Zadanie 1f, Paper 1B, listopad 2025, TZ1, Chemia IB HL

7 pktobliczeniowetrudne (szac.)Paper 1B, listopad 2025, TZ1

Wzorowane na: Chemistry HL Paper 1B, November 2025, TZ1, pytanie 1(f). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.

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Zadanie

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For a titration, a student had to make up 250.0 cm3 of aqueous potassium hydroxide, KOH, with a concentration of 0.1500 mol dm−3, starting from the solid compound.

Polecenie

The student then used the potassium hydroxide solution standardized in 1c to titrate ethanoic acid, CH3COOH, whose concentration was not known.

(i) Neutralizing 25.00 cm3 of the acid required 40.00 cm3 of the standardized potassium hydroxide solution. Calculate the concentration of the ethanoic acid. If you have no value for the concentration of potassium hydroxide from 1c, take it to be 0.1380 mol dm−3 (this is not the correct value).

(ii) The pKa of ethanoic acid is 4.76. Determine the pH of the ethanoic acid solution whose concentration you found in (i). If you have no value from (i), take the concentration to be 0.1980 mol dm−3 (this is not the correct value).

(iii) Evaluate the use of a pH probe instead of an indicator to detect the end point of the titration of ethanoic acid with potassium hydroxide.

(iv) State the name of an indicator from section 18 of the data booklet that would be suitable for the titration of ethanoic acid with potassium hydroxide.

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Odpowiedź z klucza

(i) «CH3COOH + KOH → CH3COOK + H2O; c(CHX3COOH)=40.00 cm3×0.1485 mol dm−325.00 cm3=c(\ce{CH3COOH}) = \dfrac{40.00\ \mathrm{cm^3} \times 0.1485\ \mathrm{mol\,dm^{-3}}}{25.00\ \mathrm{cm^3}} =» 0.2376 «mol dm−3» ✔

(ii) «Ka = 10−4.76 = 1.74 × 10−5; [HX+]=Kac=1.74×10−5×0.2376=[\ce{H+}] = \sqrt{K_\mathrm{a} c} = \sqrt{1.74 \times 10^{-5} \times 0.2376} =» 2.03 × 10−3 «mol dm−3» ✔

«pH = −log10(2.03 × 10−3) =» 2.69 ✔

(iii) Any three of:

the probe gives more precise/accurate readings «of pH» ✔

no personal judgement of a colour change is needed with the probe ✔

the probe has to be calibrated before use/its readings can drift/it loses sensitivity with time «an indicator needs no calibration» ✔

the probe is fragile ✔

the probe can be reused for many titrations ✔

the probe is more expensive ✔

the probe works whatever the pH at the equivalence point, so no indicator has to be matched to the equivalence point ✔

the probe monitors the pH continuously, so the whole pH curve can be recorded ✔

the probe adds no chemical to the reaction mixture ✔

(iv) phenolphthalein ✔

OR

phenol red ✔

Schemat punktowania

(i) [1]: 0.2376 mol dm-3; accept 0.2208 mol dm-3 if 0.1380 mol dm-3 is used; ECF from 1c. (ii) M1: [H+] = 2.03 × 10-3 mol dm-3 (accept 1.85 to 1.86 × 10-3 for 0.1980 mol dm-3); M2: pH = 2.69 (accept 2.73 for 0.1980 mol dm-3); award [2] for a correct final answer; ECF from (i). (iii) [3 max], one mark per point; accept the converse statements written about the indicator; do not accept 'the indicator can give a wrong end point' without a reason. (iv) [1]: phenolphthalein or phenol red; an indicator whose range lies below pH 7.6 (methyl orange, bromophenol blue, bromocresol green, methyl red, bromothymol blue) earns nothing.

Komentarz. At the equivalence point the flask contains potassium ethanoate, whose anion is a weak base, so the pH is about 9 and the indicator must change colour in the basic range; the weak-acid approximation [H+] = sqrt(Ka c) is valid here because the degree of ionization is under 1 %.

Umiejętności w zadaniu

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