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OCR A Jun 2015 Paper 1 Q4
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![10 4 A student was given 200 cm3 of solution X in which sodium hydroxide, NaOH, and sodium hydrogencarbonate, NaHCO3, had both been dissolved. The student carried out two different titrations on samples of solution X using 0.100 mol dm3 sulfuric acid, H2SO4.In the first titration, both NaOH and NaHCO3 were neutralised. In the second titration, only NaOH was neutralised. The students results for the titrations of 25.0 cm3 samples of solution X are shown. volume of H2SO4 needed to neutralise both NaOH and NaHCO3 volume of H2SO4 needed to neutralise only NaOH 29.50 cm3 18.00 cm3 2NaOH(aq) + H2SO4(aq) Na2SO4(aq) + 2H2O(l) 2NaHCO3(aq) + H2SO4(aq) Na2SO4(aq) + 2H2O(l) + 2CO2(g) (a) (i) Calculate the amount, in mol, of H2SO4 used to neutralise only the NaOH in 25.0 cm3 of solution X. (ii) Calculate the concentration, in mol dm3, of NaOH in solution X. Amount =mol [1] Concentration =mol dm3 [1]OCR 2015<br />
(b) (i) Calculate the amount, in mol, of NaHCO3 in the 200 cm3 of solution X. 11 Amount =mol [2] (ii) Calculate the mass of NaHCO3 in the 200 cm3 of solution X. Give your answer to three significant figures. Mass =g [1] [Total: 5]OCR 2015 Turn over<br />](https://learnah.org/wp-content/uploads/2018/08/OCR_1_QP_JUN15_question_4-0.jpg)
![F321 Mark Scheme June 2015 Question 4 (a) (i) Mol of H2SO4 = 0.100 x 18.00/1000 = 1.80 x 103 mol (cid:57) Answer (ii) Mol of NaOH in = 1.80 x 103 x 2 x 1000/25.0 = 0.144 mol dm3 (cid:57) (i) Check the answer line. If answer = 0.0184 mol award 2 marks Mol of NaHCO3 in 25.0 cm3 = [0.100 x 11.50/1000] x 2 = 0.00230 mol (cid:57) Mol of NaHCO3 in 200 cm3 = 0.00230 x 200/25.0 = 0.0184 mol (cid:57) Mark Guidance ALLOW calculator value or rounding to 2 significant figures or more but IGNORE trailing zeroes throughout Q4. eg 0.200 is allowed as 0.2 ALLOW ECF for (a)(i) x 2 x 1000/25 If there is an alternative answer, check to see if there is any ECF credit possible using working below. ALLOW for an alternative method for M1 Total mol of H2SO4 used = [0.100 x 29.50/1000] = 0.00295 mol Mol of H2SO4 reacting with NaHCO3 = 0.00295answer to (a)(i) Expected answer = .002950.00180 = 0.00115 mol Mol of NaHCO3 in 25.0 cm3 = 0.00115 x 2 = 0.00230 mol ALLOW ECF for mol of NaHCO3 x 200/25.0 For ECF in M2 titration values of 11.50 or 29.50 must have been used in M1 Second marking point is for scaling up number of mol of NaHCO3 by 200/25.0 (Usually seen as 8) ALLOW ECF for (b)(i) x 84.0 correctly calculated and rounded to three significant figures. (b) (ii) Mass of NaHCO3 = 0.0184 x 84.0 = 1.55 g (cid:57) (must be three significant figures) Total 10<br />](https://learnah.org/wp-content/uploads/2018/08/OCR_1_QP_JUN15_answer_4.jpg)