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Zadanie 1c

9 pktobliczeniowetrudne (szac.)Paper 2, listopad 2022

Wzorowane na: Chemistry HL Paper 2, November 2022, pytanie 1(c). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.

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Informacja do zadań 1a-1f.

Ammonium chloride, NH4Cl, is a white crystalline solid used as a nitrogen fertilizer for rice crops and as an electrolyte in dry cells.

Polecenie

A 0.50 mol dm−3 solution of ammonium chloride is prepared.

(i) Calculate the pH of an ammonium chloride solution with [H3O+] = 1.68 × 10−5 mol dm−3. Use section 1 of the data booklet.

(ii) Write the equation for the neutralization of the ammonium chloride solution by sodium hydroxide solution.

(iii) A 25.00 cm3 sample of the 0.50 mol dm−3 solution of ammonium chloride is titrated with a 0.50 mol dm−3 solution of sodium hydroxide. The pKb of ammonia is 4.75. Determine the pH at the equivalence point, to two decimal places. Use section 1 of the data booklet.

(iv) Sketch the pH curve for the titration in (iii), up to a total of 50 cm3 of sodium hydroxide solution added, on the axes provided.

empty grid with pH from 0 to 14 on the vertical axis and the volume of NaOH(aq) added from 0 to 50 cm3 on the horizontal axis

(v) State, with a reason, if phenol red is an appropriate indicator for this titration. Use section 18 of the data booklet.

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Odpowiedź z klucza

(i) «pH = −log(1.68 × 10−5) =» 4.77 ✔

(ii) NH4+(aq) + OH−(aq) → NH3(aq) + H2O(l)

OR

NH4Cl(aq) + NaOH(aq) → NH3(aq) + H2O(l) + NaCl(aq) ✔

(iii) «n(NH4+) = 0.50 mol dm−3 × 0.02500 dm3 =» 0.0125 «mol» ✔

«at the equivalence point 25.00 cm3 of NaOH has been added, so the total volume is 50.00 cm3»

«[NHX3]=0.0125 mol0.05000 dm3[\ce{NH3}] = \dfrac{0.0125\ \mathrm{mol}}{0.05000\ \mathrm{dm^3}} =» 0.25 «mol dm−3» ✔

«Kb = 10−4.75 = 1.78 × 10−5»

«[OHX−]=Kb[NHX3]=1.78×10−5×0.25[\ce{OH-}] = \sqrt{K_\mathrm{b}[\ce{NH3}]} = \sqrt{1.78 \times 10^{-5} \times 0.25} =» 2.1 × 10−3 «mol dm−3» ✔

«pOH = −log(2.1 × 10−3) = 2.68»

«pH = 14.00 − 2.68 =» 11.32 ✔

(iv) non-symmetrical S-shaped curve starting at a pH between 2 and 7 «about 4.8» AND levelling off at a pH above 12 ✔

equivalence point at a pH of approximately 11 AND at a volume of 25 cm3 ✔

(v) no AND the equivalence point «pH 11.3» does not lie within the pH range of the indicator «6.8–8.4»

OR

no AND there is no sharp rise in pH within the colour-change range of phenol red ✔

Schemat punktowania

(i) 1 mark. (ii) 1 mark; accept NH4OH instead of NH3 + H2O. (iii) Award [4] for the correct final answer (accept 11.31–11.33). M1: amount of ammonium ion; M2: concentration of ammonia at the equivalence point (total volume 50.00 cm3); M3: [OH−] from Kb; M4: pOH converted to pH. ECF from M1 and M2. (iv) M1: curve shape with start and end pH; M2: equivalence point at 25 cm3 and pH about 11 (accept 10–12). (v) 1 mark; the conclusion and the reason are both needed.

Komentarz

In (iii) the ammonium ion is the weak acid and ammonia the product: use Kb of ammonia for the conjugate base, not Ka of the ammonium ion.

Umiejętności w zadaniu