Activated carbon is a form of carbon with the ability to adsorb solutes from solutions. The adsorption of ethanedioic acid molecules, (COOH)2, from dilute ethanedioic acid, (COOH)2(aq), onto the surface of activated carbon is studied in a series of experiments. Before beginning the experiment, 500.0 cm³ of 0.139 moldm⁻³ (COOH)2(aq), solution X, is made. The activated carbon is placed in an oven at 120 °C for three hours. The experimental procedure involves the following steps. step 1 Prepare 100.0 cm³ of 0.00556 moldm⁻³ (COOH)2(aq) from solution X. step 2 Transfer 50.0 cm³ of the 0.00556 moldm⁻³ (COOH)2(aq) into a conical flask. step 3 Add 0.500 g of activated carbon to the conical flask and start a stopwatch. step 4 Shake the flask and immediately remove a 5.0 cm³ sample from the flask. Determine the mass of non-adsorbed (COOH)2 in the sample using chromatography. step 5 Repeat step 4 at suitable time intervals until there is no further change in the mass of non-adsorbed (COOH)2 in the sample.
- (a)(i)Calculate the mass of (COOH)2•2H2O(s) required to prepare 500.0 cm³ of 0.139 moldm⁻³ (COOH)2(aq), solution X. mass = .............................................. g[1]
- (a)(ii)The mass of (COOH)2•2H2O(s) calculated in (a)(i) is placed into a 100 cm³ beaker. Describe the steps taken to prepare 500.0 cm³ of 0.139 moldm⁻³ (COOH)2(aq). Give the name and capacity of any apparatus used. Write your answer using a series of numbered steps.[3]
- (b)(COOH)2•2H2O(s) is corrosive. Other than wearing eye protection and a lab coat, state one safety precaution that should be taken when using (COOH)2•2H2O(s).[1]
- (c)Suggest why the activated carbon is placed in the oven before use.[1]
- (d)In step 1, solution X is diluted to a concentration of 0.00556 moldm⁻³. Calculate the volume of solution X needed to make 100.0 cm³ of 0.00556 mol dm⁻³ (COOH)2(aq). volume of solution X = ...................................... cm³[1]
- (e)Identify the most appropriate piece of equipment to transfer 50.0 cm³ of diluted (COOH)2(aq) in step 2.[1]
- (f)Suggest a reason why the conical flask is shaken in step 4.[1]
- (g)(i)The results of the experiment are shown in Table 1.1. Table 1.1 time, t/min | mass of non-adsorbed (COOH)2 in 5.0 cm³ sample /mg | mass of (COOH)2 adsorbed by activated carbon in 5.0 cm³ sample /mg ---|---|--- 0 | 2.50 | 0.00 5 | 1.90 | 15 | 1.52 | 30 | 1.43 | 45 | 0.99 | 60 | 0.88 | 90 | 0.81 | 120 | 0.80 | 150 | 0.80 | Complete Table 1.1.[1]
- (g)(ii)Plot a graph on the grid in Fig. 1.1 to show the relationship between mass of (COOH)2 adsorbed by activated carbon in 5.0 cm³ sample and time, t. Use a cross (×) to plot each data point. Draw a curved line of best fit. [Figure 1.1][2]
- (g)(iii)Identify the independent variable.[1]
- (g)(iv)Suggest one variable that needs to be controlled that is not stated in the experimental procedure.[1]
- (g)(v)Circle the one point on the graph that you consider to be most anomalous. Explain the error in timing that may have led to this anomalous point.[1]
- (h)(i)The experiment is repeated using different concentrations of (COOH)2(aq) prepared in step 1. The results are shown in Table 1.2. Table 1.2 1 | 2 | 3 | 4 | 5 mᵢ/mg | mₑ/mg | (mᵢ - mₑ)/mg | qₑ /mg g⁻¹ | mₑ/g qₑ ---|---|---|---|--- 2.5 | 0.80 | 1.70 | 34.0 | 0.0235 5.0 | 2.10 | 2.90 | 58.0 | 0.0362 7.5 | 4.03 | 3.47 | 69.4 | 0.0581 10.0 | 6.41 | 3.59 | 71.8 | 0.0893 12.5 | 8.87 | 3.63 | 72.6 | 0.1222 mᵢ is the initial mass of (COOH)2 in 5.0 cm³ sample. mₑ is the mass of non-adsorbed (COOH)2 in the final 5.0 cm³ sample at the end of step 5. qₑ is the mass of (COOH)2 adsorbed per gram of activated carbon. A plot of mₑ/qₑ against mₑ gives the line of best fit shown in Fig. 1.2. [Figure 1.2] Use Fig. 1.2 to determine the gradient of the line of best fit. State the coordinates of both points you used in your calculation. coordinates 1 .................................................... coordinates 2 .................................................... Include units in your answer. gradient = .................................................... units = ....................................................[3]
- (h)(ii)The adsorption parameter, q₀, is the maximum mass of (COOH)2 adsorbed per gram of activated carbon. The equation of the line of best fit plotted in Fig. 1.2 is shown. mₑ/qₑ = mₑ/q₀ + constant Use the gradient determined in (h)(i) to calculate the adsorption parameter, q₀. [If you were unable to determine an answer to (h)(i), then use the value 0.0136 for the gradient. This is not the correct answer.] q₀ = .............................................. mg g⁻¹[1]
- (h)(iii)The experiment is repeated and the value of q₀ is calculated to be 78.1 mgg⁻¹. The total percentage error from the experimental procedure is 6.5%. The data book value of q₀ is 86.0 mgg⁻¹. Use this information to determine whether the error in the repeated experiment could be accounted for by experimental errors or is caused by other factors. Show your working.[1]
Mark scheme
Marking points from the official mark scheme (May/June 2025). The mark or mark type is shown in brackets.
- [1] 8.76 (g)
- [M1] dissolve the solid / ethanedioic acid (in the beaker) using (a small volume of) (distilled) water
- [M2] transfer / add to a (500 cm³ volumetric) flask AND rinse (with distilled water)
- [M3] top the 500 cm³ volumetric flask up to mark (distilled water) AND (stopper and) invert the flask
- [1] wear chemically resistant gloves
- [1] to remove water from (the surface of) the activated carbon
- [1] 4.0 (cm³)
- [1] burette
- [1] to ensure a uniformly mixed sample
- [1] 0.60, 0.98, 1.07, 1.51, 1.62, 1.69, 1.70, 1.70
- [M1] all points plotted correctly
- [M2] smooth curved line of best fit passing close to all points except marked anomaly
- [1] time (of sample removal)
- [1] temperature
- [1] anomalous point circled (at 30 mins) AND the actual time (of removing sample) was less than recorded time (30 minutes)
- [M1] co-ordinates correctly recorded from line given in Fig 1.2
- [M2] gradient calculated to at least two significant figures
- [M3] units = g mg⁻¹
- [1] q₀ = 1 / gradient in (h)(i)
- [1] percentage error of experimental result compared to data book value = (86.0 - 78.1) / 86.0 × 100% = 9.2% AND 9.2% > 6.5% AND other factors contributed towards error
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