Number of significant figures to use in reporting the concentration of the titrand in acid-base titration [closed]








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If you perform an acid-base titration yourself, how do you determine the number of significant figures to use when reporting on the concentration of the titrand?


Assume you perform seven titration trials and the titrand concentration varies between $\pu{0.495 mol l^-1}$ and $\pu{0.505 mol l^-1}.$ You use a $\pu{50 ml}$ burette, class A glassware, and you know the concentration of the titrant with many significant figures.


It is incorrect to say that the concentration of the titrand is $c = \pu{(0.500 ± 0.005) mol l^-1}$ since more trials would increase the interval of values and thus increase the uncertainty in the experiment, which is not true.




You said that you have done seven titrations, i.e., you have the set of concentrations $\{c\_1, c\_2, \ldots, c\_7\}$. I offer you the following standard statistical approach to inform your concentration and *know how many significant figures to use*:


1. Calculate the mean value of the concentration given the set of your concentration values:
$$\mu = \sum\_{k = 1}^N c\_k \tag{1}$$
2. Calculate the standard deviation $(N = 7$ in your case):
$$s = \sqrt{\frac{\displaystyle\sum\_{k = 1}^N(c\_k - \mu)^2}{N - 1}} \tag{2}$$
3. Inform your concentration given by the confidence interval
$$c = \mu \pm \Delta c = \mu \pm t\_{N - 1}^\alpha \frac{s}{\sqrt{N}}, \tag{3}$$
where $t\_{N - 1}^\alpha$ is the $t$-value of the Student’s probability distribution with a confidence interval of $\alpha$ with $N - 1$ degrees of freedom. In your case, this will correspond to $N - 1 = 6$, so looking up in the table you will have to choose within the corresponding row (data source: Harris’ *Quantitative Chemical Analysis*, Table 4-4. Values of Student’s $t$ [1, p. 72]):


$$
\begin{array}{rrrrrrrr}
\hline
N - 1 & 50~\% & 90~\% & 95~\% & 98~\% & 99~\% & 99.5~\% & 99.9~\% \\
\hline
 1 & 1.000 & 6.314 & 12.706 & 31.821 & 63.656 & 127.321 & 636.578 \\
 2 & 0.816 & 2.920 & 4.303 & 6.965 & 9.925 & 14.089 & 31.598 \\ 
 3 & 0.765 & 2.353 & 3.182 & 4.541 & 5.841 & 7.453 & 12.924 \\
 4 & 0.741 & 2.132 & 2.776 & 3.747 & 4.604 & 5.598 & 8.610 \\ 
 5 & 0.727 & 2.015 & 2.571 & 3.365 & 4.032 & 4.773 & 6.869 \\
\to 6 & 0.718 & 1.943 & 2.447 & 3.143 & 3.707 & 4.317 & 5.959 \\
 7 & 0.711 & 1.895 & 2.365 & 2.998 & 3.500 & 4.029 & 5.408 \\ 
 8 & 0.706 & 1.860 & 2.306 & 2.896 & 3.355 & 3.832 & 5.041 \\ 
 9 & 0.703 & 1.833 & 2.262 & 2.821 & 3.250 & 3.690 & 4.781 \\ 
10 & 0.700 & 1.812 & 2.228 & 2.764 & 3.169 & 3.581 & 4.587 \\ 
\hline 
\end{array}
$$


*Now we can answer your question*. Say, your data with some chosen $\alpha$ gave you $\mu = \pu{0.50145 M},$ and $\Delta c = \pu{0.00627 M}.$ Then, the *position of the first non-zero decimal number* in $\Delta c$ informs you how many you must use:


$$c = \pu{(0.501 \pm 0.006) M}\tag{4}$$


*Always* inform $\Delta c$ with only one number, never put something like


$$c = \pu{(0.50145 \pm 0.00627) M}\tag{5}$$


Some people like rounding up the absolute uncertainty $\Delta c$, and state that $\Delta c = \pu{0.01 M},$ and will inform


$$c = \pu{(0.50 \pm 0.01) M}\tag{6}$$


Rounding up the absolute uncertainty is just a matter of personal like, some do, some do not. Of course, this enlarges the confidence interval, now between $\pu{0.49 M}$ and $\pu{0.51 M}.$


If you want more details, you can check the lovely *Quantitative Chemical Analysis* by *Harris* in the sections on error or statistics.


### Reference


1.Harris, D. C. *Quantitative Chemical Analysis*, 9th ed.; W. H. Freeman: New York, **2015**. ISBN 978-1-4641-3538-5.


 4

There are two approaches to this.


1. Estimate the error from the estimated error of measurement (how good are you at reading a buret?).
2. Estimate the error from the variance of multiple measurements.


You can even do both and see if they match (do a chi-square test). If the variance is higher than what you expect from your error model, either the experimenter is having a bad day or the error model is unrealistic. If the variance is lower than what you expect from your error model, it might be beginner's luck or some kind of bias (titrating to the same volume while looking at the volume rather than looking at the color of the indicator).


Neither analysis tells you about systematic errors (like an incorrect assumption about the purity of the substance you make the titrant from, or using the glassware outside of the prescribed temperature range - there is glassware appropriate for the tropics, for example).



> 
> [OP] ... since more trials would increase the interval of values and thus increase the uncertainty in the experiment, which is not true.
> 
> 
> 


We distinguish between the standard deviation, which is roughly independent of sample size, and the standard error, which decreases with sample size. You would have to specify which one you are calculating or estimating.


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