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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?

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  • $\begingroup$ @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. $\endgroup$ – Martin - マーチン Jan 27 '15 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 '15 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. $\endgroup$ – Martin - マーチン Jan 27 '15 at 3:39
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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.

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  • $\begingroup$ 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 '15 at 1:59
  • $\begingroup$ Okay so I actually figured it out, see my comment to my question $\endgroup$ – user109992 Jan 27 '15 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. $\endgroup$ – Martin - マーチン Jan 27 '15 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 '15 at 3:48

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