Skip to content
Edexcel Jun 2011 Paper 3 Q3
Answers available below
Show answer
Question
1
2
3
4
5
![PMT 3 A titration was carried out in order to investigate the neutralization reaction shown below. HCl(aq) + NaOH(aq) (cid:111) NaCl(aq) + H2O(l) Procedure 1. Using a pipette, transfer 50.0 cm3 of sodium hydroxide solution, concentration 1.00 mol dm3, to a polystyrene cup. Allow to stand for a few minutes. 2. Record the temperature of the solution. 3. From a burette, add 5.00 cm3 of dilute hydrochloric acid to the solution in the cup. 4. Stir the mixture with the thermometer and record the temperature. 5. Add successive 5.00 cm3 portions of hydrochloric acid, stirring the mixture and recording the temperature after each addition. 6. Continue adding hydrochloric acid until a total of 50.00 cm3 of the acid has been added. Results Volume of HCl(aq) added / cm3 0.00 5.00 10.00 15.00 20.00 25.00 30.00 35.00 40.00 45.00 50.00 Temperature / C 22.2 23.7 25.1 26.6 28.0 29.5 29.2 28.4 27.6 26.8 26.0 A graph of the temperature (y-axis) against the volume of hydrochloric acid added (x-axis) enables the maximum temperature rise and the volume of acid required for neutralization to be determined. From this information it is possible to calculate (cid:3) (cid:3) (cid:120)(cid:3) (cid:120)(cid:3) the concentration of the hydrochloric acid the enthalpy change for the reaction. *P37682A0416*<br />
(a) (i) Plot a graph of temperature against volume of acid added on the axes below. Draw two straight lines on your graph and extrapolate the lines until they intersect. (2) PMT 323028Temperature / (cid:113)C 262422200 10 20 30 40 50 Volume of HCl(aq) / cm3 (ii) Use the extrapolated lines on your graph to read off the maximum temperature reached in the neutralization reaction. (2) Maximum temperature(cid:113)C (iii) The point at which the two extrapolated lines meet corresponds to the volume of hydrochloric acid required for neutralization. Read off this volume from your graph. (1) Volume of hydrochloric acidcm3 *P37682A0516* Turn over<br />
(iv) Calculate the number of moles of sodium hydroxide in 50.0 cm3 of a 1.00 mol dm3 solution. (v) HCl(aq) + NaOH(aq) (cid:111) NaCl(aq) + H2O(l) Use this equation and your answers to (iii) and (iv) to calculate the concentration of the hydrochloric acid in mol dm3. (b) (i) Use your graph and answer to (a)(ii) to calculate the maximum temperature change, (cid:507)T, for the reaction. (cid:507)T =C PMT (1) (2) (1) *P37682A0616*<br />
(ii) Use your value for the temperature rise, (cid:507)T, to calculate the heat energy produced when 50.0 cm3 of sodium hydroxide is exactly neutralized by the volume of hydrochloric acid you obtained in (a)(iii). Use the expression energy = total mass ofspecific heat capacity produced (J) solution of solution rise temperature[Assume the specific heat capacity of the solution to be 4.2 J g1 C1 and the density of the solution to be 1.0 g cm3] (iii) Use your answers to (a)(iv) and (b)(ii) to calculate the enthalpy change, in kJ mol1, for this reaction. Give your answer to two significant figures and include a sign. PMT (2) (3) (cid:507)H =kJ mol1 (Total for Question 3 = 14 marks) *P37682A0716* Turn over<br />](https://learnah.org/wp-content/uploads/2018/08/l_smalljpgJune_2011_QP_-_Unit_3_Edexcel_Chemistry_A-level_3-0.jpg)
![Question Number 3 (a) (i) Acceptable Answers Points correctly plotted (1) [allow one slip] Two best fit straight lines (1) Mark Acceptable Answers Reject If the two straight lines do not meet above 29.5 oC Extrapolation (1) Maximum temperature Read off candidates graph Expect value between 29.630.0(oC) (1) ALLOW a stated value within 0.1 (oC) of graphs max temp 6CH0701 1106 Question Number 3 (a) (ii)<br />
Question Number 3 (a) (iii) Question Number 3 (a) (iv) Question Number 3 (a) (v) Mark Mark Mark Acceptable Answers Reject Volume of HCl(aq): Read off candidates graph Expect value between 25.526.5 (cm3) ALLOW: a stated volume within 0.5 cm3 of graphs volume of HCl(aq) ALLOW: whole numbers (e.g. 27 if intersection is at 27.0) Acceptable Answers Reject 0.05(00) Acceptable Answers Reject Mark consequentially on (a)(iii) and (iv): Working e.g. 0.05(00)candidates volume of HCl(aq) in dm3 (1) Answer (1) e.g. for a volume of HCl(aq) = 26.0 cm3 [NOTE: ALLOW 26 for 26.0] 0.05(00)candidates volume of HCl(aq) in dm3 = 0.05(00) 0.026 = 1.92 (mol dm-3) Correct answer only using suitable volume of HCl(aq) scores both marks IGNORE s.f. including 1 s.f. (e.g. 2 (mol dm-3)) 6CH0701 1106<br />
Mark Mark Question Number 3 (b) (i) Read off candidates Acceptable Answers Reject Reject Incorrect rounding of figures graph: Expect T = (29.822.2) = 7.6 (oC) IGNORE s.f. IGNORE any signs NOTE: Initial temp is that from the graph Question Number 3 (b) (ii) Mark consequentially on Acceptable Answers (a)(iii): Working Answer (1) (1) ( e.g. [volume of HCl(aq) + 50(.00) x 4.2 x T] = 76 x 4.2 x 7.6 = 2426 (J) ) Assume 2.426 (or similar) is the answer in kJ NOTE: 2nd mark can be awarded cq on a wrong mass and/or a wrong T ( e.g. use of 50(.00) x 4.2 x 7.6 = 1596 (J) scores (1) ) IGNORE sig figs (except one sig fig) IGNORE any signs ALLOW use of 4.18 instead of 4.2 6CH0701 1106<br />
Question Number 3 (b) (iii) Acceptable Answers Reject Working Dividing as follows:- () answer to (b)(ii) in kJ or J answer to (a)(iv) Negative sign Answer to 2 s.f. (1) (1) (1) NOTE: The 2 s.f. mark is CQ on dividing kJ or J by mol e.g.2.426 0.05(00) =49 (kJ mol1) NOTE: must be to 2 s.f. Correct answer to correct s.f. gets 3 marks even with no working Correct answer to 2 s.f. with negative sign in J mol-1 (e.g.49000) scores max (2) Mark 6CH0701 1106<br />](https://learnah.org/wp-content/uploads/2018/08/June_2011_MS_-_Unit_3_Edexcel_Chemistry_A-level_3-0.jpg)
