_a) = 9.23657}$$ and $${pH = pK_a + log (\frac{0.122}{0.200}) = 9.23657 + (-0.21467) = 9.02190}$$ I don't see how that helps... $\endgroup$
    – MaxW
    Commented Mar 29, 2016 at 23:32
  • $\begingroup$ @Steve Lantz, I did rework it with that value for the Ka and got the same answer as MaxW. I am in taking a general chemistry course in college, so I think the method is correct. The system must be wrong. Completely aside from the answer to the problem, your post was very interesting. I am taking an intro level general chemistry course, and I had no idea that there were still debates about the way molecules behave; I'd just assumed we knew everything already. Thank you for taking the time to answer my question $\endgroup$
    – Jenny Ann
    Commented Mar 30, 2016 at 2:58

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