You perform a titration of an acid (let's say $\pu{20 ml}$ $\ce{CH_3COOH}$) with a base (let's say $\pu{0.5 M}$ $\ce{NaOH}$). The burette is marked at $\pu{0.10 ml}$ intervals. From what I read in a previous thread, the largest source of error in acid-base titrations comes from endpoint determination (i.e. adding one drop too much or little). However, I want to investigate if this is the case and compare this error to the indicator error. Below are my reasoning that I need help to correct and fill in.

One drop has approximately a volume of $\pu{0.05 ml}$. In the example titration given above, adding one extra drop would lead to overestimating the amount of substance of $\ce{CH_3COOH}$ with $\pu{0.05ml} \cdot \pu{0.5 M} = \pu{0.05E-3} \cdot \pu{0.5 mol} = \pu{2.5E-5 mol}$. This would increase the concentration with $\frac{\pu{2.5E-5 mol}}{\pu{0.020 l}} = \pu{0.00125 M}$. This error appears very small.

For the phenolphthalein indicator, we only know the effective pH range ($8.3$ - $10.0$ according to this source) with two significant figures. Furthermore, the pH value of the solution of $\ce{CH_3COOH}$ and $\ce{NaOH}$ (which is obtained after the titration) depends on the concentration of $\ce{CH_3COOH}$. Let's say the $c_{\ce{CH_3COOH}} = \pu{0.1 M}$. When this reaction

$$\ce{CH3COOH + OH^- ⇌ CH3COO^- + H2O}\tag1\label{eq:one}$$

is in equilibrium, the concentration of $\ce{CH_3COOH}$ and $\ce{OH^-}$ will have decreased with $x$:

$$x/(0.1-x)^2 = 1.7\times10^9$$

The pH of this solution will thus be $8.9$ (two significant figures). If instead $c_{\ce{CH_3COOH}} = \pu{0.2 M}$ then the pH of the solution at the equivalence point would have been $9.0$ (two significant figures). This already presents a difficulty since we don't know $c_{\ce{CH_3COOH}}$ before the titration. Combined with the broad effective range of the phenolphthalein indicator (which starts to turn pink already at pH $8.3$), the indicator error appears larger to me.

Let's investigate this claim. Assume that $c_{\ce{CH_3COOH}} = \pu{0.1 M}$. At the equivalence point, the pH will be $8.9$. Assume that you saw the faintest colour of pink when the pH was $8.3$ and you stopped titrating. In this solution, we have $c_{\ce{OH^-}} = 10^{-(14-8.3)}$. The equilibrium expression for reaction \eqref{eq:one} is:

$$\frac{x}{((0.1-x) \cdot c_{\ce{OH^-}})} = 1.7 \times 10^9,$$

where $x$ is the decrease in the concentration of $\ce{CH_3COOH}$ and $\ce{OH^-}$ compared to the initial concentrations. The initial concentration of $\ce{OH^-}$ is $10^{-(14-8.3)} + 0.09997\ldots = \pu{0.09997\dots M}$. Thus one would determine $c_{\ce{CH_3COOH}} = \pu{0.09997\dots M}$ instead of $c_{\ce{CH_3COOH}} = \pu{0.1 M}$, which is an error of $\pu{0.000027 M}$. Apparently, based on this example, the indicator error is less than the error of adding one drop too much. If this conclusion generalizable?

So if the endpoint determination error is the biggest source of error, the error in the above titration should be $\pu{0.00125 M}$. It seems like the error in titration should be larger.

  • $\begingroup$ Well, assume you are the error. Are you sure you can read the burette the same way every time? How accurate is the burette anyway (without reading errors)? What about the volume of the solution you are going to determine? $\endgroup$ Commented Feb 28, 2023 at 20:56
  • $\begingroup$ A "standard drop" is usually considered as 0.03 mL, with a pointy plastic(=hydrophobic) burette ending even 0.02 mL. // It seems you have not considered the 2nd indicator error, i.e. titration of the indicator itself ( which is assumed very small too, but worthy to estimate it.). $\endgroup$
    – Poutnik
    Commented Feb 28, 2023 at 21:15
  • $\begingroup$ @Martin yes, I was also thinking that after these calculations. So then errors due to endpoint determination and indicator error are smaller compared to the error sources you listed? $\endgroup$
    – UserE
    Commented Mar 1, 2023 at 7:08
  • $\begingroup$ @Poutnik ok that will then lower the error of adding one drop too much if the volume is lower than 0.05 ml. What is the 2nd indicator error? I tried googling it but found nothing useful. Could you show me how to calculate that? $\endgroup$
    – UserE
    Commented Mar 1, 2023 at 7:10
  • 2
    $\begingroup$ The indicator is an additional acid or base, added to the titrated system, that changes its color by being neutralized to the respective conjugate base or acid. Typical amount is 1-2 drops of 0.1 % indicator solution(water, ethanol). The exact degree of neutralization is difficult to quantify due subjective nature of color shift detection. But the rough estimate is possible. // Quick calculation for 1 drop of 0.1 % Phenolphthalein gives it is equivalent to roughly 0.001 mL of 0.1 M NaOH, i.e. 1/30 of the standard 0.03 mL drop. for the first color appearance, perhaps 1/10 of that ). $\endgroup$
    – Poutnik
    Commented Mar 1, 2023 at 9:16


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