Mean Titre Calculation from Titration Data
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(b) Ethanoic acid, $CH_3COOH$, is found in some descalers to soften hard water. A student carries out a titration with a standard solution of sodium carbonate, $Na_2CO_3$, to determine the percentage composition by mass of $CH_3COOH$ in a descaler. The equation is shown below. $$2CH_3COOH + Na_2CO_3 \rightarrow 2CH_3COONa + CO_2 + H_2O$$ (i) The method is outlined below:
• Dissolve 6.50g of the descaler in distilled water.
• Transfer the solution into a 250.0 $cm^3$ volumetric flask.
• Make up to the mark with distilled water and invert several times.
• Pipette 25.0 $cm^3$ of this solution into a conical flask and add a few drops of indicator.
• Titrate this solution with 0.200 $mol dm^{-3}$ $Na_2CO_3(aq)$ in the burette. The student carries out a trial titration, followed by further titrations. The results are shown in the table below. The trial titration has been omitted. [A table is provided with columns for Titration 1, 2, 3 and rows for Final reading, Initial reading, and Titre]. Complete the table by adding the titres. [1] (ii) Calculate the mean titre, to the nearest 0.05 $cm^3$, that the student should use for analysing these results.
This question includes visual content: The image includes a chemistry table listing three titration attempts with columns for 'Final reading / cm^3', 'Initial reading / cm^3', and 'Titre / cm^3'. The table is filled with experimental data values. Below the table, there is a hand-written calculation showing the average of two titres: (24.15 + 24.25) / 2 = 24.2.
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Hi Lilia, let's look at how to find the mean titre from this titration data.
Titration Analysis
First, we need to complete the table by calculating the titre for each run. We do this by subtracting the initial reading from the final reading.
For the first titration, forty-eight point nine five minus twenty-four point five five gives us twenty-four point four zero cubic centimeters.
For the second titration, twenty-four point one five minus zero equals twenty-four point one five cubic centimeters.
And for the third titration, forty-eight point three five minus twenty-four point one zero gives twenty-four point two five cubic centimeters.
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