Question 19M.2.HL.TZ2.5
| Date | May 2019 | Marks available | [Maximum mark: 17] | Reference code | 19M.2.HL.TZ2.5 |
| Level | HL | Paper | 2 | Time zone | TZ2 |
| Command term | Calculate, Discuss, Distinguish, Identify, Predict, Sketch, State | Question number | 5 | Adapted from | N/A |
Carbonated water is produced when carbon dioxide is dissolved in water under pressure. The following equilibria are established.
Equilibrium (1) CO2 (g) 
CO2 (aq)
Equilibrium (2) CO2 (aq) + H2O (l) 
H+ (aq) + HCO3− (aq)
Carbon dioxide acts as a weak acid.
Soda water has sodium hydrogencarbonate, NaHCO3, dissolved in the carbonated water.
Distinguish between a weak and strong acid.
Weak acid:
Strong acid:
[1]
Weak acid: partially dissociated/ionized «in aqueous solution/water»
AND
Strong acid: «assumed to be almost» completely/100 % dissociated/ionized «in aqueous solution/water» [✔]
As expected, many candidates were able to distinguish between strong and weak acids; some candidates referred to “dissolve” rather than dissociate.
The hydrogencarbonate ion, produced in Equilibrium (2), can also act as an acid.
State the formula of its conjugate base.
[1]
CO32- [✔]
More than half the candidates were able to deduce that carbonate was the conjugate base but a significant proportion of those that did, wrote the carbonate ion with an incorrect charge.
When a bottle of carbonated water is opened, these equilibria are disturbed.
State, giving a reason, how a decrease in pressure affects the position of Equilibrium (1).
[1]
shifts to left/reactants AND to increase amount/number of moles/molecules of gas/CO2 (g) [✔]
Note: Accept “shifts to left/reactants AND to increase pressure”.
Many students gave generic responses referring to a correct shift without conveying the idea of compensation or restoration of pressure or moles of gas. This generic reply reflects the difficulty in applying a theoretical concept to the practical situation described in the question.
At 298 K the concentration of aqueous carbon dioxide in carbonated water is 0.200 mol dm−3 and the pKa for Equilibrium (2) is 6.36.
Calculate the pH of carbonated water.
[3]
«Ka =» 10–6.36/4.37 × 10–7 =
OR
«Ka =» 10–6.36/4.37 × 10–7 = [✔]
[H+] « » = 2.95 × 10–4 «mol dm–3» [✔]
«pH =» 3.53 [✔]
Note: Award [3] for correct final answer.
Most candidates calculated the pH of the aqueous CO2. Some candidates attempted to use the Henderson-Hasselback equation and others used the quadratic expression to calculate [H+] (these two options were very common in the Spanish scripts) getting incorrect solutions. These answers usually ended in pH of approx. 1 which candidates should realize cannot be correct for soda water.
Identify the type of bonding in sodium hydrogencarbonate.
Between sodium and hydrogencarbonate:
Between hydrogen and oxygen in hydrogencarbonate:
[2]
Between sodium and hydrogencarbonate:
ionic [✔]
Between hydrogen and oxygen in hydrogencarbonate:
«polar» covalent [✔]
This was an easy question, especially the identification of the type of bond between H and O, yet some candidates interpreted that the question referred to intermolecular bonding.
Predict, referring to Equilibrium (2), how the added sodium hydrogencarbonate affects the pH.(Assume pressure and temperature remain constant.)
[2]
«additional HCO3-» shifts position of equilibrium to left [✔]
pH increases [✔]
Note: Do not award M2 without any justification in terms of equilibrium shift in M1.
A significant number of candidates omitted the “equilibrium” involved in the dissolution of a weak base.
100.0cm3 of soda water contains 3.0 × 10−2g NaHCO3.
Calculate the concentration of NaHCO3 in mol dm−3.
[2]
«molar mass of NaHCO3 =» 84.01 «g mol-1» [✔]
«concentration = =» 3.6 × 10–3 «mol dm-3» [✔]
Note: Award [2] for correct final answer.
This is another stoichiometry question that most candidates were able to solve well, with occasional errors when calculating Mr of hydrogen carbonate.
The uncertainty of the 100.0cm3 volumetric flask used to make the solution was ±0.6cm3.
Calculate the maximum percentage uncertainty in the mass of NaHCO3 so that the concentration of the solution is correct to ±1.0 %.
[1]
«1.0 – 0.6 = ± » 0.4 «%» [✔]
Mixed responses, more attention should be given to this simple calculation which is straightforward and should be easy as required for IA reports.
The reaction of the hydroxide ion with carbon dioxide and with the hydrogencarbonate ion can be represented by Equations 3 and 4.
Equation (3) OH− (aq) + CO2 (g) → HCO3− (aq)
Equation (4) OH− (aq) + HCO3− (aq) → H2O (l) + CO32− (aq)
Discuss how these equations show the difference between a Lewis base and a Brønsted–Lowry base.
Equation (3):
Equation (4):
[2]
Equation (3):
OH- donates an electron pair AND acts as a Lewis base [✔]
Equation (4):
OH- accepts a proton/H+/hydrogen ion AND acts as a Brønsted–Lowry base [✔]
This was a good way to test this topic because answers showed that, while candidates usually knew the topic in theory, they could not apply this to identify the Lewis and Bronsted-Lowry bases in the context of a reaction that was given to them. In some cases, they failed to specify the base, OH- or also lost marks referring just to electrons, an electron or H instead of hydrogen ions or H+ for example.
Aqueous sodium hydrogencarbonate has a pH of approximately 7 at 298 K.
Sketch a graph of pH against volume when 25.0cm3 of 0.100 mol dm−3 NaOH (aq) is gradually added to 10.0cm3 of 0.0500 mol dm−3 NaHCO3 (aq).
[2]
S-shaped curve from ~7 to between 12 and 14 [✔]
equivalence point at 5 cm3 [✔]
Note: Accept starting point >6~7.
Most students that got 1mark for this titration curve was for the general shape, because few realized they had the data to calculate the equivalence point. There were also some difficulties in establishing the starting point even if it was specified in the stem.
