Zadanie 2e

4 pktobliczenioweśrednie (szac.)Paper 2, maj 2022, TZ1

Wzorowane na: Chemistry SL Paper 2, May 2022, TZ1, pytanie 2(e). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.

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Informacja do zadań 2a-2e.

Methanol, CH3OH, is an important industrial chemical used to make plastics, solvents and fuels. It is also the starting material for methylamine, CH3NH2, a gas used in the manufacture of pesticides and medicines.

Polecenie

Methylamine is a gas that dissolves in water to form an alkaline solution:

CH3NH2(aq) + H2O(l) ⇌ CH3NH3+(aq) + OH−(aq)

(i) State the relationship between CH3NH3+ and CH3NH2 in terms of the Brønsted-Lowry theory.

(ii) A sample of methylamine gas, volume 120.0 dm3 at 295.0 K and 102.0 kPa, dissolves in water and the solution is made up to 2.50 dm3. Determine the concentration of the solution, in mol dm−3. Use sections 1 and 2 of the data booklet.

(iii) Calculate the concentration of hydroxide ions in a methylamine solution with pH = 11.6. Use sections 1 and 2 of the data booklet.

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Odpowiedź z klucza

(i) conjugate acid and base / conjugate acid-base pair «CH3NH3+ is the conjugate acid of CH3NH2» ✔

(ii) amount of methylamine n=PVRT=102.0 kPa×120.0 dm38.31 J K−1 mol−1×295.0 Kn = \dfrac{PV}{RT} = \dfrac{102.0\ \mathrm{kPa} \times 120.0\ \mathrm{dm^3}}{8.31\ \mathrm{J\,K^{-1}\,mol^{-1}} \times 295.0\ \mathrm{K}} = 4.99 «mol» ✔

concentration c=nV=4.99 mol2.50 dm3c = \dfrac{n}{V} = \dfrac{4.99\ \mathrm{mol}}{2.50\ \mathrm{dm^3}} = 2.00 «mol dm−3» ✔

(iii) [OH−] «=Kw[HX+]=1.00×10−1410−11.6= \dfrac{K_w}{[\ce{H+}]} = \dfrac{1.00 \times 10^{-14}}{10^{-11.6}}» = 10−2.4 = 4.0 × 10−3 «mol dm−3» ✔

Schemat punktowania

(i) [1] for “conjugate” acid-base pair; the acid and the base must be correctly assigned if stated. (ii) M1: amount of gas from the ideal gas equation; M2: concentration. Award [2] for a correct final answer. ECF from M1. (iii) [1].

Umiejętności w zadaniu