Question

7. A solid sample X contains zinc and silver only. An experiment was carried out to determine the percentage by mass of zinc in the sample X. 1.80 g of the sample X was dissolved in 40.0 cm of 1.50 M hydrochloric acid. The excess acid required 13.60 cm of 1.00 M sodium hydroxide solution for complete neutralization. (a) Write an equation for the reaction between the sample X and HCl(aq). (1 mark) (b) How can we know that the reaction in (a) had been completed? (1 mark) (c) Suggest a suitable indicator for the titration. State the colour change at the end point. (2 marks) (3 marks) (d) Calculate the percentage by mass of zinc in the sample X. (Relative atomic mass: Zn = 65.4)

Answer 1

a. As Sample zn can react with Hee but Contain zn & Ag mly. Ay cannot. so ng present as such. so ant and lat.) anch + H219) b so no H2 gas since In reaction H₂ gas evolved When evolved it means hole en react with her and now no more reaction completed. then an left so C. Phenophtalein le Used for acid-base Phenophthalein indicator can be CHU+ NaOH) reaction colour change - penketa ecola Colourless to pink

1.80g of sample x was disolved in 40.0 cm3 of 1-50m Hal Molarity of Naok= 1.00M volume of Naot used to excess HU = 13.60 cm3 neutralised 1cm3= 1 mL and im L= 0.001L so 13.60 0.01360l moly NaoH used = molarity X volumell) slog oomed X 0.01360L = 0.0136 mo) Naont HU Naal +420 I mol Naou react with Imol nice 0.0136 mol Naol react with 0 0 136 mol Hu mol Juce present in excess = 0.0136 mol

mol of nice added of samplex = 0.04% X 1.Som loso mene = 0.06 mol O'036 mol mol of He present in excess= mol of Hu used to dissolve sample x = (0.06 - 0.0136) mo) 0.0464 mo) that means From reaction 2 mol 4cl react with Imol en react with 0.0464 mol en 2 0.0464 mol HCl moly? 2n = 0.0232 mol mass j zn= molx molar mass =0.0232 mot x 65.4 g mot = 105174 lan ma SS issy zn xloo many sample 1.sing 1.808 yzn = 84.3-4 x 100 = 84.3 . Annver