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

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  1. Study the reaction below and answer the questions that follow
    NH3 (g)+ H2O (l) â‡Œ 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
    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
    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 22oC 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 and 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 50oC, what will be the total mass of crystals formed if the solution was cooled to 20oC?
  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
    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 o 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. 44oC ………………………………………….
        2. 62oC ………………………………………….
      4. What mass of crystals of the salt will be formed if the solution was cooled from 62oC to 44oC
      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
    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 47oC
        2. Determine the concentration in moles per litre of potassium chlorate at 47o(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 62oC to 45oC
    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)
      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
    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 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 and 22g of 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)
    changes in conductivity
      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:
    analysis of mixture R
      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

      flowchart analysis of mixture R
      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
    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
      AgNO3(aq) + NaCl (aq) → AgCl(s) + NaNO3(aq)

      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 cm3of 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 45oC
    1. A solution of the salt in 100g water contains 63g at 95oC. 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
    preparation of oxygen
    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:-
    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
    1. Substance responsible for:
      1. Light blue precipitate…………………………………………………………………
      2. Deep blue solution …………………………………………………………….
    2. Gas evolved in test III above …………………………………………………………
    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.
    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 30oC 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 30oC
    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)
    1. What would happen if a sample of a saturated solution of the substance at 30oC is heated to 70oC. Explain.
    2. What is the most likely state of substance K..................................................................
  29. In the equilibrium given below:-
    3+(aq)SCN(aq) â‡Œ [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)
    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
    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




    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
    1. Ethylbutanoate
    2. CH3CH2CH2
    3. Esters
    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
    = 6.0895
    So solubility = 6.09g of FeSO4 in 100g of water
  5.                                heat
    1. Ca(HCO3)2(aq) → CaCO3(s) + CO2 + H2O(l)
      or:- Mg(HCO3) → MgCO3(s) + CO2(g) + H2O(s) (award 1mk for any)
    2. - Addition of Na2CO3(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
    1. The acid is water H2O
      Reason H2O has donated a proton (H+)
    2. 2H+(g) + CO32-(aq) → CO2(g) + H2O(l)
    1. - Magnesium carbonate reacts with rain water
      - Containing caborn (iv) oxide dissolved.
      - Forming magnesuin hydrogencarbonate
    2. Or MgCO3(s) + CO2(g) + H2O(l) → Mg (HCO3)2(aq)

    1. Lead ions
    2. Lead (II) hydroxide
    3. [Pb(OH)4]2-
    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
    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
      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
        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.
      2. 72g/100g water ✓ 1.0
      3. 100cm3 dissolve 72g
        1000cm3 dissolve = (1000 x 72
        = 720g/l
        KClO3 = 39 + 35.5 + 3 x 16 = 122.5
        molarity = 720g/l
        = 5.878mol/l
      4. Mass dissolved at 62o = 116g
        Mass dissolved at 42o= 66g
        mass crystallized out = 50g
      1. (25 x 0.2M) = 0.005mol
      2. 0.005mol (mole ration Acid: Base = 1:1)
      3. 20cm3 contain 0.005mol
        25cm3 contain = (250cm3 x 0.005mol)
        = 0.0625mol
      4. Mass = (0.0625x 4ogmol-1) = 2.5g
      5. Mass of solvent = 28g – 2.5g = 25.5g
        solubility = (100 x 2.5)
        = 9.804g/100g water
    1. Fractional crystallization
    2. Scale = 1 mk
      Plotting = 1 mk
      Curve L = 1 mk
      Curve M = 1 mk
      1. 26 C
      2. 18g 
    4. 1 mole of salt M = 132g
      18x1/132 = 0.13863636 moles
      Concentration = 1000 x 0.13863636
      = 1.386M
    5. L = 20g M= 19g
      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
      1. Conductivity decreases wince H+ ions form he acid are neutralized by OHions 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) + Na2CO3(aq) → 2HCOONa(aq) + H2O(l) + CO2(g)
      moles of HCOOH = 50 x 0.1
      = 0.005moles
      mole ration acid : base
                           2 : 1
      moles of Na2CO3 = 0.005/2
      = 0.0025
      Molarity of Na2CO3 = 0.0025 x 1000
      = 0.125M
        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
      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 / CO32- √1
        It produces CO2 on reaction with H+
      3. Zn2+√1 Al3+√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)
      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
        = 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
    1. 139g of solution contains 39g solute
      ∴ 90kg of solution contains 39 x 90 = 25.25g
      Mass of solvent = 90 – 25= 64.75g
    2. 80oC
    1. Calcium hydrogen carbonate/Magnesium hydrogen carbonate;
    2. Water boils off and is condensed leaving the salt;
    3. Provides minerals used to strengthen bones
    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
    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
      1. - Cu(OH)2 or copper (II)hydroxide√1
      2. Cu(NH3)42+√1
    2. Hydrogen sulphide or H2Sg√1
    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
      = 60g/100g√1
    1. Diagrammatical presentation on how to prepare an aqueous solution of hydrogen
    2. Ammonia gas *MAT
    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
      35g per 100g of water
    2. – Used is fractional crystallization of salt mixture.
    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.
    1. They became a white powder
    2. Efflorescency
    1. calcium hydrogen carbonate/ magnesium hydrogen carbonate
    2. Ca(HCO3)2(aq) + Na2CO3(aq) → CaCO3(g) + 2NaHCO3(aq)
      Mg(HCO3)2(aq) + Na2CO3(aq) → CaCO3(g) + 2NaHCO3(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
    1. Weak acid ✓1
    2. Has few free H+ (Hydrogen) ions
    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) + H2(g)
      H2(g) + O2(g) → H2O(g)
    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

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