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I hope this is the right place to ask. We have this exercise without solution

One litre of diluted sodium hydroxide solution (c = 0.15mol/litre) is to be prepared in the laboratory. The titer from the finished NaOH solution is then determined. For this purpose, 0.62 g potassium hydrogen phthalate (M = 204.2 g/mol, $pK_a = 5.4$) are dissolved in 0.05 litre of water and titrated with the prepared sodium hydroxide solution until turnover. The consumption is 0.022 litre. Calculate the titer of the prepared sodium hydroxide solution.

On Wikipedia, I found "In titration, the titer is the ratio of actual to nominal concentration of a titrant" https://en.wikipedia.org/wiki/Titer

So if I understand correctly, the titer is just a fraction of the form $ t = \frac {c_{real}} {c_{ideal}} $, so a fraction of the real concentration over the ideal concentration, is this correct ?

In our case, we would simply need to use the titration equation at equivalence point: $$V_1 \cdot M_1 = V_2 \cdot M_2$$ where $V$ is the volume and $M$ the molarity

So we would solve

$$M_2 = \frac{V_1 \cdot M_1}{V_2} $$

and then the titer is $$t = \frac{c_{real}}{c_{ideal}} = \frac{M_2}{0.15} $$ because 0.15 mol/litre is given as ideal concentration in the exercise

$V_1$ and $V_2$ are given as 0.05 litre and 0.022 litre, and $M_1$ could be found by dividing 0.62 g by the molar mass (M = 204.2 g/mol), and then divide again by the volume of 0.05 litre


I am really unsure because there aren't a lot of results when searching on the Internet, so it seems to me that this isn't a frequently searched topic. Is my understanding above correct ? Why are there relatively few search results for the titer in Chemistry ? Even on this site, I did not find a similar question. Is this topic considered "not important" or am I missing something ?

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2 Answers 2

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Your approach is correct, although there is no information about which solution is considered as ideal, and which is real. It is also a pity you are not able to be more precise than $0.022$ liter as a final result. Two significant figures ($2$ and $2$) are not really reliable. Usually three or four significant figures are easy to obtain with ordinary burettes. The real volume should look like $22.02$ mL and not $22$ mL. Anyway the final titer should be equal to $0.935$.

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  • $\begingroup$ Thanks a lot for your feedback ! Now that you say it, it's in fact strange to have two significant digits, this exercise comes actually from an old mock exam they gave us. I also wonder why there are few results for what a titer is, maybe I used the wrong terms for searching. And yes in fact now that you say it, there is no indication about which solution is real and which is ideal, I wonder why $\endgroup$
    – wengen
    Dec 28, 2022 at 22:45
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    $\begingroup$ @wengen. Usually a first solution is made to be "about 0.1 M" by dissolving an approximate amount of solute. Later on, a sample of this first solution is introduced in a conical flask and carefully titrated with a reference solution coming out of a burette. The volume so obtained can be known to ± 1 drop, so ±0.03 or ±0.05 mL. This titration allows you to calculate the concentration of the first solution with 3 - 4 significant figures. This concentration can be written down on the label (like a title = "titre" in French) on the bottle of the first solution. Titration comes from a French word. $\endgroup$
    – Maurice
    Jan 2, 2023 at 10:35
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Wikipedia is not the ultimate resource for classical volumetric titrations. This is why textbooks still exist. The definition of titer given in Wikipedia has nothing to do with the classical meaning of titer. There is no ideal or real solution in volumetric titrations. Titer in volumetric analysis means an expression of type:

1 mL (Given NaOH conc. X) = Y (weight) of analyte.

In your case, it will be,

1 mL (Given NaOH conc. X) = mg of KHP.

It is a quick way of analyzing solutions. Once an analytical chemist determines the titer of a titrant, all he/she need to know is the burette volume, and we directly get the weight of the analyte in one step. I will give you hints:

Write balanced equations of titration reaction. Calculate the actual molarity. This will give you X. From X, calculate the number moles of NaOH in 1 mL of that X NaOH. From the moles determine the moles of your target analyte based on the equation stoichiometry.

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  • $\begingroup$ Interesting. That's almost how I did it in my lab courses, but never how I learned it. Do you have a source for this? $\endgroup$ Dec 30, 2022 at 0:43
  • $\begingroup$ @Martin-マーチン, It would be good to start a new post on this topic with an example. I would be curious to learn the approach that was taught to you. I do not have a web reference but I think this was discussed in Vogel Quant. Analysis (1960s edition) if I recall correctly. $\endgroup$
    – AChem
    Dec 30, 2022 at 1:08

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