Zadanie 1b, Paper 2, maj 2025, TZ2, Chemia IB HL
Wzorowane na: Chemistry HL Paper 2, May 2025, TZ2, pytanie 1(b). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.
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Zadanie
Wspólne informacje do wiązki zadań
Hydrogen cyanide, HCN, is an extremely poisonous compound that is nevertheless manufactured on a large scale, because it is a starting material for polymers such as nylon and poly(methyl methacrylate).
Polecenie
In aqueous solution hydrogen cyanide behaves as a weak acid.
(i) Write an equation for the ionization of hydrogen cyanide in water.
(ii) Outline two ways in which a solution of 0.050 mol dm−3 HCN could be distinguished experimentally from a solution of 0.050 mol dm−3 hydrobromic acid, HBr.
(iii) Calculate the pH of a 0.160 mol dm−3 solution of HCN, given that Ka = 6.25 × 10−10 at this temperature.
(iv) Sketch, on the axes below, how the pH changes when 25.0 cm3 of 0.250 mol dm−3 sodium hydroxide solution is added a little at a time to 25.0 cm3 of 0.160 mol dm−3 HCN.
Powiększ(v) Deduce where on your sketch the solution consists of NaCN(aq) alone, and mark this point with the label "Salt".
(vi) The enthalpy of neutralization of HCN by NaOH is −12.1 kJ mol−1. Calculate the rise in temperature that occurs when 25.0 cm3 of 0.250 mol dm−3 NaOH is mixed with 25.0 cm3 of 0.160 mol dm−3 HCN. Use sections 1 and 2 of the data booklet.
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Pokaż rozwiązanieUkryj rozwiązanieKlucz i punktacja
Odpowiedź z klucza
(i) HCN(aq) ⇌ H+(aq) + CN−(aq) ✔
OR
HCN(aq) + H2O(l) ⇌ H3O+(aq) + CN−(aq) ✔
(ii) Any two of:
measure the pH «with a meter or universal indicator» AND the HCN solution has the higher pH ✔
measure the electrical conductivity AND the HCN solution is the poorer conductor ✔
add a reactive metal / a metal carbonate / magnesium AND the HCN solution reacts more slowly «fewer bubbles per second» ✔
titrate with a strong base AND the HCN titration has an equivalence point above pH 7 / a smaller temperature rise ✔
(iii) «mol dm−3» ✔
pH «= −log10(1.00 × 10−5)» = 5.00 ✔
(iv) sigmoidal curve AND equivalence point above pH 7 ✔
curve starts between pH 4 and 6 «pH 5.0, from (iii)» AND finishes between pH 11 and 14 «about 12.7» ✔
steep, almost vertical rise at 16.0 cm3 «n(HCN) = 0.0250 × 0.160 = 4.00 × 10−3 mol, V = 4.00 × 10−3/0.250» ✔
(v) equivalence point «on the vertical section at 16.0 cm3» labelled "Salt" ✔
(vi) amount neutralized «= n(HCN) = 0.0250 × 0.160» = 4.00 × 10−3 mol «HCN is limiting; n(NaOH) = 6.25 × 10−3 mol» ✔
= 0.232 «K / °C» ✔
Schemat punktowania
(i) [1] either equation with the equilibrium arrow. (ii) [2 max] one mark per method together with its observation; award [1] for two correct methods without observations; answers must refer to rates, not amounts. (iii) [2] M1 [H+]; M2 pH; award [2] for a correct final answer. (iv) [3] M1 shape and equivalence pH above 7; M2 starting and finishing pH; M3 vertical section at 16.0 cm3. (v) [1] the label at the equivalence point of the student's own curve. (vi) [2] M1 amount of HCN neutralized; M2 temperature rise; award [2] for a correct final answer; ECF from M1.
Komentarz. In (vi) the acid is the limiting reactant, so the excess 2.25 × 10^-3 mol of NaOH releases no heat, while the whole 50.0 g of solution is warmed.