Zadanie 5a

(i) ALTERNATIVE 1:

«[OH⁻] = 0.150 mol dm⁻³; pOH = −log10 0.150 =» 0.824 ✔

«pH = 14.00 − 0.824 =» 13.18 ✔

ALTERNATIVE 2:

«$[\ce{H+}] = \dfrac{1.00 \times 10^{-14}}{0.150}$ =» 6.67 × 10⁻¹⁴ «mol dm⁻³» ✔

«pH = −log10(6.67 × 10⁻¹⁴) =» 13.18 ✔

(ii) CH₃COOH(aq) + KOH(aq) → CH₃COOK(aq) + H₂O(l) ✔

(iii) «n(KOH) = 0.150 × 0.0216 = 3.24 × 10⁻³ mol = n(acid); $[\text{acid}] = \dfrac{3.24 \times 10^{-3}}{0.0250}$» = 0.130 «mol dm⁻³» ✔

(iv) ALTERNATIVE 1:

«K_a = 10^−4.76 = 1.74 × 10⁻⁵»

«$[\ce{H+}] = \sqrt{K_\mathrm{a} \times [\text{acid}]} = \sqrt{1.74 \times 10^{-5} \times 0.130}$» = 1.50 × 10⁻³ «mol dm⁻³» ✔

pH «= −log10(1.50 × 10⁻³)» = 2.82 ✔

ALTERNATIVE 2:

«pH = ½(pK_a − log10[acid]) = ½(4.76 − log10 0.130)» ✔

2.82 ✔

(v) phenolphthalein ✔

«colour change range 8.3 to 10.0» contains the pH at the equivalence point, which is above 7 «a salt of a weak acid and a strong base; pH about 8.8» ✔

(i) Award [2] for the correct final answer. (ii) Accept the ionic or net ionic equation. (iv) Award [2] for the correct final answer; if 0.250 mol dm-3 was used, 2.68 scores [2]. (v) M1: phenolphthalein; M2: the colour change range includes the pH at equivalence, or equivalence pH above 7. Phenol red is not accepted because its range (6.8 to 8.4) ends below the equivalence pH of about 8.8. (v) Also accept phenol red when the answer links its range to the steep, vertical section of the titration curve (as the original markscheme did); phenolphthalein is the expected answer.