Question

Question B2. Answer ALL parts from (a) to (c). (a) A base is added to an acid: ) Sketch the pH curves for both a strong base added to a weak acid, and for a strong base added to a strong acid. Note the differences between these two curves and briefly account for why they would be different. (4 marks) (ii) The end point of a titration was reached when 16 cm of a 0.25 mol dm-1 solution of sodium hydroxide (NaOH) was titrated against 100 cm of an unknown, monoprotic acid. Calculate the concentration of the unknown acid, and therefore the pH of the original acid solution. (4 marks) The unknown acid solution was made by dissolving 1.6 g of a white powder Into 500 cm3 of water. Identify the white powder knowing that it is either hydrogen chloride (HCI), hydrogen bromide (HBr) or hydrogen iodide (HI). (4 marks) (b) The Henderson equation for buffer solutions is given by: pH = pKq + log A solution was made up by adding 35 cm of 0.1 mol dm-3 ethanoic acid, and 15 cm3 of 0.1 mol dm-3 sodium ethanoate. Use the Henderson equation to calculate the pH of this solution knowing that the pka of ethanoic acid is 4.8. (c) (5 marks) (ii) With reference to the Henderson equation, above, briefly account for why the solution made up in part (i) would resist changes in pH better than a solution of only that contained only an acid. (3 marks) The following question concerns solutions of benzoic acid, which has a mass of 122.12 g mol-1, and a Ka of 6.3 x 10-5 mol dm-3. (0) Write an equation that can be used to determine Ka from the concentration(s) of benzoic acid. (3 marks) (II) Calculate the amount of benzoic acid required to make a 500 cm3 solution with pH 3.3. (5 marks) (400467) Page 2 of 2

Answer 1

acid Base (ii) at the end point (equivalent) = (equivalent) eq = NV for MoDo Basicaid, Mono acidic Base N i eg=MV Mava = Molo → Max100 = 0.25 x 16 Ma=4x10² [H] = 4x10 pH = -log[H] = log [Hx102) – 2-1094 = lol GD M = W Mwl 500 mW COM > 4x10= 6 x 100 =MW=30 Molecular weight go is For HBY Hence the unknown acid is HBr 6. CH₃COOH + CH3COONa it is an acidic Butter wearaid pH = pka + log h (Moes = A. Vit Ma V2 Tacid) Vit pH = 4.8+ log (15 xool 4.8+log) = 4.13 Salt SOIL 15+35 135×Dol "15+35 Не се р* of Bukes іѕ ц.43 () Explamantion for Resisting of change in lu CH3COOH + CHCOONa CH₂COOH CH₃Coot HT (H ₂ COUNa Hzroo + Nat Due to common in effect, degree of dissociation of CH₂COOH is neglizable, Hence all CH3100 are ions are produced due to CH3COONa Chacoo +H+ + CH3COOH (Undissociated + Hlions added to the above 'B H + oH CH3cO0+th Cundissociated OVC Solution they combine with CH3C00 to form undissociated acid, hence no change in pH →is on ions added to the above solution they combine with CH3COOH to form undissociated water, hence no change in ptt

(c) i phoot Phoo + H+ C coca 0 ca 0 ca Ka= (CA) (C4) (C) (( 2 ) [phcoo] [H) [phCOoh) (12) C-CL for an weakadd x441=1-421 I-a ka=CK [ht] = ca cka = (²x ². (ka= (C4) = [ht] [[HT] =Vcka 109(ax 10°) 3 = 10910 + 1092 - - CC ii) pt = 3.3 = 3+0.3 = 10910° +1092 = 1 pH = -109[*+] = log [t)) = log(2x103) → [H+] = 0.5X10" = 5x107 We know [4t) = Vicka 5x104 = 11 cx 6.3x1050 Squaning on Both side >> 25x108 = CX 63x10° (= 3.96x103 = w x 1000 - 3.96410 mo Vmi) 12v2 100 = 3.96X163 W = 241.79 x 10 g at W = 0.241grams (approximately)

me completly ionized. ally pl curve of Strongacid and strong Base ler (Hel) (NaOH) → In this case. Both the titoants are completly on Helt Naot NocltHO penantiano > 5 the Oh added to an add solution, Hjacca and [0w) increases + pH goes on increasing as the equivalence point is reached, [Ht rapidly rede Above this point [of increases rapidly, so that pe ( Ge MA rapidly reduced Hincreases Faisly e equivalence point Let to 5 10 15 15 20 25 pt Cuove of weakacid and Stoong Base Let (CH3COOH) (NAOH) Intially Just by adding Few drops of on are added to the Kleakacid, it forms an acidic Buffer ((1600H(CH3COONa), thus pH of the solution changes slightly, almost constant at Equivalence point solution will be alkaline, Pt OH addition is continued after the equivalence point, [ouincreases sapidly, thus pit Pncreases rapidly- somation se equivalence point COTTOLO + Butterfomation 2 pH const let 1 II 10 15 20