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Edexcel Jan 2013 Paper 4 Q16
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![16 Methanoic acid, HCOOH, is present in ant stings. A scientist analyzed 25.0 cm3 of an aqueous solution of methanoic acid, solution Z, by titrating it with dilute sodium hydroxide, NaOH(aq).20.0 cm3 of sodium hydroxide was required to neutralize the methanoic acidThe equation for the neutralization of methanoic acid is HCOOH(aq) + NaOH(aq) o HCOONa(aq) + H2O(l) (a) (i) Give the expression for Kw, the ionic product of water. (ii) The concentration of the sodium hydroxide, NaOH(aq), used in the titration was 0.00750 mol dm3. Calculate the pH of the sodium hydroxide solution. [Kw = 1.00 x 1014 mol2 dm6 ] (b) Use the equation for the reaction and the data from the titration to show that the concentration of the methanoic acid in solution Z was 6.00103 mol dm3. 14 *P41214A01428* (1) (2) (2)<br />
(c) Methanoic acid is a weak acid. (i) Explain the term weak acid. (2) WeakAcid(ii) The equation for the dissociation of methanoic acid in aqueous solution is shown below. HCOOH(aq) U HCOO(aq) + H+(aq) Write the expression for the acid dissociation constant, Ka, for methanoic acid. (1) *P41214A01528* 15 Turn over<br />
*(iii) At 298 K, the acid in ant stings has a concentration of 6.00103 mol dm3 and a pH of 3.01. Calculate the value of Ka for methanoic acid at 298 K. State clearly any assumptions that you have made. Calculation: (4) Assumption(s):(Total for Question 16 = 12 marks) 16 *P41214A01628*<br />](https://learnah.org/wp-content/uploads/2018/08/l_smalljpgJanuary_2013_QP_-_Unit_4_Edexcel_Chemistry_A-level_16-0.jpg)
![Mark Mark Question Number 16(a)(i) Kw = [H+] x [OH] Acceptable Answers OR Kw = [H3O+] x [OH] State symbols are not required IGNORE any incorrect state symbols Reject Inclusion of [H2O] Question Number 16(a)(ii) FIRST, CHECK THE FINAL ANSWER Acceptable Answers Reject IF answer pH = 11.875 / 11.88 / 11.9 /12 award 2 marks IGNORE sf except 1 sf [H+] = Kw = 1.00 x 1014 [OH] 0.00750 = 1.3333 x 1012 = 1.33 x 1012 (1) (mol dm3) ALLOW first mark for just [H+] = Kw [OH] pH = log10 [H+] = 11.875 = 11.88 / 11.9 (1) OR pOH = log10 [OH] = 2.12 (1) pH = pKwpOH pH = 11.88 / 11.9 (1) Second mark only awarded CQ if pH between 8 and 14<br />
Reject Mark Acceptable Answers First mark Moles NaOH = 0.00750 x 20.0 1000 = 1.50 x 104 (mol) (1) (Since HCOOH : NaOH ratio is 1:1) Second mark [HCOOH(aq)] = 1.50 x 104 0.0250 OR = 1.50 x 104 x 1000 25.0 (1) (= 6.00 x 103 mol dm3) ALTERNATIVE APPROACH: Use of an expression such as 0.00750 x 20.0 = 25 x y (1) y = 0.00750 x 20.0 25 (1) Mark Acceptable Answers Reject not easily dissociated (Weak) dissociates / ionizes to a small extent OR dissociate / ionizes partially OR dissociates / ionizes incompletely OR does not fully dissociate / ionize OR forms an equilibrium when reacted with water (1) (Acid) proton donor ALLOW proton donator OR produces / releases H+ ions OR produces / releases H3O+ ions (1) Ignore reference to typical acid reactions Question Number 16(b) Question Number 16(c)(i)<br />
Mark Mark Acceptable Answers Reject (Ka =) [HCOO] [H+] [HCOOH] State symbols are NOT required IGNORE any incorrect state symbols (Ka = ) [H+]2 [HCOOH] Inclusion of [H2O] Acceptable Answers Reject If incorrect units max 1 Just partial / incomplete Orno dissociation IGNORE sf except 1 sf THROUGHOUT FIRST, CHECK THE FINAL ANSWER IF answer Ka = 1.59 x 10-4 (mol dm-3) award the first two 2 marks [H+] (= 10-pH = 10-3.01) = 9.77 x 10-4 (mol dm-3) (1) Ka = [H+]2 [HCOOH] Ka = (9.77 x 10-4)2 6.00 x 103 = 1.59 x 10-4 (mol dm-3) (1) Assumption 1 [H+] = [HCOO] OR no H+ from the (ionization of) water OR H+ only from the acid (1) Assumption 2 Ionization of the (weak) acid is negligible / very small / insignificant OR [HCOOH]initial x = [HCOOH]eqm OR [HCOOH]eqm = [HCOOH]initial OR [HCOOH]eqm = 6.00 x 103 (mol dm3) OR [H+] << [HA] (1) Assumptions can be in either order Question Number 16(c)(ii) Question Number 16(c)(iii)<br />
(Total 12 marks) 16(c)(iii) contd OR [H+] (= 10-pH = 10-3.01) = 9.77 x 10-4 (mol dm-3) (1) Ka = [H+]2 [HCOOH] Ka = (9.77 x 10-4)2 (6.00 x 1039.77 x 10-4) (1) = 1.90 x 10-4 (mol dm-3) (1) Assumption [H+] = [HCOO] OR no [H+] from the (ionization of) water OR H+ only from the acid (1) Ignore references to constant temperature<br />](https://learnah.org/wp-content/uploads/2018/08/January_2013_MS_-_Unit_4_Edexcel_Chemistry_A-level_16-0.jpg)
