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Zadanie 4c

5 pktobliczeniowetrudne (szac.)Paper 2, maj 2022, TZ1

Wzorowane na: Chemistry HL Paper 2, May 2022, TZ1, pytanie 4(c). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.

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Informacja do zadań 4a-4c.

Methylamine, CH3NH2, is a gas that dissolves in water to form an alkaline solution:

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

Polecenie

The base dissociation constant of methylamine at 298 K is Kb = 4.4 × 10−4.

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

(ii) Calculate the concentration, in mol dm−3, of methylamine molecules in the solution with pH = 11.6.

(iii) A solution with high concentrations of both CH3NH2 and CH3NH3+ is a buffer because of the equilibrium:

CH3NH3+(aq) ⇌ CH3NH2(aq) + H+(aq)

Use this equilibrium to outline why adding a small volume of strong base hardly changes the pH of the buffer.

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Odpowiedź z klucza

(i) [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» ✔

(ii) Kb=[CHX3NHX3X+][OHX−][CHX3NHX2]K_b = \dfrac{[\ce{CH3NH3+}][\ce{OH-}]}{[\ce{CH3NH2}]} AND [CH3NH3+] = [OH−] ✔

[CH3NH2] «=(3.98×10−3)24.4×10−4= \dfrac{(3.98 \times 10^{-3})^2}{4.4 \times 10^{-4}}» = 0.036 «mol dm−3» ✔

(iii) «added» OH− reacts with H+ so the equilibrium shifts to the right / CH3NH3+ dissociates to replace the H+ ✔

«as both components are in large excess,» the ratio [CH3NH2] : [CH3NH3+] «and hence [H+] and the pH» is almost unchanged ✔

Schemat punktowania

(i) [1]. (ii) Award [2] for a correct final answer; accept 0.036-0.037 mol dm^-3 and other correct methods. M1: expression for Kb with the two products equal; M2: value. (iii) M1: shift of the equilibrium with the reaction of OH- and H+; M2: ratio of the buffer components almost unchanged. Accept “strong base converted to the weak base / water”.

Umiejętności w zadaniu