Chemistry 9701/53 — October/November 2016

Background Concept

In a precipitation titration using the Mohr method, silver nitrate is used to titrate chloride ions. The indicator is potassium chromate(VI). Silver chloride is highly insoluble, while silver chromate is sparingly soluble. This difference in solubility ensures that AgCl precipitates first. Once all the chloride ions are consumed, the next drop of silver nitrate reacts with the chromate(VI) ions to form a visible red precipitate of silver chromate(VI), signalling the endpoint.

Understanding the Question

The question asks for the identity of the red precipitate formed at the endpoint and the chemical equation for its formation. The stem provides the solubility table and states that the indicator forms a red precipitate in the presence of excess silver ions.

Approach

Identify the ions present that can form a precipitate with excess silver ions: $\text{Ag}^+$ and $\text{CrO}_4^{2-}$. Use the solubility data to confirm it is insoluble (0.022 g dm$^{-3}$ is very low). Write the net ionic equation for the formation of the precipitate.

Step-by-Step Reasoning

• The indicator is $\text{K}_2\text{CrO}_4$, which provides $\text{CrO}_4^{2-}$ ions. Excess titrant provides $\text{Ag}^+$ ions.
• From the table, $\text{Ag}^+$ and $\text{CrO}_4^{2-}$ have a very low solubility (0.022 g dm$^{-3}$), so they form a precipitate. The compound is silver chromate(VI), $\text{Ag}_2\text{CrO}_4$.
• The net ionic equation combines the silver ions and chromate(VI) ions: $2\text{Ag}^+ + \text{CrO}_4^{2-} \rightarrow \text{Ag}_2\text{CrO}_4$. State symbols (aq) and (s) are required for full marks.

Key Takeaways

The Mohr method relies on the selective precipitation of the analyte ion before the indicator ion. Recognising the indicator's role and writing the correct net ionic equation for the endpoint colour change is essential.

Common Mistakes

• Writing the molecular equation with spectator ions (e.g., including $\text{K}^+$ or $\text{NO}_3^-$) when the net ionic equation is expected or preferred.
• Forgetting state symbols, particularly (s) for the precipitate and (aq) for the ions.
• Naming the precipitate incorrectly (e.g., 'silver chromate' without the (VI) or 'barium chromate').

Things to Be Careful About

Ensure the equation is balanced (2 Ag$^+$ for every CrO$_4^{2-}$). The mark scheme accepts the net ionic equation or full molecular/ionic equations as long as they are balanced and correct.

Background Concept

The indicator must remain in solution until the endpoint. If another cation present in the analyte solution can form an insoluble salt with the indicator anion, it will precipitate out prematurely. This consumes the indicator, preventing a clear colour change at the true endpoint, or creates a background precipitate that masks the red silver chromate(VI) colour.

Understanding the Question

The question asks how $\text{Ba}^{2+}$ ions would interfere with the potassium chromate(VI) indicator. We must use the provided solubility table to determine if a precipitate forms between $\text{Ba}^{2+}$ and $\text{CrO}_4^{2-}$.

Approach

Look at the intersection of the $\text{Ba}^{2+}$ row and the $\text{CrO}_4^{2-}$ column in the solubility table. A very low solubility value indicates an insoluble compound would form.

Step-by-Step Reasoning

• The table shows the solubility of $\text{BaCrO}_4$ is 0.0028 g dm$^{-3}$, which is extremely low (insoluble).
• Therefore, $\text{Ba}^{2+}$ ions in the solution would react with the $\text{CrO}_4^{2-}$ indicator ions to form a solid precipitate of barium chromate(VI).
• This would remove the indicator from solution before the endpoint is reached, interfering with the titration.

Key Takeaways

Always check the solubility table for competing precipitation reactions when designing a titration. The analyte cation must not form an insoluble salt with the indicator anion.

Common Mistakes

• Stating that $\text{Ba}^{2+}$ reacts with $\text{Ag}^+$ (they are both cations and do not react).
• Failing to name the specific interfering product (barium chromate(VI)).
• Saying 'it would form a precipitate' without specifying which precipitate or why it interferes.

Things to Be Careful About

The mark scheme specifically looks for 'insoluble / solid barium chromate(VI)'. Simply saying 'a precipitate forms' is not enough; you must identify the correct salt.

Background Concept

To prevent $\text{Ba}^{2+}$ from interfering, it must be removed from the solution without introducing new interfering ions. We need an anion that forms a highly insoluble salt with $\text{Ba}^{2+}$ but does not interfere with the silver nitrate titration or the chromate(VI) indicator.

Understanding the Question

The question asks how adding sulfuric acid ($\text{H}_2\text{SO}_4$) prevents $\text{Ba}^{2+}$ interference. We must look at the solubility of barium sulfate in the provided table.

Approach

Check the solubility of $\text{BaSO}_4$ in the table. Sulfuric acid provides $\text{SO}_4^{2-}$ ions. If $\text{BaSO}_4$ is insoluble, it will precipitate out, removing $\text{Ba}^{2+}$ from the solution.

Step-by-Step Reasoning

• The table shows the solubility of $\text{BaSO}_4$ is 0.00245 g dm$^{-3}$, which is very low (insoluble).
• Adding sulfuric acid introduces $\text{SO}_4^{2-}$ ions.
• These ions react with $\text{Ba}^{2+}$: $\text{Ba}^{2+} + \text{SO}_4^{2-} \rightarrow \text{BaSO}_4(\text{s})$.
• The formation of the insoluble barium sulfate precipitate removes $\text{Ba}^{2+}$ ions from the aqueous phase, so they are no longer available to react with the chromate(VI) indicator.

Key Takeaways

Adding a reagent to 'mask' or remove an interfering ion by precipitation is a common technique in analytical chemistry. The precipitate formed must be sufficiently insoluble and not interfere with the main reaction.

Common Mistakes

• Stating that sulfuric acid 'neutralises' the barium ions (incorrect terminology).
• Forgetting to mention that barium sulfate is insoluble/solid.
• Not explaining that this removes $\text{Ba}^{2+}$ from solution.

Things to Be Careful About

The mark scheme accepts 'insoluble / solid barium sulfate is formed'. Ensure you link the formation of this precipitate to the removal of the interfering $\text{Ba}^{2+}$ ions.

Background Concept

A titration experiment requires precise measurement of volumes. The standard volumetric apparatus used in acid-base or precipitation titrations includes a burette (for delivering the titrant), a pipette (for transferring a known volume of the analyte), and a volumetric flask (for preparing a standard solution of known concentration).

Understanding the Question

The question asks for three pieces of volumetric apparatus and their capacities in cm$^3$ that would be used to plan this titration.

Approach

List the three essential pieces of volumetric glassware for a titration and state their standard capacities.

Step-by-Step Reasoning

• Volumetric flask: Used to prepare the standard barium chloride solution. A standard capacity is $250\text{ cm}^3$.
• Pipette: Used to transfer a known, precise volume of the barium chloride solution into the conical flask. A standard capacity is $25\text{ cm}^3$.
• Burette: Used to deliver the silver nitrate titrant. A standard capacity is $50\text{ cm}^3$.
• All three must be named correctly and paired with a realistic capacity. (2 marks for all three correct, 1 mark for two correct).

Key Takeaways

Know the standard capacities and uses of volumetric flasks, pipettes, and burettes. A volumetric flask is for preparing solutions, a pipette is for transferring fixed volumes, and a burette is for variable delivery of titrant.

Common Mistakes

• Naming 'measuring cylinder' instead of 'volumetric flask' or 'pipette'. Measuring cylinders are not sufficiently accurate for volumetric analysis.
• Giving incorrect capacities (e.g., 'burette: 100 cm$^3$' or 'pipette: 50 cm$^3$'). While 50 cm$^3$ pipettes exist, 25 cm$^3$ is standard; 100 cm$^3$ burettes are not standard.
• Forgetting to include the units (cm$^3$).

Things to Be Careful About

The mark scheme requires 'volumetric / graduated flask', 'pipette (graduated)', and 'burette'. Ensure the apparatus names match the expected terminology. Capacities must be in cm$^3$.

Background Concept

To perform a titration, you need a standard solution of known concentration. When starting from a solid hydrate, the procedure involves dissolving the solid in a solvent (usually distilled water), transferring it quantitatively to a volumetric flask, and diluting to the exact mark to achieve a precise concentration.

Understanding the Question

The question asks how to make a solution of barium chloride suitable for use in the titration. You have $3.00\text{ g}$ of hydrated barium chloride. You must describe the preparation of the standard solution.

Approach

Outline the standard procedure for making a solution from a solid using a volumetric flask. Key steps: dissolving, transferring, and making up to volume. Emphasise the use of distilled water.

Step-by-Step Reasoning

• Dissolving: Measure the known mass of $\text{BaCl}_2 \cdot x\text{H}_2\text{O}$ and dissolve it in a beaker with a small amount of distilled water. Stir until fully dissolved.
• Transferring: Pour the solution into a $250\text{ cm}^3$ volumetric flask (the capacity chosen in part (c)(i)). Rinse the beaker and stirring rod with distilled water and add the rinsings to the flask to ensure all solute is transferred.
• Making up to volume: Add distilled water to the volumetric flask until the bottom of the meniscus rests exactly on the graduation mark. Stopper and invert to mix.
• Mark scheme points: 'Dissolve/mix known mass in distilled water' (1 mark). 'Transfer to volumetric flask, make to mark with distilled water' (1 mark). Distilled water must be mentioned for both marks.

Key Takeaways

The preparation of a standard solution requires precise language: 'dissolve', 'transfer', 'volumetric flask', 'make up to the mark', and 'distilled water'. Omitting 'distilled water' or 'make up to the mark' will cost marks.

Common Mistakes

• Saying 'add water to the flask until it is full' instead of 'make up to the mark'.
• Forgetting to mention 'distilled water' (tap water contains ions that could interfere).
• Not mentioning that the entire mass of the solid is dissolved.
• Using a measuring cylinder to add water to the flask.

Things to Be Careful About

The mark scheme explicitly states: 'Water must be mentioned at least once for one mark... Distilled / deionised / purified water must be mentioned for 2 marks.' Ensure 'distilled water' is used. Also, reference the volumetric flask capacity from part (c)(i) if possible, though the general description 'volumetric flask' may suffice for the second mark if the capacity is stated.

Background Concept

In the Mohr method for chloride determination, the order of addition is critical. The analyte (chloride) is in the conical flask. The indicator (chromate) must be present. The interfering ion (barium) must be removed before the indicator is added, otherwise the barium will precipitate with the chromate and ruin the indicator. Finally, the titrant (silver nitrate) is added last to begin the titration.

Understanding the Question

A known volume of barium chloride solution is already in the conical flask. You must state the order in which the remaining three solutions (sulfuric acid, potassium chromate(VI), silver nitrate) should be added.

Approach

Logical sequence: 1. Remove interference. 2. Add indicator. 3. Add titrant.

Step-by-Step Reasoning

• First: Sulfuric acid. This must be added first to react with and precipitate out the $\text{Ba}^{2+}$ ions as insoluble $\text{BaSO}_4$. If added after the indicator, the $\text{Ba}^{2+}$ would react with the chromate(VI) indicator and destroy it.
• Second: Potassium chromate(VI). This is the indicator. It is added after the interference ($\text{Ba}^{2+}$) has been removed, so it remains in solution and is available to signal the endpoint.
• Third: Silver nitrate. This is the titrant. It is added last, typically from a burette, to titrate the chloride ions.

Key Takeaways

The order of addition in analytical procedures is dictated by chemical compatibility. Masking/removing interfering ions always precedes adding the indicator, and the titrant is always added last.

Common Mistakes

• Adding the silver nitrate before the indicator (this would start the titration without an indicator, or cause premature precipitation of Ag2CrO4 if added together).
• Adding the indicator before the sulfuric acid (this would cause BaCrO4 to precipitate, consuming the indicator).
• Swapping the order of sulfuric acid and silver nitrate.

Things to Be Careful About

The mark scheme is strict on the order: first = sulfuric acid, second = potassium chromate(VI), third = silver nitrate. Any other order scores 0.

Background Concept

Reliability (or repeatability) in an experiment is ensured by performing multiple trials and checking for consistency. In a titration, this means obtaining 'concordant results'—typically titres that are within $0.10\text{ cm}^3$ of each other.

Understanding the Question

The question asks how to ensure the titration result is reliable. This is a standard evaluation point for practical experiments.

Approach

State the standard method for improving reliability in a titration: repeating the experiment and using concordant results.

Step-by-Step Reasoning

• Perform the titration more than once (at least 2-3 times).
• Compare the titres obtained.
• Calculate the mean titre using only the concordant results (those within $0.10\text{ cm}^3$ of each other), discarding any anomalous results.
• This reduces the effect of random errors.

Key Takeaways

'Repeat and get concordant results' is the standard, expected answer for ensuring reliability in titrations. 'Doing it twice' is not enough; you must mention concordance or repeating.

Common Mistakes

• Saying 'be more careful' or 'use better equipment' (these improve accuracy, not necessarily reliability, and are not specific enough).
• Saying 'repeat until you get the same result' (results will rarely be exactly the same; they must be concordant within a small range).
• Confusing reliability with accuracy.

Things to Be Careful About

The mark scheme accepts 'experiment / titration is repeated to get concordant titre'. Ensure you use the word 'concordant' or specify the range (e.g., within 0.10 cm$^3$).

Background Concept

The formula of the hydrated salt is $\text{BaCl}_2 \cdot x\text{H}_2\text{O}$. The molar mass of the anhydrous salt ($\text{BaCl}_2$) is $137.3 + 2(35.5) = 208.3\text{ g mol}^{-1}$. The molar mass of water is $18.0\text{ g mol}^{-1}$. By finding the mass of the anhydrous salt and the mass of the water in the sample, we can find the mole ratio and determine $x$.

Understanding the Question

A student dissolved $3.13\text{ g}$ of $\text{BaCl}_2 \cdot x\text{H}_2\text{O}$ to make $1.00\text{ dm}^3$ of solution. The concentration of $\text{Ba}^{2+}$ is $0.0128\text{ mol dm}^{-3}$. We need to find $x$.

Approach

1. Calculate moles of $\text{Ba}^{2+}$ (and thus $\text{BaCl}_2$) in the solution.
2. Calculate the mass of anhydrous $\text{BaCl}_2$.
3. Subtract this from the total mass of the hydrated salt to find the mass of water.
4. Calculate moles of water.
5. Find the simplest whole number ratio of water to $\text{BaCl}_2$ to get $x$.

Step-by-Step Reasoning

• Moles of BaCl2: $n = c \times V = 0.0128\text{ mol dm}^{-3} \times 1.00\text{ dm}^3 = 0.0128\text{ mol}$.
• Mass of BaCl2: $m = n \times M_r = 0.0128\text{ mol} \times 208.3\text{ g mol}^{-1} = 2.66624\text{ g}$. (Use 2.67 g for intermediate steps, but keep more digits if possible to avoid rounding errors: $2.666\text{ g}$).
• Mass of H2O: Total mass - mass of BaCl2 = $3.13\text{ g} - 2.66624\text{ g} = 0.46376\text{ g}$. (Rounding to 2.67 gives $3.13 - 2.67 = 0.46\text{ g}$, which is acceptable and matches the mark scheme).
• Moles of H2O: $n = \frac{m}{M_r} = \frac{0.46376\text{ g}}{18.0\text{ g mol}^{-1}} = 0.02576\text{ mol}$. (Using 0.46 g gives $0.46 / 18.0 = 0.02556\text{ mol}$).
• Ratio x: $x = \frac{n(\text{H}_2\text{O})}{n(\text{BaCl}_2)} = \frac{0.02576}{0.0128} = 2.0125$. (Using 0.46 g gives $0.02556 / 0.0128 = 1.997 \approx 2$).
• The value of $x$ is 2. The hydrated salt is $\text{BaCl}_2 \cdot 2\text{H}_2\text{O}$.

Key Takeaways

When determining the number of water molecules of crystallisation, always calculate the mass of the anhydrous salt first, then find the mass of water by subtraction. Use the mole ratio to find $x$.

Common Mistakes

• Using the molar mass of the hydrated salt to find moles (circular reasoning).
• Forgetting to subtract the mass of BaCl2 from the total mass to find the mass of water.
• Rounding intermediate values too early, leading to a ratio like 1.9 or 2.1 instead of exactly 2.
• Calculating the ratio backwards ($\text{BaCl}_2$ / $\text{H}_2\text{O}$) and getting $0.5$.

Things to Be Careful About

The mark scheme shows the working: $0.0128 \times 208.3 = 2.67\text{ g}$; $3.13 - 2.67 = 0.46\text{ g}$; $x = (0.46 / 18.0) \div 0.0128 = 2$. Follow this exact calculation path to ensure you hit the method marks (M1) and accuracy marks (A1).