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Zadanie 1d

9 pktobliczenioweśrednie (szac.)Paper 2, listopad 2023, TZ2

Wzorowane na: Chemistry HL Paper 2, November 2023, TZ2, pytanie 1(d). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.

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Informacja do zadań 1a-1d.

Ethanoic acid, CH3COOH, is the carboxylic acid that gives vinegar its sour taste.

Polecenie

Ethanoic acid is a weak monobasic acid. The acid dissociation constant of ethanoic acid at 298 K is Ka = 1.74 × 10−5.

(i) A student dissolved a sample of ethanoic acid in water and made the solution up to 2.00 dm3. Exactly 19.4 cm3 of 0.100 mol dm−3 aqueous sodium hydroxide was needed to neutralize 25.0 cm3 of this solution, forming sodium ethanoate, CH3COONa. Calculate the mass of ethanoic acid in the sample.

(ii) Determine the pH of the 2.00 dm3 solution of ethanoic acid.

(iii) Using an equation, predict whether a solution of sodium ethanoate has a pH above 7, below 7 or equal to 7.

(iv) Explain why the two C–O bonds in the ethanoate ion have the same length, and state how that length compares with the C–O bonds in ethanoic acid.

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(i) «[CHX3COOH]=0.100 mol dm−3×19.425.0=[\ce{CH3COOH}] = 0.100\ \mathrm{mol\,dm^{-3}} \times \dfrac{19.4}{25.0} =» 0.0776 «mol dm−3» ✔

mass «= 2.00 dm3 × 0.0776 mol dm−3 × 60.06 g mol−1 =» 9.32 «g» ✔

(ii) «Ka=[HX+][CHX3COOX−][CHX3COOH]=[HX+]20.0776K_\mathrm{a} = \dfrac{[\ce{H+}][\ce{CH3COO-}]}{[\ce{CH3COOH}]} = \dfrac{[\ce{H+}]^2}{0.0776}» ✔

«[HX+]=1.74×10−5×0.0776=[\ce{H+}] = \sqrt{1.74 \times 10^{-5} \times 0.0776} =» 1.16 × 10−3 «mol dm−3» ✔

pH = 2.93 ✔

(iii) CH3COO−(aq) + H2O(l) ⇌ CH3COOH(aq) + OH−(aq) ✔

the ethanoate ion acts as a «Brønsted–Lowry» base AND pH > 7

OR

OH⁻ ions make the solution alkaline AND pH > 7 ✔

(iv) the π electrons of the C=O bond are delocalized over both C–O bonds

OR

the ion is a resonance hybrid of two equivalent structures ✔

each C–O bond is shorter than the single C–O bond AND longer than the C=O bond in ethanoic acid

OR

intermediate in length between a single and a double bond ✔

Schemat punktowania

(i) [2]: M1 concentration of the acid, M2 mass. Accept 9.3 g. The mass in the 25.0 cm3 sample only (0.117 g) earns [0]. (ii) [3]: M1 for the Ka expression with [H+] = [CH3COO−] and the acid concentration substituted, M2 for [H+], M3 for the pH. Award [3] for a correct final answer; accept 2.90 to 2.96 and other methods (solving the quadratic gives 2.94). ECF from (i). (iii) [2]: M1 for an equation showing the ethanoate ion (or sodium ethanoate) reacting with water, M2 for the base or alkaline reasoning together with pH > 7; the reason may be implied by the equation. Accept “formed from a strong base and a weak acid” for M2. The equation for the titration reaction does not earn M1. (iv) [2]: accept drawings of resonance structures for M1.

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