Chemistry (9-1) 0971 | Paper 5 Mock 01 | Practical Test
Question 1 Report
Solution AA is an aqueous alkali. Solution BB is a dilute acid. In this experiment you follow the pH of the alkali as the acid is added to it.
Use a pipette to place 25.0 cm3 of solution AA into a small beaker standing on a white tile. Rinse the pH probe with distilled water and stand it in the alkali. Read the meter before any acid is added. Fill a burette with solution BB. Run 5.0 cm3 of solution BB into the beaker, stir the mixture with a glass rod, wait until the meter reading is steady and read it. Repeat until a total of 40.0 cm3 of solution BB has been added. Record every reading in Table 9.1.
Table 9.1
| total volume of solution BB added / cm3 | pH of the mixture |
|---|---|
| 0.0 | |
| 5.0 | |
| 10.0 | |
| 15.0 | |
| 20.0 | |
| 25.0 | |
| 30.0 | |
| 35.0 | |
| 40.0 |
(a) Record, in the first row of Table 9.1, the pH of solution AA before any acid is added. [1]
(b) Record in Table 9.1 the pH of the mixture after each 5.0 cm3 portion of solution BB. [4]
(c) Plot your pH values against the total volume of solution BB added on the grid provided, and draw a smooth curve through your points. [4]
(d) Use your curve to deduce the volume of solution BB that just neutralises 25.0 cm3 of solution AA. [2]
(e) Describe how the pH changed over the three portions added around this volume. [2]
(f) State the colour a drop of universal indicator solution would give in the beaker at the volume you deduced in (d). [1]
(g) After 40.0 cm3 of solution BB the pH is still above 1, although solution BB itself has a pH of 1. Suggest why. [2]
(h) Plan how you would use the volume you deduced in (d) to prepare a pure dry sample of the salt formed, without using an indicator. [3]
This is a pH-monitoring titration: a fixed 25.0 cm3 of an aqueous alkali (AA) has a dilute acid (BB) run into it 5.0 cm3 at a time, and the pH probe follows how the mixture changes from strongly alkaline to strongly acidic. The whole question tests careful recording to a fixed precision, plotting a titration curve and reading the neutralisation point from its steepest part.
(a) Starting pH [1]. Before any acid is added the beaker holds only the alkali, so the reading is high. Record it to one decimal place, the precision of the meter, for example pH 13.0. Any value of 12 or above earns the mark; a whole number such as "13" without a decimal place would lose the precision requirement carried in part (b).
(b) The eight further readings [4]. One mark is for entering all eight further values, one for keeping every value to the same one-decimal-place precision, and the remaining marks reward the correct shape: the pH falls only slowly at first while excess alkali is still present, drops steeply as the last of the alkali is neutralised, then falls slowly again as excess acid builds up, and no reading lies outside the 0 to 14 range. A representative correctly recorded table is:
| total volume of BB / cm3 | pH of the mixture |
|---|---|
| 0.0 | 13.0 |
| 5.0 | 12.8 |
| 10.0 | 12.5 |
| 15.0 | 12.1 |
| 20.0 | 11.4 |
| 25.0 | 7.0 |
| 30.0 | 2.2 |
| 35.0 | 1.8 |
| 40.0 | 1.6 |
(c) The graph [4]. Volume goes on the horizontal axis and pH on the vertical axis, each labelled with its quantity and unit; the scales are chosen so the points fill more than half the grid; all nine points are plotted; and a single smooth curve is drawn through them, not straight lines joining point to point. Plotting the values above gives the characteristic titration curve, almost flat at each end with a steep fall in the middle:
(d) Neutralisation volume [2]. Neutralisation is the mid-point of the steep, near-vertical part of the curve, where the pH passes through 7. Reading down from that point on the curve above gives about 25.0 cm3 of solution BB. In an examination the candidate's own value, read correctly from the centre of their own steep section, is accepted.
(e) How the pH changed around this volume [2]. For the portions before and after this volume the pH changed only slightly (from about 12.1 at 15.0 cm3 to 11.4 at 20.0 cm3, and from about 2.2 at 30.0 cm3 to 1.8 at 35.0 cm3) [1], but over the single portion that spans the neutralisation point the pH fell very steeply, by several pH units (from about 11.4 down to about 2.2) [1]. This is because near the end point almost all the remaining alkali is used up by one small addition of acid.
(f) Colour of universal indicator at neutralisation [1]. At pH 7 universal indicator is green.
(g) Why the pH stays above 1 [2]. Although solution BB has pH 1 on its own, the excess acid added past the end point is mixed with, and so diluted by, the 25.0 cm3 of neutralised mixture already in the beaker [1]. Diluting it lowers the concentration of hydrogen ions, so the pH stays higher than the pH 1 of the undiluted acid [1].
(h) Preparing a pure dry salt without an indicator [3]. Because part (d) tells you exactly how much acid neutralises the alkali, you no longer need an indicator to find the end point:
- Pipette a fresh 25.0 cm3 portion of solution AA into a clean flask and run in exactly the volume of solution BB deduced in (d), giving a neutral salt solution [1].
- Pour this neutral mixture into an evaporating basin and heat it to the point of crystallisation (until crystals just begin to form at the edge) [1].
- Leave it to crystallise as it cools, then filter off the crystals and dry them between filter papers [1].
Answer Details
This is a pH-monitoring titration: a fixed 25.0 cm3 of an aqueous alkali (AA) has a dilute acid (BB) run into it 5.0 cm3 at a time, and the pH probe follows how the mixture changes from strongly alkaline to strongly acidic. The whole question tests careful recording to a fixed precision, plotting a titration curve and reading the neutralisation point from its steepest part.
(a) Starting pH [1]. Before any acid is added the beaker holds only the alkali, so the reading is high. Record it to one decimal place, the precision of the meter, for example pH 13.0. Any value of 12 or above earns the mark; a whole number such as "13" without a decimal place would lose the precision requirement carried in part (b).
(b) The eight further readings [4]. One mark is for entering all eight further values, one for keeping every value to the same one-decimal-place precision, and the remaining marks reward the correct shape: the pH falls only slowly at first while excess alkali is still present, drops steeply as the last of the alkali is neutralised, then falls slowly again as excess acid builds up, and no reading lies outside the 0 to 14 range. A representative correctly recorded table is:
| total volume of BB / cm3 | pH of the mixture |
|---|---|
| 0.0 | 13.0 |
| 5.0 | 12.8 |
| 10.0 | 12.5 |
| 15.0 | 12.1 |
| 20.0 | 11.4 |
| 25.0 | 7.0 |
| 30.0 | 2.2 |
| 35.0 | 1.8 |
| 40.0 | 1.6 |
(c) The graph [4]. Volume goes on the horizontal axis and pH on the vertical axis, each labelled with its quantity and unit; the scales are chosen so the points fill more than half the grid; all nine points are plotted; and a single smooth curve is drawn through them, not straight lines joining point to point. Plotting the values above gives the characteristic titration curve, almost flat at each end with a steep fall in the middle:
(d) Neutralisation volume [2]. Neutralisation is the mid-point of the steep, near-vertical part of the curve, where the pH passes through 7. Reading down from that point on the curve above gives about 25.0 cm3 of solution BB. In an examination the candidate's own value, read correctly from the centre of their own steep section, is accepted.
(e) How the pH changed around this volume [2]. For the portions before and after this volume the pH changed only slightly (from about 12.1 at 15.0 cm3 to 11.4 at 20.0 cm3, and from about 2.2 at 30.0 cm3 to 1.8 at 35.0 cm3) [1], but over the single portion that spans the neutralisation point the pH fell very steeply, by several pH units (from about 11.4 down to about 2.2) [1]. This is because near the end point almost all the remaining alkali is used up by one small addition of acid.
(f) Colour of universal indicator at neutralisation [1]. At pH 7 universal indicator is green.
(g) Why the pH stays above 1 [2]. Although solution BB has pH 1 on its own, the excess acid added past the end point is mixed with, and so diluted by, the 25.0 cm3 of neutralised mixture already in the beaker [1]. Diluting it lowers the concentration of hydrogen ions, so the pH stays higher than the pH 1 of the undiluted acid [1].
(h) Preparing a pure dry salt without an indicator [3]. Because part (d) tells you exactly how much acid neutralises the alkali, you no longer need an indicator to find the end point:
- Pipette a fresh 25.0 cm3 portion of solution AA into a clean flask and run in exactly the volume of solution BB deduced in (d), giving a neutral salt solution [1].
- Pour this neutral mixture into an evaporating basin and heat it to the point of crystallisation (until crystals just begin to form at the edge) [1].
- Leave it to crystallise as it cools, then filter off the crystals and dry them between filter papers [1].
Question 2 Report
You are planning an experiment to find how fast a carbonate reacts with dilute hydrochloric acid by collecting the gas given off. Fig. 1.1 shows the two main pieces of apparatus, A and B, set out on the bench before they are joined together.
Fig. 1.1
(a) Name the piece of apparatus labelled A and the piece of apparatus labelled B. [2]
(b) Describe how you would complete the assembly of this apparatus so that none of the gas escapes. [2]
(c) State the two quantities you would record during the experiment, and state how often you would record them. [2]
This question tests recognition of standard gas-collection apparatus and how to assemble it so no gas leaks before it is measured. The reaction of a carbonate with hydrochloric acid, \( \text{carbonate} + \text{HCl} \rightarrow \text{salt} + \text{H}_2\text{O} + \text{CO}_2 \), gives off carbon dioxide, and the rate is found from how the gas volume grows with time.
(a) Naming the apparatus [2]. Piece A is the wide-based vessel that holds the reactants, a conical flask [1]. Piece B, the graduated barrel with a moving plunger, is a gas syringe, which measures the volume of gas collected [1].
(b) Completing the assembly gas-tight [2]. The join must be sealed at both ends:
- Push a bung carrying the delivery tube firmly into the neck of the conical flask so no gas escapes at the top [1].
- Connect the delivery tube to the nozzle of the gas syringe, clamp the syringe horizontally, and check the plunger is pushed fully in and reads zero before the solid is added [1].
Starting from zero with an airtight seal matters because any gas lost before or during collection makes the measured volume too low.
(c) What to record and how often [2]. Record two quantities: the volume of gas in the syringe and the time on the stop-clock [1]. Take the volume at regular intervals, for example every 30 s (any stated regular interval from 10 s to 60 s is acceptable), continuing until no more gas is collected [1]. Fixed intervals are needed so the readings can be plotted as a smooth volume-time graph from which the rate is read.
Exam takeaway: for any gas-collection rate experiment, name the two variables as volume and time, and always state a regular time interval rather than just saying you would time it.
Answer Details
This question tests recognition of standard gas-collection apparatus and how to assemble it so no gas leaks before it is measured. The reaction of a carbonate with hydrochloric acid, \( \text{carbonate} + \text{HCl} \rightarrow \text{salt} + \text{H}_2\text{O} + \text{CO}_2 \), gives off carbon dioxide, and the rate is found from how the gas volume grows with time.
(a) Naming the apparatus [2]. Piece A is the wide-based vessel that holds the reactants, a conical flask [1]. Piece B, the graduated barrel with a moving plunger, is a gas syringe, which measures the volume of gas collected [1].
(b) Completing the assembly gas-tight [2]. The join must be sealed at both ends:
- Push a bung carrying the delivery tube firmly into the neck of the conical flask so no gas escapes at the top [1].
- Connect the delivery tube to the nozzle of the gas syringe, clamp the syringe horizontally, and check the plunger is pushed fully in and reads zero before the solid is added [1].
Starting from zero with an airtight seal matters because any gas lost before or during collection makes the measured volume too low.
(c) What to record and how often [2]. Record two quantities: the volume of gas in the syringe and the time on the stop-clock [1]. Take the volume at regular intervals, for example every 30 s (any stated regular interval from 10 s to 60 s is acceptable), continuing until no more gas is collected [1]. Fixed intervals are needed so the readings can be plotted as a smooth volume-time graph from which the rate is read.
Exam takeaway: for any gas-collection rate experiment, name the two variables as volume and time, and always state a regular time interval rather than just saying you would time it.
Question 3 Report
Solution E is aqueous copper(II) sulfate. In this experiment you pass a current through solution E using two carbon electrodes and find the mass of metal deposited.
Clean both carbon electrodes with emery paper, wash them with distilled water, dry them and weigh the one you will use as the negative electrode. Pour 100 cm3 of solution E into a beaker. Clamp the two electrodes so that they dip about 4 cm into the solution without touching. Connect them through an ammeter to the power supply, close the switch and start the stopclock at once. Keep the current steady for fifteen minutes, then switch off. Lift out the negative electrode, rinse it gently with distilled water, dry it in a warm oven, let it cool and weigh it again.
Table 5.1
| measurement | value |
|---|---|
| mass of negative electrode before the experiment / g | |
| current / A | |
| time for which the current passed / s | |
| mass of negative electrode after washing and drying / g | |
| increase in mass / g |
(a) Record in Table 5.1 the mass of the clean dry negative electrode before the experiment. [1]
(b) Record in Table 5.1 the current shown on the ammeter and the time for which the current passed. [2]
(c) Record in Table 5.1 the mass of the negative electrode after it has been washed, dried and cooled. [1]
(d) Use your readings to work out the increase in mass of the negative electrode. [1]
(e) Record your observations at the positive electrode during the fifteen minutes. [2]
(f) Describe the test you would use on the gas collected from the positive electrode, and give the result you expect. [2]
(g) Record what happens to the blue colour of solution E during the experiment, and state which ion is being removed. [2]
(h) Suggest why the negative electrode is rinsed with distilled water and dried before the second weighing. [2]
(i) Plan how you would use the same apparatus to find out whether doubling the current doubles the mass of metal deposited in fifteen minutes. [3]
This is an electrolysis experiment: a current is passed through aqueous copper(II) sulfate using carbon electrodes, copper deposits on the negative electrode (cathode) and oxygen is released at the positive electrode (anode). The mass gained by the cathode is measured.
(a) Starting mass of the negative electrode [1]. Record it to 0.01 g, a sensible value of about 2 g to 10 g [1] (for example 5.42 g).
(b) Current and time [2]. Record the current with the unit A, about 0.2 A to 1.0 A [1]; record the time as 900 s (15 minutes) [1].
(c) Mass after washing and drying [1]. Record the second mass to the same precision and larger than the first [1] (for example 5.63 g), because copper has been added.
(d) Increase in mass [1]. Increase = final mass minus starting mass, correctly subtracted, a small positive value with unit g [1] (for example \(5.63 - 5.42 = 0.21\) g).
(e) Observations at the positive electrode [2]. Bubbles of a colourless gas form on the positive electrode [1]; the bubbling is steady, the gas has no smell and the carbon electrode may crumble slightly [1].
(f) Test for the gas [2]. Collect the gas in an inverted test tube held over the electrode and put a glowing splint into the mouth of the tube [1]; the splint relights, showing the gas is oxygen [1].
(g) The blue colour and the ion removed [2]. The blue colour becomes paler as the experiment goes on [1]; this is because the blue copper(II) ions, \(\text{Cu}^{2+}\), are being removed from the solution and deposited as copper [1].
(h) Why rinse and dry before reweighing [2]. Any solution left on the electrode would dry to leave solid copper(II) sulfate, which would be weighed along with the copper [1]; and water left on the electrode would itself add to the mass, so the increase would come out too large [1].
(i) Testing whether doubling the current doubles the mass [3]. Repeat the whole experiment with a fresh 100 cm3 portion of solution E and a freshly cleaned, dried and weighed electrode [1]; use the variable resistor to set the current at twice the first value while keeping the time at fifteen minutes and keeping the temperature, the depth of the electrodes and the concentration the same [1]; compare the two increases in mass, repeating each current twice to check the results agree [1].
Answer Details
This is an electrolysis experiment: a current is passed through aqueous copper(II) sulfate using carbon electrodes, copper deposits on the negative electrode (cathode) and oxygen is released at the positive electrode (anode). The mass gained by the cathode is measured.
(a) Starting mass of the negative electrode [1]. Record it to 0.01 g, a sensible value of about 2 g to 10 g [1] (for example 5.42 g).
(b) Current and time [2]. Record the current with the unit A, about 0.2 A to 1.0 A [1]; record the time as 900 s (15 minutes) [1].
(c) Mass after washing and drying [1]. Record the second mass to the same precision and larger than the first [1] (for example 5.63 g), because copper has been added.
(d) Increase in mass [1]. Increase = final mass minus starting mass, correctly subtracted, a small positive value with unit g [1] (for example \(5.63 - 5.42 = 0.21\) g).
(e) Observations at the positive electrode [2]. Bubbles of a colourless gas form on the positive electrode [1]; the bubbling is steady, the gas has no smell and the carbon electrode may crumble slightly [1].
(f) Test for the gas [2]. Collect the gas in an inverted test tube held over the electrode and put a glowing splint into the mouth of the tube [1]; the splint relights, showing the gas is oxygen [1].
(g) The blue colour and the ion removed [2]. The blue colour becomes paler as the experiment goes on [1]; this is because the blue copper(II) ions, \(\text{Cu}^{2+}\), are being removed from the solution and deposited as copper [1].
(h) Why rinse and dry before reweighing [2]. Any solution left on the electrode would dry to leave solid copper(II) sulfate, which would be weighed along with the copper [1]; and water left on the electrode would itself add to the mass, so the increase would come out too large [1].
(i) Testing whether doubling the current doubles the mass [3]. Repeat the whole experiment with a fresh 100 cm3 portion of solution E and a freshly cleaned, dried and weighed electrode [1]; use the variable resistor to set the current at twice the first value while keeping the time at fifteen minutes and keeping the temperature, the depth of the electrodes and the concentration the same [1]; compare the two increases in mass, repeating each current twice to check the results agree [1].
