CHEMISTRY WAEC QUESTION AND ANSWERS FOR 2013

1. VOLUMENTRIC ANALYSIS QUESTION 1. 'A' is a solution of tetraoxosulphate(vi). B is a solution coomtaining !.4g of potassiumb hydroxide per 25.0cm3. (a.) Put A into burette and titrate with 20.0cm3 or 25.0cm3 portion into of B using Methyl Orange as an indicator. Record the volume of your pipette. Tabulate your burette readings and calculate the average volume of acid used. (b.) From your result and information provided. Calculate the (i.) Concentration of B in mol/ dm3 (ii.) Concentration of A inmol/ dm3 (iii.) Number of Hydrogenions in 1.0dm^2 of A. The equation for the reaction is: H2SO4 + 2KOH --> k2SO4 + 2H2O. {H=1, O=16, K=39} SOLUTION Burrete reading(cm3)| Rough | 1st | 2nd | 3rd Final(cm3) | 18.60 | 18.00| 20.30| 28.00 Initial(cm3) | 0.00 | 0.00 |2.30 | 10.00 Vol. Acid used (cm3)| 18.60 |18.00| 18.00| 18.00 The average volume of acid used = 18.00 + 18.00 + 18.00/3 = 18.00cm3. (bi.) Mass of KOH per dm3 of B = 39+16+1= 56. Conc. of B = 5.6/56 = 0.01mol/ dm3. (bii.) The equation for the reaction is: H2SO4 + 2KOH --> k2SO4 + 2H2O. From the equation ==> Conc of A * Volume of A* Mole ratio = CA*18.00/0.10*2 = 1/2.'. CA= 0.10 * 25/2*18' CA= 0.0694mol/ dm3. (biii.) 1dm3 of 1.0mol/ dm3H2SO4 contains (2*6.0*10^23). Hydrogen atom ==> 1dm3 of 0.694mol/ dm3 H2SO4 contains 2*6.0*10^23 = 8.328x10^22.