Acids, Bases and Salts Questions and Answers - Chemistry Form 4 Topical Revision

Questions

1. Study the reaction below and answer the questions that follow
   NH_{3 (g)} + H₂O _(l) ⇌ NH₄⁺_(aq) + OH^-_(aq)
   1. Define the term acid
   2. Identify an acid in the above reaction
   3. Explain your answers in (b) above
2. A student mixed equal volumes of Ethanol and butanoic acid. He added a few drops of concentrated Sulphuric (VI) acid and warmed the mixture
   1. Name and write the formula of the main products
      Name………………………………….
      Formula……………………………………..
   2. Which homologous series does the product named in (i) above belong?
3. A sample of water from a village in Trans Mara East District was divided into equal portions and each mixed with equal volume of soap solution. The observations made are tabulated below:
   

   Sample of water	Treatment before adding soap	Observations made on shaking with soap
   I	Boiled	Lather form immediately
   II	No treatment	Slight lather form slowly
   III	Treatment with washing soda	Lather formed immediately

   1. What type of hardness is present in water from the village. Explain
   2. State one advantage of hard water
4. The solubility of Iron (II) Sulphate crystals are 22^oC is 15.65g per 100g of water. Calculate the mass of iron(II) sulphate crystals in 45g of saturated solution at the same temperature
5. Hardness of water may be removed by either boiling or addition of chemicals:
   1. Write an equation to show how boiling removes hardness of water
   2. Name two chemicals that are used to remove hardness of water
6. State one advantage of drinking hard water rather than soft water.
7. Given this reaction;
   RNH₂ + H₂O ⇌ RNH₃⁺ + OH^-
   1. Identify the acid in the forward reaction .Explain
   2. bDilute nitric acid can react with a solution of sodium carbonate. Write an ionic equation for the reaction
8. Magnesium hydrogen carbonate is responsible for the temporary hardness of water.
   This type of hardness can be removed by addition of ammonia solution
   1. Describe how temporarily hard water is formed
   2. Write an equation to show the softening of temporarily hard water by the addition of aqueous ammonium solution
9. When 2M potassium hydroxide solution was added to solution R, a white precipitate T was formed which dissolved in excess potassium hydroxide solution to form solution L. solution R forms a white precipitate with sodium chloride solution:
   1. Identify the cation in solution R ......................................................................
   2. Name precipitate T
   3. Write the molecular formula of the compound in solution L
      
      
10. Below is a table showing the solubilities of salts Q and R at different temperatures.
    

    Temperature in oC		0	10	20	30	40	50
    Solubility in grammes per 100g of water	Salt Q	3.0	5.0	7.4	10.0	14.0	19.0
    	Salt R	15.0	17.0	20.7	25.7	28.7	33.0

    1. Define the term “Solubility of salt”
    2. If both salts Q and R are present in 100cm³ of saturated solution at 50^oC, what will be the total mass of crystals formed if the solution was cooled to 20^oC?
11. The following results were obtained during an experiment to determine the solubility of potassium chlorate(V)in water at 30ºC.
    Mass of evaporating dish =15.86g
    Mass of evaporating dish + saturated solution at 30ºC = 26.8g
    Mass of evaporation dish +solid potassium chlorate (V) after evaporation to dryness=16.86g
    Calculate the mass of the saturated solution containing 60.0g of water at 30ºC
12.   
    1. What is meant by the term solubility of salts?
    2. Calculate the solubility of salt given that 15g of the salt can saturate 25cm³ of water
    3. The table below gives the solubility of salt X in grams per 100g of water at different temperatures
       

       Temp oC	10	20	30	40	50	60	70	80	90	100
       Solubility (g/100g) water	5.0	7.5	10.5	14.0	18.5	24.0	30.0	38.0	46.0	50.1

       1. Plot a solubility curve for salt X (solubility in g /100g water Y- axis) (temp ^oC (X –axis)
       2. What is meant by the points plotted in (i) above?..................................................................
       3. From your graph determine the solubility of salt X at the following temperatures
          1. 44^oC ………………………………………….
          2. 62^oC ………………………………………….
             
       4. What mass of crystals of the salt will be formed if the solution was cooled from 62^oC to 44^oC
       5. Name two areas where knowledge of solubility curves is applied
          
13. You are given a mixture of Lead (II) Chloride, Iodine, ammonium chloride and sodium chloride.
    Explain how you would separate all the four solids using methylbenzene, a source of heat and water
14.    
    1. The table below shows the solubility of potassium chlorate at different temperatures
       

       Temperature (oC )	10o	20o	30o	40o	50o	60o	70o
       Solubility g/100g water	27	30	36	55	80	110	140

       1. Plot a graph of solubilities of potassium chlorate against temperature
       2. Using your graph:
          1. Determine the solubility of potassium chlorate at 47^oC
          2. Determine the concentration in moles per litre of potassium chlorate at 47^oC (K= 39, Cl = 35.5, O= 16) density of solution = 1g/cm³
          3. Determine the mass of potassium chlorate that would crystallize if the solution is cooled from 62^oC to 45^oC
    2. In an experiment to determine the solubility of sodium hydroxide, 25cm³ of a saturated solution of sodium hydroxide weighing 28g was diluted in a volumetric flask and the volume made to 250cm³ mark. 20cm³ of this reacted completely with 25cm³ of 0.2M hydrochloric acid according to the equation.
       NaOH_(aq) + HCl_(aq) → NaCl_(aq) + H₂O_(l)
       Calculate:
       1. The number of moles of hyrdrochloric acid used
       2. The number of moles of sodium hydroxide in 20cm3
       3. The moles of sodium hydroxide in 250cm3 of solution
       4. The mass in grams of sodium hydroxide in 250cm3 of solution
       5. The solubility of sodium hydroxide in g/100g water
15.   
    1. Define the term solubility of a substance
    2. The table below shows the solubilities of two salts L and M at different temperatures.
       

       Temperature (oC)		10	20	30	40	50
       Solubility in g/100g of water	L	11.0	14.0	20.1	28.0	36.0
       	M	15.0	17.0	19.0	21.2	25.0

       1. Name the method that can be used to separate the two salts
       2. Plot on the same axes a graph of solubilities of L and M against temperature
       3. From the graph determine:-
          1. The temperature at which solubilities are equal
          2. The solubility at the temperature mentioned above
       4. If the relative formula mass of M is 132, determine the concentration of M in moles per litre in (iii) II above
       5. A solution contains 38g of L and 22g of M at 50°C. Calculate the total mass of crystals obtained in cooling this solution to 30°C.
16. The graph below shows the changes in conductivity when 50cm³ of 0.1M Nitric (V) acid is titrated with potassium hydroxide (curve I) and when 50cm³ of 0.1M methanoic acid is reacted with the same potassium hydroxide solution (curve II)
    
    
    1.   
       1. Explain the changes in conductivity in the regions:
          AB…………… BC………………………………………………….
       2. Using curve (I), explain why the conductivity does not have a value of zero at end-point
       3. Calculate the concentration of KOH with reference to curve II
       4. Explain why the two curves shows different trends in conductivity
    2. 50cm³ of 0.1M methanoic acid was reacted with 20cm³ of a solution of sodium carbonate of unknown concentration. Work out the concentration of the carbonate
17. The flow charts below show an analysis of a mixture R that contains two salts. Study the analysis and answer the questions that follow:
    
    
    1.  
       1. State:-
          1. The condition in step I
          2. The process in step II
       2. A small portion of mixture R is added to dilute nitric (V) acid in a test-tube. What would be observed?
       3. Write an equation for the reaction between the cation in filtrate X and sodium hydroxide solution
       4. Explain how water vapour in step I could be identified
          
          
          
          
          
    2.  
       
       
       1. State and explain the conclusion that can be made from step IV only
       2. Name the anion present in residue U. Explain
       3. From the flow chart in (a) and (b); Write the formulae of cations present in mixture R
18.   
    1. Define:
       1. A saturated solution.
       2. Solubility of a solute.
    2. In an experiment to determine solubility of sodium chloride, 10.0 cm3 of a saturated solution of sodium chloride weighing 10.70g were placed in a volumetric flask and diluted to a total of 500 cm³. 25.0 cm³ of the diluted solution of sodium chloride reacted completely with 24.0 cm³ of 0.1M silver nitrate solution. The equation for the reaction is
       AgNO_3(aq) + NaCl _(aq) → AgCl_(s) + NaNO_3(aq)
       
       Calculate;
       1. Moles of silver nitrate in 24.0 cm³ of solution.
       2. Moles of NaCl in 25.0 cm³ of solution.
       3. Moles of NaCl in 500 cm³ of solution.
       4. Mass of NaCl in 10.0 cm³ of saturated sodium chloride (Na = 23, Cl = 35.5)
       5. Mass of water in 10.0cm³ of saturated solution.
       6. The solubility of NaCl in g/100g of waters.
19. Describe how you would prepare a dry sample of crystals of potassium sulphate starting with 100cm³ of 1M sulphuric (VI) acid.
20. The table shows solubility of potassium chlorate (V)
    

    Temp (oC)	45	80
    Solubility	39	63

    Calculate the mass of solute and solvent in 90g of the saturated solution of the salt at 45^oC
    1. A solution of the salt in 100g water contains 63g at 95^oC. At what temperature will the solution start forming crystals when cooled
21. Two samples of hard water C and D were boiled. When tested with drops of soap, sample D formed lather easily while C did not:-
    1. Name the possible salt that caused hardness in sample D
    2. Explain how distillation can remove hardness in sample C
    3. Give one advantage of hard water
       
       
       
22. A student attempted to prepare a gas using the set-up below. She could not collect any gas
    
    
    1. Give two reasons why no gas was collected
    2. Which gas did the student intend to prepare?
23. Water from a town in Kenya is suspected to contain chloride ions but not sulphate ions.
    1. Describe how the presence of chloride ions in the water can be shown
    2. State one advantage of drinking hard water rather than soft water
24. Study the following tests and observation and answer the questions that follow:-
    

    	TEST	OBSERVATION
    I	- Add few drops of acqueous ammonia to copper (II) nitrate solution	- Light blue precipitate is formed
    II	- Add excess of ammonia to copper (II) nitrate	- Deep blue solution
    III	- Add cold dilute hydrochloric acid to substance E1 and warm gently	- Gas evolved, smells of rotten eggs and blackens lead acetate paper

    Identify:-
    1. Substance responsible for:
       1. Light blue precipitate…………………………………………………………………
       2. Deep blue solution …………………………………………………………….
    2. Gas evolved in test III above …………………………………………………………
25.   
    1. What is meant by the term solubility of salts?
    2. Calculate the solubility of a salt given that 15g of the salt can saturate 25cm³ of water.
26.   
    1. Draw a well labeled diagram to show how to prepare an acqueous solution of hydrogen chloride gas
    2. Name one other gas whose aqueous solution can be prepared in the same way
27. In an experiment to determine the solubility of solid Y in water at 30^oC the following results
    Mass of evaporating dish + saturated solution = 42.4g
    Mass of evaporating dish + dry solid Y = 30.4g
    1. Use the data to calculate the solubility of solid Y at 30^oC
    2. State one application of solubility curves and values
28. Study the table below showing the solubility of substance K at various temperatures
    

    Temperature (oC)	Solubility (g/100g water)
    0  30  70  100	30  24  19  14

    1. What would happen if a sample of a saturated solution of the substance at 30^oC is heated to 70^oC. Explain.
    2. What is the most likely state of substance K..................................................................
       
29. In the equilibrium given below:-
    Fe³⁺_(aq) + SCN_(aq) ⇌ [Fe(SCN)]²⁺ _(aq)
    Brown                             Red
    What would be observed when Iron (III) Chloride is added to the equilibrium mixture. Explain
30. Sodium Carbonate Decahydrate crystals were left exposed on a watch glass for two days.
    1. State the observations made on the crystals after two days.
    2. Name the property of salts investigated in the above experiment
31. The label on a bottle of mineral; water had the information below.
    

    Ions present	Concentration (g/litre)
    Ca2+ Mg2+ Na+ K+ SO4 HCO3	0.10 0.20 0.01 0.01 0.14 0.26

    1. Name the compound that causes temporary hardness in the mineral water.
    2. Using an equation, describe how the water can be made soft by adding sodium carbonate solution.
    3. Give one advantage of drinking mineral water such as the one above
32. A solution of hydrogen chloride gas in methylbenzene has no effect on calcium carbonate. A solution of hydrogen chloride in water reacts with calcium carbonate to produce a gas. Explain
33.   
    1. Is concentrated sulphuric acid a weak acid or a strong acid?
    2. Explain your answer in (i) above.
34. When water reacts with potassium metal the hydrogen produced ignites explosively on the surface of water.
    1. What causes this ignition?
    2. Write an equation to show how this ignition occurs
35. In an experiment, soap solution was added to three samples of water. The results below show the volume of soap solution required to lather with 500cm3 of each water sample before and after boiling
    

    	Sample 1	Sample 2	Sample 3
    Volume of water used before water boiled	26.0	14.0	4.0
    Volume of soap water after water boiled	26.0	4.0	4.0

    1. Which water samples are likely to be soft?
    2. Explain the change in volume of soap solution used in sample 2
36. How does the pH value of 0.25M KOH_(aq) compare with that of 0.25M ammonia solution

Answers

1.  
   1. Proton donor/electron acceptor/a substance which when dissolved in water dissociates/break to hydrogen ions as the only positive ion.
   2. Water/ H₂O
   3. It is a proton donor/electron acceptor
2.  
   1. Ethylbutanoate
   2. CH₃CH₂CH₂
   3. Esters
3.  
   1. Temporary water hardness . This is because hardness is removed by boiling
   2. - Provide Ca²⁺ ions needed in formation of strong teeth and bones
      - Hard water form a layer of carbonate of lead which prevent water coming in contact with lead which cause poisoning (award 1mk for any one)
4. Let x be the mass of FeSO₄ crystals in saturated solution
   ∴ Mass of water = 45 – x
   X g of FeSO₄ dissolves in (45 − x)g of water
   
   100x of FeSO₄ dissolves in 100g of water
   45 - x
   So, solubility is 100x = 15.65
                         45 – x
   100x = 15.56 (45 – x)
   100x + 15.65x = 15.65 x 45
   115.65x = 15.65 x 45
   x = 15.65 x 45
            115.65
   = 6.0895
   So solubility = 6.09g of FeSO₄ in 100g of water
5.                                _heat
   1. Ca(HCO₃)_2(aq) → CaCO_3(s) + CO₂ + H₂O_{(l)
                            heat}
      or:- Mg(HCO₃) → MgCO_3(s) + CO_2(g) + H₂O_(s) (award 1mk for any)
   2. - Addition of Na₂CO_3(s)
      - Addition of Ca(OH)_2(s)
      - Addition of aqueous ammonia (award 1mk each for any two; Total =2mks)
6. – Provides essential minerals e.g. Ca²⁺ for strong bornes and teeth ✓1
   - It has a better taste
7.  
   1. The acid is water H₂O
      Reason H₂O has donated a proton (H⁺)
   2. 2H⁺_(g) + CO₃^2-_(aq) → CO_2(g) + H₂O_(l)
8.  
   1. - Magnesium carbonate reacts with rain water
      - Containing caborn (iv) oxide dissolved.
      - Forming magnesuin hydrogencarbonate
   2. Or MgCO_3(s) + CO_2(g) + H₂O_(l) → Mg (HCO₃)_2(aq)
      
      
9.  
   1. Lead ions
   2. Lead (II) hydroxide
   3. [Pb(OH)₄]^2-
10.  
    1. Solubility of a salt is mass of a salt that dissolves in 100g of water at a given temperature. √1
    2. Mass of Q that crystallizes out = 19.0 – 7.4 √½ = 11.6 g.
       Mass of R that crystallizes out = 33 – 20.7√½ = 12.3g.
       Total mass of crystals = 12.3 + 11.6√½ = 23.9g √½
11. Mass of dry salt = 16.86 – 15.86 √ ½
    = 1.00g √ ½
    Mass of water = 26.86 – 16.86 = 10g√ ½
    Mass of salt in 60g of water = 60x1 = 6 g √ ½
                                                 10
12.  
    1. This is the maximum mass of a salt that will dissolve in 100g of water of a given temperature
    2. 15g dissolve in 25cm³ water
       ? dissolve in 2100cm³ water
       = 15 x 100 = 60g/100g water
              25
    3.  
       1. in graph paper
       2. Every point on the solubility curve is a saturated point of a solution which contains a maximum amount of salt X at a graph temperature
       3.  
          1. 16g
          2. 25g
       4. 25 – 16 = 9g/100g water
       5. - Extraction of Na₂CO₃ from Lake Magadi
          - Extraction of Nacl from sea water
13. Add Methyl benzene to the mixture and stir to dissolve iodine. Filter and crystallize the filtrate to obtain sodium chloride crystals.
14.  
    1.  
       1.  
       2. 72g/100g water ✓ 1.0
       3. 100cm³ dissolve 72g
          1000cm³ dissolve = (1000 x 72) 
                                               100
          = 720g/l
          KClO₃ = 39 + 35.5 + 3 x 16 = 122.5
          molarity = 720g/l
                        122.5gmol^-1
          = 5.878mol/l
       4. Mass dissolved at 62^o = 116g
          Mass dissolved at 42^o= 66g
          mass crystallized out = 50g
    2.  
       1. (25 x 0.2M) = 0.005mol
               1000
       2. 0.005mol (mole ration Acid: Base = 1:1)
       3. 20cm³ contain 0.005mol
          25cm³ contain = (250cm³ x 0.005mol)
                                             20cm³
          = 0.0625mol
       4. Mass = (0.0625x 4ogmol^-1) = 2.5g
       5. Mass of solvent = 28g – 2.5g = 25.5g
          solubility = (100 x 2.5)
                                25.5
          = 9.804g/100g water
15.  
    1. Fractional crystallization
    2. Scale = 1 mk
       Plotting = 1 mk
       Curve L = 1 mk
       Curve M = 1 mk
    3.  
       1. 26 C
       2. 18g 
    4. 1 mole of salt M = 132g
       ^18x1/₁₃₂ = 0.13863636 moles
       Concentration = 1000 x 0.13863636
                                       100
       = 1.386M
    5. L = 20g M= 19g
       38-20=18
       22-19= 3+
       Total  21 g
    1. Solubility refers to the maximum mass of solute dissolving in a 100g of a solvent at a particular temperature
    2.  
16.  
    1.  
       1. Conductivity decreases wince H+ ions form he acid are neutralized by OH^- ions from the base. This reduces the concentration of ions available for conductivity.
       2. Conductivity increases since the OH^- ions accumulate after complete neutralization of the acid OH^- increases conductivity.
       3. Neutralization leads to the formation of a slat. The ions in the salt are responsible for conducting of electricity.
       4. They yield different concentration of H⁺ ions
          For HNO₃ – dissociates completely hence more H⁺ ions
          HCOOH – dissociates partially hence less H+ ions
    2. 2HCOOH_(aq) + Na₂CO_3(aq) → 2HCOONa_(aq) + H₂O_(l) + CO_2(g)
       moles of HCOOH = 50 x 0.1
                                    1000
       = 0.005moles
       mole ration acid : base
                            2 : 1
       moles of Na₂CO₃ = ^0.005/₂
       = 0.0025
       Molarity of Na₂CO₃ = 0.0025 x 1000
                                                20
       = 0.125M
17.  
    1.  
       1.  
          1. Heating √1
          2. Filtration. √1
       2. Effervescence √1 / Bubles.
       3. Zn²⁺_(aq) + 2OH^-_(aq) → Zn(OH)_2(s) √1
       4. Pass the water vapour over white anhydrous√1 Copper (II) suplhate. It turns blue. √½
    2.  
       1. R is a mixture of sulphur √½ and insoluble√½ salt. It forms √1 a filtrate and residue in filtration of mixture
       2. Carbonate √1 / CO₃^2- √1
          It produces CO₂ on reaction with H⁺
       3. Zn²⁺√1 Al³⁺√1
18.  
    1.  
       1. A saturated solution is one which cannot dissolve more solute at that particular temperature. ✓1 (1 mk)
       2. Solubility of a soluble is the amount of grams of solute present in 100g of water at that particular temperature. ✓1 (1 mk)
    2.  
       1. Mole = M x ^V/₁₀₀₀
          0.1 x ²⁴/₁₀₀₀ ✓1 = 0.0024 moles✓1 (2 mks)
       2. Moles of NaCl in 25cm³
          Mole ratio is 1 : 1
          Moles of NaCl = 0.0024 moles✓1 (1 mk)
       3. Moles of NaCl in 500 cm³
          If 25cm³ = 0.0024 moles
          ∴ 500 cm³ = ?
          = 500 cm³ ✓1 x 0.0024 moles
              25 cm³
          = 0.048 moles ✓1 (2 mks)
       4. Mass of NaCl in 10cm3
          Mass = moles x R.F.M.
          = 0.048 x 58.5 = 2.808g
       5. Mass of water = mass of solution – mass of NaCl
          = (10.70 – 2.808)g ✓1
          = 7.892 g ✓1 (2 mks)
       6. If 7.892 of H₂O →2.808g ✓1
          100g of H₂O    →     ?
          100g x 2.808 ✓1
          7.892g
          = 35.6g /100g of H₂O✓1
19. Add 100cm³ of 2M √ potassium hydroxide or 200cm³ of 1M potassium hydroxide to the acid.
    Heat the solution until it is saturated and cool to obtain crystals. Dry the crystals between filter papers
20.  
    1. 139g of solution contains 39g solute
       ∴ 90kg of solution contains 39 x 90 = 25.25g
                                                 139
       Mass of solvent = 90 – 25= 64.75g
    2. 80^oC
21.  
    1. Calcium hydrogen carbonate/Magnesium hydrogen carbonate;
    2. Water boils off and is condensed leaving the salt;
    3. Provides minerals used to strengthen bones
22.  
    1. Delivery tube should not dip into solution
       - Thistle funnel should did into the solution
       - Gas jar was no water/ little water in trough ( 1 each max 2)
    2. Oxygen
23.  
    1. acidity water with Nitric add aqueous lead nitrate or
       - silver nitrate formation of white precipitates shows presence penalize fully for uric acid 1 ½ mk of chloride ions
    2. provide essentials minerals e.g. Ca2+ ions
24.  
    1.  
       1. - Cu(OH)₂ or copper (II)hydroxide√1
       2. Cu(NH₃)₄²⁺√1
    2. Hydrogen sulphide or H2Sg√1
25.  
    1. this is the maximum mass of a salt that will dissolve in 100g of water at a given temperature √1
    2. 15g dissolve in 25cm³ water
       xg dissolve in (15 x 100)g√1
                                25
       = 60g/100g√1
26.  
    1. Diagrammatical presentation on how to prepare an aqueous solution of hydrogen
       
    2. Ammonia gas *MAT
27.  
    1. Mass of saturated soln. = 42.4 – 26.2 = 16.2
       Mass of dry solid Y = 30.4 – 26.2 = ^4.2g/_12.0
       Solubility of Y = 4.2 x 100
                                  12.0
       35g per 100g of water
    2. – Used is fractional crystallization of salt mixture.
28.  
    1. 24 -19 = 5g of substance K will be produced
       Reason: Solubility decreases with increase in temperature
    2. Gaseous state
29. Deep red solution will be formed. Equilibrium shifts to the right/forward reaction is favoured since Fe³⁺ ions favours forward reaction.
30.  
    1. They became a white powder
    2. Efflorescency
31.  
    1. calcium hydrogen carbonate/ magnesium hydrogen carbonate
    2. Ca(HCO₃)_2(aq) + Na₂CO_3(aq) → CaCO_3(g) + 2NaHCO_3(aq)
       Mg(HCO₃)_2(aq) + Na₂CO_3(aq) → CaCO_3(g) + 2NaHCO_3(aq)
    3. Contains Ca²⁺ ions needed to harden teeth and bones
32. HCl_(g) in water ionizes to produce H⁺ _(aq) and Cl ^– _(aq)
    HCl_(g) in methylbenzene remain as moles hence no H⁺ ion
33.  
    1. Weak acid ✓1
    2. Has few free H⁺ (Hydrogen) ions
34.  
    1. The reaction is too exothermic that alot of heat is produced causing ignition of hydrogen in presence of oxygen
    2. K_(s) + H₂O_(g) → KOH_(aq) + H_2(g)
       H_2(g) + O_2(g) → H₂O_(g)
35.  
    1. Sample 1 and 2
    2. Sample 2 contained ions that caused temporary hardness therefore required large ( volume of soap solution before boiling, but after boiling the temporary hardness was removed, hence requiring very little volume ( ½mk) of soap solution to lather.
36. - KOH has higher pH value than ammonia
    - KOH is a stronger base; dissociates fully
    - Ammonia solution is a weak base; dissociates partially