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My research question is investigating how temperature affects the rate constant for the reaction between ethyl ethanoate (acetate) and sodium hydroxide. I understand that I must titrate sodium ethanoate with an acid e.g. $\ce{HCl}$, but that will just yield me the concentration of sodium ethanoate in the sample. How would I go about calculating the rate constant?

Possibilities:

  1. Would I use the integrated rate law, with a concentration against time graph?

  2. Would I use the rate law expression? For this one I am unsure what I would use as the "rate".

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  • $\begingroup$ @Poutnik Would there be an issue in titrating sodium ethanoate with hydrochloric acid, using bromocresol green as indicator (equivalence point about 5.2) $\endgroup$
    – aayush
    Oct 22, 2022 at 16:11
  • $\begingroup$ @Poutnik google tells me equivalence point for a reaction between weak base and strong acid is between 5 - 5.5. Is that wrong or am I misinterpreting the situation? $\endgroup$
    – aayush
    Oct 22, 2022 at 17:27
  • $\begingroup$ @Poutnik That makes sense, thanks. What would you recommend I do instead? I need to titrate the sodium acetate to find the concentration change. $\endgroup$
    – aayush
    Oct 22, 2022 at 18:07
  • $\begingroup$ @Poutnik Could you please rephrase your initial comment? I don't understand it. $\endgroup$
    – aayush
    Oct 23, 2022 at 4:28
  • $\begingroup$ There are unknown concentrations of NaOH and CH3COONa, equal in sum to initial NaOH concentration. Titration by HCl until discolouring phenolphthaleinum at pH 8.2 titrates NaOH, while all (>99.97%) CH3COONa still remains. Concentration of CH3COONa is then the difference between the initial and current NaOH concentrations. $\endgroup$
    – Poutnik
    Oct 23, 2022 at 6:17

1 Answer 1

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The general process of trying to solve this problem:

  1. Determine starting molarities of each reactant. Let's say 0.1 M
  2. Have a set of sample collection tubes (test tubes) pre-measured with 0.1 M of standardized HCl that would yield a slight excess of moles of NaOH in the starting mixture. This will be used to STOP the reaction (more on this). I would choose at least 5.
  3. Setup your reaction batch and mix the reactants with enough volume. Probably 100mL of each would suffice. As well as make sure the reactants and mixture maintain a temperature desired.
  4. During set intervals, withdrawal 10mL of the reaction mixture at periodic intervals and place them into one of the sample tubes. Placing it in the tube will immediately quench the NaOH in the reaction sample and convert any remaining acetate ions to acetic acid.
  5. After each sample tube has been used, you can titrate the remaining HCl + acetic acid with a standardized NaOH solution and an appropriate indicator such as bromthymol blue to obtain the amount of acid present.

You can now determine the amount of NaOH extracted from the reaction by using the number of moles of HCl present in the tube to start subtracted by the number of moles of acid that remain (step 5) to determine the amount of NaOH.

Since we now know the amount of NaOH extracted in a 10 mL sample of the reaction batch, you can determine the concentration of the NaOH at time of extraction within the reaction batch AND the time it took to reach this concentration.

Repeating this for each tube, we now have the change of the molarity of NaOH over time. This is also allows us to plot the change of the molarity of ethyl acetate since their reaction is a 1 mol : 1 mol ratio.

Using the integrated rate laws, a plot of these concentrations over time, you can determine the slope of the plot which creates a linear graph and thus find the reaction constant $(k)$.

Thanks to @Poutnik for the help on refining the answer.

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  • $\begingroup$ A very nice post. But it seems you have not mentioned acetic acid being titrated together with HCl. There will remain the constant amount of HCl, that would neutralize either NaOH either sodium acetate. And the variable part will be acetic acid on constant HCl baseline, titrated to first colour on phenolphthalein. With the equivalent amount of HCl Vs NaOH, it would be solely titration of acetic acid. $\endgroup$
    – Poutnik
    Oct 24, 2022 at 6:42
  • $\begingroup$ Without considering it,, one could be tempted to use e.g. methyl orange or other indicators for acidic pH ranges, what would be a big mistake. $\endgroup$
    – Poutnik
    Oct 24, 2022 at 11:00
  • $\begingroup$ @Poutnik Since the molar relationship between acetic acid and $\ce{HCl}$ are equal in the test tube, you could titrate using phenolphthalein to find the amount of both substances and halve it. But a better indicator probably could be chosen, either methyl violet or thymol blue. I can update the answer with the indicator option as that is probably the more appropriate experimental answer. $\endgroup$
    – Avogadro ♦
    Oct 24, 2022 at 19:45
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    $\begingroup$ 2 mmol NaOH + 3 mmol HCl -> 1 mmol HCl // 1 mmol NaOH + 1 mmol AcONa + 3 mmol HCl -> 1 mmol AcOH + 1 mmol HCl // With just 2 mmol HCl, just 0 or 1 mmol AcOH // What dividing by 2? $\endgroup$
    – Poutnik
    Oct 24, 2022 at 19:54
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    $\begingroup$ @Poutnik My apologies, it took me a little to understand what was being communicated, but I do get it now after sitting with an ICE chart. I updated the answer to mention that you will be back titrating both excess HCl and acetic acid. $\endgroup$
    – Avogadro ♦
    Oct 25, 2022 at 0:09

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