Titration problem answer 1

1.837 g of impure potassium hydrogencarbonate was dissolved in water and made up to 100 cm³ in a volumetric flask. 10.0 cm³ of this solution was then titrated with 0.100 M hydrochloric acid using methyl orange indicator. The titres were 17.40, 17.10 and 17.20 cm³. What is the percentage purity of the original sample? Evaluate the accuracy and reliability of this experiment.

The equation for the reaction is:

KHCO₃(aq) + HCl(aq) ⇒ KCl(aq) + H₂O(l) + CO₂(g)

This calculation has a different format to the example shown, and consequently, some students find this very difficult. Always get started (the beginning is the same) and see where things lead to.

Average titre (ignoring anomalous 17.40) = (17.10 + 17.20)/2 = 17.15 cm³

\number of moles of HCl = 0.100 × 17.15/1000

From the equation KHCO₃ º HCl

\number of moles of KHCO₃ in 10.0 cm³ = 0.100 × 17.15/1000

This is where we have to head in a different direction. We could calculate the concentration of the potassium hydrogencarbonate as we did before, but we haven't been asked for this. We need to know the percentage purity of the original sample. That is how much potassium hydrogencarbonate is in the original 1.837 g. We know how many moles are present in 10.0 cm³ of solution, but the original sample was dissolved in 100 cm³. There must be more in 100 cm³ than there was in 10.0 cm³, we have to scale up:

\number of moles of KHCO₃ in 100 cm³ = 0.100 × 17.15/1000 × 100/10

KHCO₃ = 100

\ mass of KHCO₃ in 100 cm³ = 0.100 × 17.15/1000 × 100/10 × 100 = 1.715 g

This looks promising as it's a bit less than the original total mass.

Percentage purity = 1.715/1.837 × 100 = 93%

Error evaluation

We'll start with all the measurement errors. The tolerances of all the items of equipment can be found in the error analysis section.

Weighing error = 1.837 ± 0.001 g = ± 0.05%

Volumetric flask error = 100 ± 0.16 cm³ = ± 0.16%

Pipette error = 10 ± 0.04 cm³ = ± 0.4%

Burette error = 17.15 ± 0.15 cm³ = ± 0.87%

The burette error is made up of a ± 0.1 cm³ tolerance for a class B burette and a ± 0.05 cm³ error for the reading of the end-point.

This gives a total measurement error of just under 1.5%. You should have noticed that I gave the final answer to the nearest %. We can now see that we were not justified in putting in any more significant figures. The burette reading is easily the biggest error here. We could reduce it by using a class A burette (but these are expensive!). The cheaper way to half the error would have been to use twice as much sample so that the titre was twice the size.

There are other non-measurement errors. A popular one with the exam boards is lack of homogeneity (or in simpler words lack of even mixing) or even lack of complete solution. I don't like this one, because if you have made your solution up properly you will have checked to see that it is fully dissolved and should have mixed the solution thoroughly by several inversions of the volumetric flask. If this is not done the solution will be more concentrated near the bottom of the flask, and less concentrated at the top where the water was added to make up to the mark. To solve this problem always ensure thorough mixing. If time permits, the solution could be left stoppered overnight. In this time any very small undissolved particles should dissolve.

The biggest problem may be the technique itself. We are assuming that the acid only reacts with the potassium hydrogencarbonate and not with any impurities. Quite a likely impurity would be sodium hydrogencarbonate. This would also react with the acid, giving a larger titre and suggesting the sample is more pure than it actually is.