You prepare $1.0~\mathrm{L}$ of a $0.25~\mathrm{M}$ acetic acid solution with a final $\ce{pH}$ of $6.0$. What are the molar concentrations of all relevant acetic acid species ($[\ce{HA}]$ and $[\ce{A-}]$) given that the $K_a$ for acetic acid is $1.74 \cdot 10^{-5}~\mathrm{M}$?

I am getting confused with this problem.

Since the $\ce{pH}$ is given, I know what the $[\ce{H^+}]$ is. So now when I try to do the ICE table

$$ \begin{array}{l|ccc} & \ce{HA} & \ce{H+} & \ce{A-} \\\hline \text{Initial} & 0.25 & 0 & 0 \\ \text{Change} & -x & +x & +x \\ \text{End} & 0.25 -x & +x & +x \\\hline \end{array}$$

And from here I begin to assume $[\ce{H^+}] = [\ce{A^-}]$, which I am not sure about. Then I set $x = 10^{-6.0} = 1\cdot 10^{-6}$ so I get $[\ce{A^-}] = 1\cdot 10^{-6}~\mathrm{M}$ and $[\ce{HA}] = 0.24999~\mathrm{M}$ which I think is incorrect, and to even further ensure my that it's incorrect, when I attempt to check the $K_a$ value with this, it does not match.

My second approach: $[\ce{HA}] = 0.25~\mathrm{M}$

In this I determine the $\mathrm{p}K_a$ from the $K_a$ which turns out to be $4.759$, which indicates that there should be more $[\ce{A-}]$ than $[\ce{HA}]$.

I now use the Henderson–Hasselbalch equation: \begin{align} 6.0 &= 4.759 + \log\left(\frac{[\ce{A-}]}{[\ce{HA}]}\right)\\ 17.40 &= \frac{[\ce{A-}]}{[\ce{HA}]}\\ 17.40 &= \frac{[\ce{A^-}]}{0.25~\mathrm{M}}\\ [\ce{A-}] &= 4.35~\mathrm{M}\\ \end{align}

I feel more confident about my second answer.

Can someone please help me out with this particular problem and perhaps tell me procedure I should use as well as what the correct answer should come out to be and why?

  • $\begingroup$ @FredSenese There is a command for chemical formula (using mhchem): \ce{...}. Find more information here and here. It would also be very nice if you could be on the lookout for buzz-words/-statements especially in titles, see also meta. $\endgroup$ Jan 27, 2015 at 3:25
  • $\begingroup$ Okay so using Henderson Hasselbalch:One will get 17.40 = [A-]/[HA]. Following this idea, one can assume that [HA] + [A-] = 0.25M, using algebra, [HA] = 0.25 - X. Now 17.40 = x/0.25-x, again with algebra one can solve that x = 0.236. So [A-] = 0.236M and [HA]=0.0136M and if you use these to find the Ka, you match the Ka if you use [H+]=10^-6.0 $\endgroup$
    – user109992
    Jan 27, 2015 at 3:27
  • $\begingroup$ @user109992 Could you write that up as an answer, please. This way you can accept it and it can be upvoted. $\endgroup$ Jan 27, 2015 at 3:39

2 Answers 2


The value given for the pH of the solution seems to be incorrect*.

As the reaction is $\ce{HA <=> H+ + A-}$, your initial assumption that $[\ce{H+}]=[\ce{A-}]$ is correct.

From your chart,

\begin{array}{l|ccc} & \ce{HA} & \ce{H+} & \ce{A-} \\\hline \text{Initial} & 0.25 & 0 & 0 \\ \text{Change} & -x & +x & +x \\ \text{End} & 0.25 -x & +x & +x \\\hline \end{array}

it follows that $$K_a=\frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]} = \frac{x^2}{0.25 -x} = 1.74\cdot10^{-5}.$$

We can solve this for $x$: \begin{align} x^2 &= 1.74\cdot 10^{-5}(0.25-x)\\ x^2 + 1.74\cdot10^{-5} x - 4.35\cdot10^{-6} &= 0\\ x &= 0.00207698~\mathrm{M}\\ \end{align}

Thus, $[\ce{HA}] = 0.25~\mathrm{M} - 0.00207698~\mathrm{M} = 0.24792302~\mathrm{M}$ and $[\ce{H+}]=[\ce{A-}] = 0.00207698~\mathrm{M}$.

Also note that in your second attempt, the value $[\ce{A-}] = 4.35~\mathrm{M}$ makes no sense, as the amount of $\ce{A-}$ present should be in a 1:1 ratio to the amount of $\ce{HA}$ consumed in the reaction, and $[\ce{HA}] < 4.35~\mathrm{M}$.

The given $\ce{pH}$ is likely incorrect because the concentration of $[\ce{H+}]$ ions for the given molarity and concentration does not match with the given $\ce{pH}$, and because the given $K_a$ is close to values of $K_a$ provided online.

  • $\begingroup$ I don't think there's an error in the question, I think that my first attempt might be wrong in the assumption that H+ = A- in this case. $\endgroup$
    – user109992
    Jan 27, 2015 at 1:59
  • $\begingroup$ Okay so I actually figured it out, see my comment to my question $\endgroup$
    – user109992
    Jan 27, 2015 at 3:25
  • $\begingroup$ I think your answer would be correct for a pure solution of acetic acid. I would read the question different. After the pure solution is obtained, the pH is adjusted to 6, thus making it a buffer. This part is however horribly implied. $\endgroup$ Jan 27, 2015 at 3:37
  • $\begingroup$ @user13006 I think your answer is great. Martin could be right about what the person had in mind writing the question too. $\endgroup$
    – DavePhD
    Jan 27, 2015 at 3:48

Ratio of protonated to deprotonated species

We know the pH and the acid dissociation constant. Using the equilibrium constant expression, we can calculate the concentration ratio of acetate to acetic acid:

$$\frac{[\ce{A-}]}{[\ce{AH}]} = \frac{K}{[\ce{H+}]} = \frac{\pu{1.74e−5}}{10^{-6}} = 17.4 $$

Ambiguity in the question

To complete the exercise, we have to understand the question. They might be saying that $\pu{0.25 M}$ is the concentration of acetic acid, or this value refers to the total concentration of acetic acid plus acetate. The latter is more likely, as they talk about "the molar concentrations of all relevant acetic acid species". This implies that acetate is viewed as a acetic acid species, and probably was not added to the acetic acid but was made in situ when setting the pH of the solution. In the lab, we would maybe label this solution "acetic acid buffer, 0.25 M, pH = 6" or more clearly "0.25 M acetic acid, adjusted to pH = 6 with NaOH".

Concentration of species

If we take the total concentration of acetic acid plus acetate to be $\pu{0.25 M}$, $\frac{1}{18.4}$ of it will be acetic acid, and $\frac{17.4}{18.4}$ of it will be acetate.

If we take the final concentration of acetic acid to be $\pu{0.25 M}$, the acetate concentration would be $17.4$ as big, i.e. more than four molar. There might not be any acetate salt that soluble in water (assuming water is the intended solvent).

Definitely not a watertight case.


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