Make your own free website on Tripod.com
 

 

 

Answers and Explanations

MENU

 

                                                                                           2003 Questions

1. What are the products of the reaction between ethanoic acid and sodium carbonate?

   I      Water

   II    Ethyl ethanoate

   III  Carbon dioxide

   IV   Sodium ethanoate

   A     I and II only

   B     II and IV only

   C     I, II and III only

      I, III and IV only                                                                                                                                 Question 16, paper, 1 2003

 

Explanation: Refer to the Test of anions at Test for carbonates (Qualitative Analysis of Salts).

                    The reaction occurs as follows:

                    Ethanoic acid (CH3COOH)  +  Sodium carbanates (Na2CO3)      Sodium ethanoate (NaCH3COO)  +  Water (H2O)  +  Carbon dioxide (CO2)

 

2. Table below shows the volume and concentration used in titration of potassium hydroxide solution with sulphuric acid.

       What is the total volume of the mixture in the conical flask at the end point of the titration?   

      A     10 cm3

      B     20 cm3

      C     30 cm3

      D    40 cm3                                                                                                                                                 Question 42, paper 1, 2003

 

Explanation: Total volume of mixture at end of titration = Initial volume of sodium hydroxide + Volume of sulphuric acid needed to neutralize sodium hydroxide,X

                     Initial volume of sodium hydroxide given = 20 cm3

                     Volume of sulphuric acid needed to neutralize sodium hydroxide can be find as follows:

                     1. Make the equation of reaction occurs:

                        2KOH   +   H2SO4        K2SO4  +   2H2O

                    2. From the equation you can see that 2 mole KOH will react with 1 mole H2SOin the titration reaction.

                    3. Mole of KOH used in reaction = (0.1 x 20)/1000 = 0.002 mole

                    4. Therefore, 0.002 mole of KOH will react with 0.001 mole H2SOin the titration reaction.

                    5. Mole of  H2SOused to neutralize KOH = 0.1X/1000 = 0.001

                    6. The volume of H2SOused to neutrali ze KOH,X = 10 cm3

                    Terefore total volume of mixture at end of titration =  20 cm3  +  10 cm3 30 cm3                        

 

             

Section C, Question 4, 2003

(a) A farmer discovered that his vegetable were not growing well because the soil was poor and acidic. As a chemistry student, you

     help the farmer.

     Suggest how the farmer can overcome the problem.                                                                                                       (2 marks)

(a) Figure 7 shows an incomplete flow chart of cation and anion tests for salt X.

                                  

 

      Use the reagents listed below and complete the flow chart to confirm the salt X contains Pb2+ ions and CO32- ions. Include your 10

      obsevations. 

                                                  

                                                                                                                                                                                      (8 marks)

(c) You are required to prepare dry magnesium chloride salt. The chemicals supplied are

     Describe a laboratory experiment to prepare the salt. In your description, include the chemical equations involved.        (10 marks)

 

     Answers and explanations: (a) Scatters chalk or calsium hydroxide and add calsium phosphate or phosphate fertilizer on the acidic soil to reduce the 

                                                       acidity of the soil and to supply nitrate ions that will be absorbed by the vegetable root when the nitrate ions dissolve in

                                                       water.

                                                  (b) Put X salt into a test tube and add nitric acid, heat and shake the test tube.

                                                  (c) Step 1 :

                                                       - 25 cm3 magnesium sulphate 1 mol dm-3 is put into a beaker and then added with 25 cm3 potassium carbonate 1 mol dm-3 

                                                       - Mixture then stirred. The white preciptates of magnesium carbonate produced  were filtered and dried  between the filter  

                                                         paper. 

                                                       - The reaction involved is:

                                                          MgSO4   +   K2CO3       MgCO3   +   K2SO4

                                                 (c) Step 2:

                                                       - Put 50 cm3 hydrochloric acid 1 mol dm-3  

                                                       - Add the precipitates of magnesium carbonate into the hydrochloric acid until excess.

                                                       - The  mnixture then are filtered to remove the excess magnesium carbonate.

                                                          MgCO3   +   2HCl          MgCl2   +   CO  +   H2O

                                                       - The product of filtration then heated till saturated

                                                       - The crystals produced are dried between two filter paper.

 

 

 

 

 

                                                                                         2004 Questions

1. Which of the following is true about acids?

   A     Acid reacts with metal to produce salt and water

   B     Acid reacts with alkali to produce salt and hydrogen gas

   C     Acid reacts with metal oxide to produce salt, water, and hydrogen gas

   D    Acid reacts with carbonate of metal to produce salt, water, and cabon dioxide gas                                    Question 4, paper 1, 2004

 

Explanation: Refer to the Test of anions at Test for carbonates (Qualitative Analysis of Salts).

                     Acid + carbonate metal          carbonate salt + water + carbon dioxide

 

2. The following equation represents a neutralization reactions?

             Acid   +    Base        Salt   +   Water  

    Which pairs are reactants in neutralization reactions?

   I      Sulphuric acid   +   Sodium hydroxide

   II    Hydrochloric acid   +   Solid copper (II) oxide

   III  Sulphuric acid   +   Solid calcium carbonate

   IV   Hydrochloric acid   +   Potassium carbonate solution

       I and II only

   B     II and IV only

   C     II and III only

   D    III and IV only                                                                                                                                     Question 19, paper 1, 2004

 

Explanation: Reactants must be acid and base all choice from I till IV have correct acid. Base is will be either hydroxides or oxides. Only choice I and II fulfill

                    this criteria.

 

3. Which of the following ions form a white precipitate that dissolves in excess sodium hydroxide solution?

   I      Al3+

   II    Mg2+

   III  Pb2+

   IV    Zn2+

   A     I and II only

   B     II and IV only

   C     I, II and III only

   D    I, III and IV only                                                                                                                               Question 20, paper 1, 2004

 

Explanation: Refer to the Test for Cations in Qualitative Analysis of Salts.

                     Mg2+ion form a white precipitate but it not dissolves in excess sodium hydroxide.

 

4. The equations shows the reaction between sulphuric acid and sodium hydroxide.

                   H2SO4   +    2NaOH       Na2SO4   +   2H2O  

    What is the volume of 1.0 mol dm-3 sodium hydroxide solution which can neutralize 25.0 cm3 of 1.0 mol dm-3 sulphuric acid?

   A     12.5 cm3

   B     25.0 cm3

   C     50.0 cm3

   D    75.0 cm3                                                                                                                                                 Question 40, paper 1, 2004

 

Explanation: 1. From the equation given, you can see that 2 mole NaOH will react with 1 mole H2SOin the neutralization reaction.

                    2. Mole of H2SOused in reaction = (1 x 25)/1000 = 0.025 mole

                    3. Therefore, 0.025 mole of H2SO4  will react with 0.050 mole KOH  in the reaction.

                    4. Mole of  KOH used to neutralize H2SO4  = 1X/1000 = 0.050

                    5. The volume of KOH used to neutralize H2SO4 ,X = 50 cm3  

 

 

5. 3.2 g of copper (II)oxide powder is reacted with excess dilute nitric acid.

    What is the mass of copper (II) nitrate formed in the reaction?

    Use the information that the relative atomic mass of N = 14, O = 16 and Cu = 64.

   A     3.76 g

   B     4.96 g

   C     5.04 g

   D     7.52 g                                                                                                                                                              Question 44, paper 1, 2004

 

Explanation: 1. Make the equation of reaction first (can refer to Soluble Salts which is reaction between acid and oxide metal):

                        CuO   +   2HNO3           Cu(NO3)2   +    H2O

                    2. Number of mole of CuO used = 3.2/(64+16) = 0.04 mole.

                    3. According to the equation, 1 mole of copper (II) oxide,CuO will produce 1 mole copper (II) nitrate, Cu(NO3)2

                    4. Therefore, 0.04 mole of CuO will produce 0.04 mole Cu(NO3)2  in the reaction.

                    5. Therefore, the mass of Cu(NO3)2 formed in the reaction is 0.04 x [64 + 2(14) + 6(16)] = 7.52 g

 

 

Section A, Question 5, 2004

Table 5  shows Experiment I and II in the preparation of a salt

 

                                                        

 

(a) State one observation in Experiment I. (1 mark)

(b) Based on Experiment II:

     (i) State the reason why copper (II) oxide powder is added in excess. (1 mark)

     (ii) State how excess copper (II) oxide powder can be separated from the products. (1 mark)

     (iii) State the chemical equation for the reaction that takes place in Experiment II. (1 mark)

     (iv) Calculate the maximum mass of the salt formed.

           Use the information that the relative atomic mass of O = 16, S = 32 and Cu = 64 (2 marks) 

(c) Experiment I is repeated. Sulphuric acid is replaced by hydrochloric acid of the same concentration. Predict the volume of

     hydrochloric acid required for a complete reaction. (1 mark)

(d) There are several steps in the preparation of the salts in each of the Experiments I and II. State one difference in the steps

     between the two experiments. (1 mark)

(e) (i) State the type of reaction in the preparation of salts in Experiment I and II. (1 mark)

     (ii) State one type of reaction in the preparation of a salt other than that in (e)(i). (1 mark)

Answers and explanations:  (a) The pink solution of phenolptalein will turns colourless

                                             (b) (i) To ensure that acid will completely reacts with copper (II) oxide

                                                  (ii) Through filtration

                                                  (iii) H2SO4    +    CuO          CuSO4    +   H2O    

                                                 (iv) 1. Number of mole of H2SO4 used = (50 x 0.1)/1000 = 0.005 mole

                                                       2. According to the equation, 1 mole of H2SO4 will produce 1 mole CuSO4

                                                      3. So, 0.005 mole of H2SO4 will produce 0.005 mole CuSO4

                                                      4. Therefore, the maximum mass of salt (CuSO4 ) formed in the reaction is 0.005 x [64 + 32 + 4(16)] = 0.8 g

                                              (c) 1. Mole of potassium hidroxide must = mole of hydrochloric acid

                                                   2. Mole of potassium hydroxide = (0.1 x 20)/1000 = 0.002 mole = mole of hydrochloric acid

                                                   3. 0.1V/1000 = 0.002 mole

                                                   4. Therefore, volume of hydrochloric acid, V = (0.002 x 1000)/0.1 = 20 cm3

                                              (d)

                                            

                                               (e) (i) Neutralization reaction

                                                    (ii) Precipitation method using the double decomposition reaction