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Zadanie 1a

15 pktobliczeniowetrudne (szac.)Paper 2, maj 2026, TZ2

Wzorowane na: Chemistry HL Paper 2, May 2026, TZ2, pytanie 1(a). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.

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Polecenie

Hydrochloric acid, HCl(aq), and hydroiodic acid, HI(aq), ionize completely in water, whereas propanoic acid, CH3CH2COOH(aq), ionizes only partly.

25.0 cm3 of 0.200 mol dm−3 HCl(aq) was titrated with sodium hydroxide solution, NaOH(aq).

(i) Calculate the pH of the hydrochloric acid before any alkali is added. Use section 1 of the data booklet.

(ii) Write the equation for the neutralization of the hydrochloric acid by the sodium hydroxide solution.

(iii) Calculate how many moles of NaOH are needed to neutralize all the hydrochloric acid in the flask.

(iv) Calculate the volume of 0.180 mol dm−3 NaOH(aq) that must be added to reach the equivalence point.

(v) Suggest one test, and the result you would expect, that would allow you to tell apart unlabelled solutions of HCl(aq) and CH3CH2COOH(aq) of the same concentration.

(vi) Deduce the formula of the conjugate base of propanoic acid and its pKb, given that pKa (CH3CH2COOH) = 4.87.

(vii) A 25.0 cm3 sample of 0.200 mol dm−3 CH3CH2COOH(aq) is exactly neutralized by 21.0 cm3 of NaOH(aq) of a different concentration. Calculate the pH of the solution formed. Assume that the volumes of dilute solutions are additive. Kb=[CHX3CHX2COOH][OHX−][CHX3CHX2COOX−]=7.41×10−10K_\mathrm{b} = \dfrac{[\ce{CH3CH2COOH}][\ce{OH-}]}{[\ce{CH3CH2COO-}]} = 7.41 \times 10^{-10}

(viii) Use section 18 of the data booklet to suggest a suitable indicator for the titration of propanoic acid with sodium hydroxide solution.

(ix) The indicator bromothymol blue can be represented as HInd; the undissociated form HInd is yellow and the anion Ind− is blue. Explain, using equations, why a few drops of the indicator turn yellow in acidic solution and blue in alkaline solution. Give the equilibrium that exists in the solution of the indicator, and then explain the effect of adding H+ and the effect of adding OH−.

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(i) «[H+] = 0.200 mol dm−3» pH = 0.699 ✔

(ii) HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) ✔

(iii) «n(NaOH) = n(HCl) = 0.200 × 0.0250 =» 5.00 × 10−3 «mol» ✔

(iv) «V=5.00×10−30.180=V = \dfrac{5.00 \times 10^{-3}}{0.180} =» 0.0278 «dm3» / 27.8 «cm3» ✔

(v) ALTERNATIVE 1: Method: measure the pH / use a pH meter / universal indicator ✔ Expected result: lower pH / red for HCl OR higher pH / orange or yellow for CH3CH2COOH ✔

ALTERNATIVE 2: Method: add a carbonate or hydrogencarbonate / a reactive metal ✔ Expected result: HCl gives bubbles of gas faster / more vigorously ✔

ALTERNATIVE 3: Method: measure the electrical conductivity ✔ Expected result: HCl has the higher conductivity ✔

ALTERNATIVE 4: Method: measure the temperature change on adding the same volume of alkali / a base / a metal oxide ✔ Expected result: HCl gives the greater temperature change ✔

(vi) Formula of conjugate base: CH3CH2COO− ✔

pKb: «14.00 − 4.87 =» 9.13 ✔

(vii) «n(CH3CH2COO−) = n(CH3CH2COOH) = 0.200 × 0.0250 = 5.00 × 10−3 mol; total volume = 25.0 + 21.0 = 46.0 cm3»

[CH3CH2COO−] «=5.00×10−30.0460== \dfrac{5.00 \times 10^{-3}}{0.0460} =» 0.109 «mol dm−3» ✔

[OH−] «=7.41×10−10×0.109=8.08×10−11== \sqrt{7.41 \times 10^{-10} \times 0.109} = \sqrt{8.08 \times 10^{-11}} =» 8.99 × 10−6 «mol dm−3» ✔

«pOH = 5.05» pH = 8.95 ✔

(viii) phenolphthalein «pH range 8.3–10.0 includes the pH at the equivalence point, 8.95» ✔

(ix) Equilibrium: HInd(aq) ⇌ H+(aq) + Ind−(aq) «yellow ⇌ blue» ✔

Adding H+: the equilibrium shifts to the left / H+ + Ind− → HInd OR more of the yellow form is present ✔

Adding OH−: the equilibrium shifts to the right / HInd + OH− → Ind− + H2O OR more of the blue form is present ✔

Schemat punktowania

(i) [1] 0.699; accept 0.70. (ii) [1] balanced symbol equation; state symbols ignored. (iii) [1]. (iv) [1] accept 0.0278 dm3 or 27.8 cm3. (v) [2] M1 a valid method; M2 the expected result, which must match the method and must compare the two acids; accept specific bases or metals in alternatives 2 and 4. (vi) [2] M1 formula, accept C2H5COO−; M2 pKb = 9.13. (vii) [3] M1 concentration of the salt using the total volume; M2 [OH−] from Kb; M3 pH. Award [3] for a correct final answer, 8.95; ECF from M1 and M2. (viii) [1] phenolphthalein only; do not accept phenol red, whose range ends at 8.4. (ix) [3] M1 the equilibrium with the two colours identified; M2 effect of adding H+; M3 effect of adding OH−.

Komentarz

In (vii) the volume to use is the total volume of the mixture, 46.0 cm3, and the concentration of the alkali is not needed; the 5.00 × 10^-3 mol of salt is diluted by the whole mixture.

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