2022 Paper 4 Ans

G reen filtrate Residue is blue ppt Blue green ppt formed upon addition of NaOH(aq) Blue green ppt remains insoluble in green solution Blue green ppt formed upon addition of NH 3 ( aq). Grey green ppt remains insoluble in excess NH 3 ( aq ) Deep blue filtrate Residue is grey green ppt 2022 A level H2 Chem Paper 4 Ans

Upon addition of Zn powder, vigorous effervescence of gas is observed. The gas "pops" with a lighted splint. A red suspension in observed. Green filtrate. Residue is a mixture of red brown solid and grey solid Cr 3+ In excess NaOH(aq), it gives a green filtrate. In excess NH 3 ( aq), grey green ppt is collected as residue. Cu 2+ In excess NaOH(aq), blue ppt is collected as residue. In excess NH 3 aq), it gives a deep blue filtrate. ( Redox reaction. Cu 2+ is reduced to red brown Cu(s), Zn is oxidized to Zn 2+ ( aq ) Note: H 2 ( g) is produced via acid-metal reaction of Zn with H 2 SO 4 aq), NOT with FA 1 (

Total Vol of FA4 added/cm3 Temp/°C 0.00 28.0 2.00 31.0 4.00 32.8 6.00 34.2 8.00 35.4 10.00 36.4 12.00 37.2 14.00 36.6 16.00 35.8 18.00 35.0 20.00 34.4 22.00 34.0 24.00 33.6

Temp/ °C Total Vol of FA4 added/ cm3 0.00 5.00 10.00 15.00 20.00 25.00 39.0 38.0 37.0 36.0 35.0 34.0 33.0 32.0 31.0 30.0 29.0 28.0 × × × × × × × × × × × × ×

28.0 C 37.5 C +9.5 C 12.50 cm3 q = maqc∆T = (25.0 + 12.50)  4.18  (+9.5) = 1489 J Note: maq is the total mass of aq solution at point of neutralisation. T in C is the same as T in K Amount of NaOH used = 25.0 1000  1.00 = 0.025 mol = Amount of H2O formed ∆Hn = − mc∆T mol of 𝐻2O formed = − 1489 0.025 = −59565 J mol−1  −59.6 kJ mol−1 Note: T is +ve, H is -ve

∆Hn is defined as H+ (aq) + OH−(aq) ⎯→ H2O(l). KOH and NaOH are strong bases that dissociates completely to give OH−(aq). HCl and H2SO4 are strong acids that dissociates completely to give H+(aq). The ionic equations for both acid-base reactions are the same. In this experiment, heat lost to the environment is not accounted for. Hence the ∆T recorded is smaller than expected and ∆H calculated would also be smaller in magnitude. Use a lid for the cup / use a draught shield to reduce heat lost to the environment. Collect more data points near Vneut (e.g. at interval of 1 cm3) so that we could better estimate the Tmax and Vneut using graphical method. This method collects data when volume of H2SO4 used is increased at 2 cm3 interval, we use graphical method to estimate the volume of neutralisation. However, in a conventional titration, we can control and add volume of H2SO4 slowly when approaching the end point, and hence obtain a more accurate volume of neutralisation.

25.50 25.50 0.00 25.40 0.00 25.40 Average volume of FA6 used = 25.50+25.40 2 = 25.45 cm3

Concentration of NaOH in FA6 = 25.0 250  1.00 = 0.100 mol dm−3 Amount of NaOH used = 25.45 1000  0.100 = 0.002545 mol Amount of acid in 25.0 cm3 = ½  0.002545 = 0.0012725 mol Concentration of acid = 0.0012725 (25 1000⁄ ) = 0.0509 mol dm−3 Mr of acid = 6.00 0.0509 = 117.9 (1 d.p.) Mr of HOOC(CH2)nCOOH = 117.9 90 + 14n = 117.9 14n = 27.9 n  2

Amount of phenol required = 75 1000  0.01 = 0.00075 mol Mass of phenol required = 0.00075  (12.0  6 + 1.0  6 + 16.0) = 0.0705 g 1) As solid phenol is corrosive and toxic by skin absorption, wear a pair of glove and lab coat before the start of the procedure. Careful to avoid any skin contact with phenol. 2) Using a 100 cm3 beaker and electronic balance, weigh out 0.07g of solid phenol. 3) Measure 75 cm3 of deionised water using a 100 cm3 measuring cylinder, add deionised water slowly into the beaker, stir with a glass rod to speed up the dissolution process. Note: Question requires us to prepare a 75 cm3 solution of approximate 0.01 mol dm−3. We do not need to use a volumetric flask for preparation as its typical size is 100 or 250 cm3