C. C. C. Heep Woh College Form 4 Chemistry

							                                   C. C. C. Heep Woh College
                                   Form 4 Chemistry Quiz 10
                              Unit 16 : Acids (Volumetric Analysis)
                                        Suggested Answer

1.   (a) Burette                                                                             1

     (b) Rinse / wash the burette with distilled water and then 2M NaOH solution.            1,1
         Fill the burette with 2M NaOH solution by filter funnel.                            1

     (c) Phenolphthalein                                                                     1

     (d) CH3COOH(aq) + NaOH(aq) ⎯→ CH3COO-Na+(aq) + H2O(aq)                                  1

     (e) Brand A:
          Volume = (24.8 + 25.2 + 25) / 3 x 1000 = 0.025 dm3                                 1
          No. of mole of NaOH = 0.025 x 2 = 0.05 mol                                         1

           Brand B:
           Volume = (26.1 + 25.9 + 26) / 3 x 1000 = 0.026 dm3                                1
           No. of mole of NaOH = 0.026 x 2 = 0.052 mol                                       1
     (f) Brand A:
         No. of mole of $1 can buy = [0.05 x (300 / 25) ] / 6 = 0.1 mol / $1                 1
         Brand B:
         No. of mole of $1 can buy = [0.052 x (500 / 25) ] / 8 = 0.13 mol / $1               1
         ∴Brand B is better buy.                                                             1


2.   (a) Al2O3(s) + 6HCl(aq) 　　 2AlCl3(aq) + 3H2O(l)                                         1,1

     (b)
                      Antacid                     A                    B              C

                   Cost per gram           30/100 =$0.3/g         $0.65/g           $0.2/g

           Volume of 2.0M HCl reacted
                                      25.0-10.0 = 15.0             13.0              6.5
             with 1g of antacid (cm3)

           Cost of antacid for complete
                                         $0.3/15 = $0.02           $0.05            $0.031
           neutralization of 1cm3 of HCl
                       Each row : 1 mark                                               1x3

     (c) Antacid A is the best-buy,                                                          1
         because the smaller the amount of NaOH required,
         the more is the amount of the HCl reacted with the antacid.                   1
(d) No. of moles of excess HCl = no. of moles of NaOH = 2 x 12/1000 = 0.024

   No. of moles of HCl reacted with Mg(OH)2 in antacid B.
   = total no. of moles of HCl – no. of moles of excess HCl
   = 2 x 25/1000 – 0.024
   = 0.026                                                                    1

   From the equation: Mg(OH)2 + 2HCl ⎯→ MgCl2 + 2H2O

   No. of moles of Mg(OH)2 = 0.5 x no. of moles of HCl
                           = 0.5 x 0.026
                           = 0.013

   Mass of Mg(OH)2 = 0.013 x [24.3 + 2x(16+1)] = 0.7579g                      1

   % by mass of Mg(OH)2 in antacid B = (0.7579/1) x 100% = 75.79%             1

                                        THE END