Multiple choice, drill bank
Which volume gives c(unknown). c(unknown) = n ÷ V(flask). The flask volume is the aliquot of the unknown solution.
Titration back-calculation, 1:1 ratio. n(HCl) = 0.200 × 0.0150 = 0.00300; ratio 1:1; c(NaOH) = 0.00300 ÷ 0.0250 = 0.120 mol/L.
Titration back-calculation, 2:1 ratio. n(NaOH) = 0.100 × 0.020 = 0.00200; ratio 2:1; n(H₂SO₄) = 0.00100; c = 0.00100 ÷ 0.025 = 0.0400 mol/L.
Averaging concordant titres. Discard rough (22.1). Average T1, T2, T3 = (23.4+23.5+23.3)÷3 = 23.4 mL.
Concentration of a primary standard solution. n = 0.212 ÷ 105.99 = 2.000×10⁻³; c = 2.000×10⁻³ ÷ 0.250 = 0.00800 mol/L.
Judging whether titres are concordant. The rough titre (18.4) must be discarded. T1 = 19.8, T2 = 19.6, T3 = 20.1. Range = 20.1 − 19.6 = 0.5 mL, which exceeds 0.1 mL, not strictly concordant. In practice, students should ideally repeat to get truly concordant results. If forced to average T1–T3: (19.8+19.6+20.1)÷3 = 19.83 mL. The student's inclusion of the rough titre in their average is an error regardless.
Which solution goes in the burette. In a back-calculation titration, the unknown solution (acetic acid) goes in the flask (aliquot) and the standard solution (standardised NaOH) goes in the burette. The titre of NaOH gives n(NaOH), then the mole ratio gives n(acetic acid), and dividing by the flask volume gives c(acetic acid). Any option that puts the unknown in the burette and the standard in the flask has the two swapped.
Short answer model answers
Q1 (5 marks):
(a) Discard rough (19.8). T1=21.28, T2=21.32, T3=21.30, all within 0.10 mL of each other (range 0.04 mL), concordant. Average = (21.28+21.32+21.30)÷3 = 21.30 mL.
(b) n(Na₂CO₃) = 0.424 ÷ 105.99 = 4.001×10⁻³; c = 4.001×10⁻³ ÷ 0.100 = 0.04001 mol/L. n(aliquot) = 0.04001 × 0.0250 = 1.000×10⁻³; ratio 1:2; n(HCl) = 2.000×10⁻³. c(HCl) = 2.000×10⁻³ ÷ 0.02130 = 0.0939 mol/L.
Q2 (4 marks):
(a) MM(AgCl) = 143.32; n = 1.148 ÷ 143.32 = 8.010×10⁻³; ratio 1:1; n(AgNO₃) = 8.010×10⁻³. c(AgNO₃) = 8.010×10⁻³ ÷ 0.0400 = 0.200 mol/L.
(b) If NaCl is not in excess, some Ag⁺ ions would remain in solution unreacted. The precipitate mass would be less than the amount corresponding to all the AgNO₃ present. The calculated n(AgNO₃) would be too small, giving an underestimate of c(AgNO₃). Excess NaCl ensures all Ag⁺ is precipitated, so the mass of AgCl accurately reflects the full quantity of AgNO₃.
Q3 (6 marks):
(a) Discard rough (24.5). T1 = 25.78, T2 = 25.82, T3 = 25.80. Range = 25.82 − 25.78 = 0.04 mL ≤ 0.10 mL, so all three are concordant. Average = (25.78+25.82+25.80) ÷ 3 = 25.80 mL.
(b) n(H₂C₂O₄) = 0.756 ÷ 126.07 = 5.997×10⁻³ mol; c = 5.997×10⁻³ ÷ 0.250 = 0.02399 mol/L. n(aliquot) = 0.02399 × 0.0250 = 5.997×10⁻⁴; ratio 1:2; n(NaOH) = 1.199×10⁻³. c(NaOH) = 1.199×10⁻³ ÷ 0.02580 = 0.04647 mol/L ≈ 0.0465 mol/L.
(c) Oxalic acid dihydrate is suitable as a primary standard because it has a high molar mass (126.07 g/mol), which reduces the relative error in weighing; it is stable, non-hygroscopic, available in high purity, and does not absorb CO₂ or water vapour from the air appreciably during weighing.
Q4 (4 marks):
(a) If 0.250 mL is used instead of 25.0 mL, the denominator is 100 times too small (0.000250 L instead of 0.025 L). Since c = n ÷ V, a smaller V gives a larger c. The calculated concentration will be 100× too high (a factor of 100 error).
(b) A chemist could detect this error by: (i) checking the units, 0.250 mL is implausibly small for a pipette measurement; a standard pipette delivers 25.0 mL not 0.250 mL; (ii) comparing the calculated concentration with expected values for a typical solution (e.g., a concentration of 40 mol/L for NaOH is physically impossible, as NaOH is only soluble to about 25 mol/L); (iii) checking the recorded experimental data against the flask volume stated in the method.
