chemistry maths questions Flashcards
In a titration, 6.25 cm3 of a 0.0532 mol dm–3 solution of EDTA reacted completely with the calcium ions in a 150 cm3 sample of a saturated solution of calcium hydroxide.
Calculate the mass of calcium hydroxide that was dissolved in 1.00 dm3 of the calcium hydroxide solution.
Moles EDTA = 6.25 × 0.0532 / 1000 = (3.325 × 10–4)
Moles of Ca2+ in 1 dm3 = 3.325 × 10–4 × 1000 / 150 = (2.217 × 10–3)
Mass of Ca(OH)2 = 2.217 × 10–3 × 74.1 = 0.164 g
k
1s2 2s2 2p6 3s2 3p6 3d10
d sub-shell / shell / orbitals / sub-level full (or not partially full)
The vanadium in 50.0 cm3 of a 0.800 mol dm−3 solution of NH4VO3 reacts with 506 cm3 of sulfur(IV) oxide gas measured at 20.0 °C and 98.0 kPa.
Use this information to calculate the oxidation state of the vanadium in the solution after the reduction reaction with sulfur(IV) oxide.
Explain your working.
The gas constant R = 8.31 J K−1 mol−1.
Stage 1: mole calculations in either order
Moles of vanadium = 50.0 × 0.800 / 1000 = 4.00 × 10–2
Extended response
Maximum of 5 marks for answers which do not show a sustained line of reasoning which is coherent, relevant, substantiated and logically structured.
1
Moles of SO2 = pV / RT = (98 000 × 506 × 10–6 ) / (8.31 × 293)
= 2.04 × 10–2
1
Stage 2: moles of electrons added to NH4VO3
When SO2 (sulfur(IV) oxide) acts as a reducing agent, it is oxidised to sulfate(VI) ions so this is a two electron change
1
Moles of electrons released when SO2 is oxidised = 2.04 × 10–2 × 2
= 4.08 × 10–2
1
Stage 3: conclusion
But in NH4VO3 vanadium is in oxidation state 5
1
4.00 × 10–2 mol vanadium has gained 4.08 × 10–2 mol of electrons
therefore 1 mol vanadium has gained 4.08 × 10–2 / 4.00 × 10 – 2 = 1 mol
of electrons to the nearest integer, so new oxidation state is 5 – 1 = 4
Describe what you would observe when dilute aqueous ammonia is added dropwise, to excess, to an aqueous solution containing copper(II) ions.
Write equations for the reactions that occur.
Blue precipitate
1
Dissolves to give a dark blue solution
1
[Cu(H2O)6]2+ + 2NH3 –>Cu(H2O)4(OH)2 + 2NH4+
1
Cu(H2O)4(OH)2 + 4NH3 –> [Cu(NH3)4(H2O)2]2+ + 2OH– + 2H2O
A student weighed out a 2.29 g sample of impure K3[Fe(C2O4)3].3H2O and dissolved it in water.
This solution was added to a 250 cm3 volumetric flask and made up to 250 cm3 with distilled water.
A 25.0 cm3 portion was pipetted into a conical flask and an excess of acid was added.
The mixture was heated to 60°C and titrated with 0.0200 mol dm–3 KMnO4 solution.
26.40 cm3 of KMnO4 solution were needed for a complete reaction.
In this titration only the C2O42– ions react with the KMnO4 solution.
Write an equation for the reaction between C2O42– ions and MnO4– ions in acidic solution.
Calculate the percentage purity of the original sample of K3[Fe(C2O4)3].3H2O.
Give your answer to 3 significant figures.
M1 5 C2O42-(aq) + 2 MnO4-(aq) + 16 H+(aq) →
10 CO2(g) + 2 Mn2+(aq) + 8 H2O(l)
Ignore state symbols
1
M2 n(MnO4–) = 26.2 x 0.02 /1000. OR n(MnO4-) = 5.28 × 10-4
1
M3 n(C2O42-) = 5/2 × 5.28 × 10-4 = 1.32 × 10-3
M3 is for M2 × 5/2
If wrong ratio used then can only score M2, M4, M5 and M6
1
M4 n(C2O42– in flask originally) = 1.32 × 10–3 × 10 = 1.32 × 10-2
M4 is for M3 × 10
1
M5 n(K3[Fe(C2O4)3].3H2O) = 1.32 x 10-2 / 3 = 4.40 × 10-3
(Mr K3[Fe(C2O4)3].3H2O = 491.1)
M5 is for M4 ÷ 3
1
M6 Mass of K3[Fe(C2O4)3].3H2O reacted = 4.40 × 10-3 × 491.1 = 2.16 g
M6 is for M5 × 491(.1)
1
M7 % purity =2.16/2.29 x100 = 94.3 or 94.4%
A 50.0 cm3 sample of solution X was added to 50 cm3 of dilute sulfuric acid and made up to 250 cm3 of solution in a volumetric flask.
A 25.0 cm3 sample of this solution from the volumetric flask was titrated with a 0.0205 mol dm−3 solution of KMnO4
At the end point of the reaction, the volume of KMnO4 solution added was 18.70 cm3.
(i) State the colour change that occurs at the end point of this titration and give a reason for the colour change.
(2)
(ii) Write an equation for the reaction between iron(II) ions and manganate(VII) ions.
Use this equation and the information given to calculate the concentration of iron(II) ions in the original solution X.
(i) Colourless / (pale) green changes to pink / purple (solution)
Do not allow pale pink to purple
Just after the end−point MnO4− is in excess / present
(ii) MnO4− + 8H+ + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+
Moles KMnO4 = 18.7 × 0.0205 / 1000 = (3.8335 × 10−4)
Process mark
Moles Fe2+ = 5 × 3.8335 × 10−4 = 1.91675 × 10−3
Mark for M2 × 5
Moles Fe2+ in 250 cm3 = 10 × 1.91675 × 10−3 = 0.0191675 moles in 50 cm3
Process mark for moles of iron in titration (M3) × 10
Original conc Fe2+ = 0.0191675 × 1000 / 50 = 0.383 mol dm−3
