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NCERT Exemplar · Class 10 Science Acids, Bases and Salts

48 questions · 48 still being checked

Long Answer Questions 43–48 (part 5 of 5)

  1. Exercise 43

    In the following schematic diagram for the preparation of hydrogen gas as shown in Figure 2.3\displaystyle 2.3, what would happen if following changes are made?
    (a)
    In place of zinc granules, same amount of zinc dust is taken in the test tube
    (b)
    Instead of dilute sulphuric acid, dilute hydrochloric acid is taken
    (c)
    In place of zinc, copper turnings are taken
    (d)
    Sodium hydroxide is taken in place of dilute sulphuric acid and the tube is heated.
    NCERT_Question_Class10_Science_Exemplar_Ch2_Q43

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    NCERT’s answer
    Hint— (a) Hydrogen gas will evolve with greater speed
    (b)
    Almost same amount of gas is evloved
    (c)
    Hydrogen gas is not evolved
    (d)
    If sodium hydroxide is taken, hydrogen gas will be evolved
    \[\begin{array}{ll} \mathrm{Zn}+2 \mathrm{NaOH} \rightarrow & \mathrm{Na}_2 \mathrm{ZnO}_2+\mathrm{H}_2 \\ & \text { Sodium zincate } \end{array} \]
    (a)
    Dust has a far larger surface area than granules \(\displaystyle \Rightarrow\) faster reaction; the total \(\displaystyle \mathrm{H_2}\) is unchanged.
    (b)
    Zinc also displaces hydrogen from dilute \(\displaystyle \mathrm{HCl}\); the gas is the same.
    \[\mathrm{Zn(s) + 2HCl(aq) \rightarrow ZnCl_2(aq) + H_2(g)} \]
    (c)
    Copper lies below hydrogen in the reactivity series \(\displaystyle \Rightarrow\) it cannot displace \(\displaystyle \mathrm{H_2}\) from dilute acid; no gas.
    (d)
    Zinc also reacts with a strong base such as \(\displaystyle \mathrm{NaOH}\) on heating, giving sodium zincate.
    \[\mathrm{Zn(s) + 2NaOH(aq) \xrightarrow{\Delta} Na_2ZnO_2(aq) + H_2(g)} \]
    Answer: (a) \(\displaystyle \mathrm{H_2}\) evolves faster; (b) \(\displaystyle \mathrm{H_2}\) still evolves, \(\displaystyle \mathrm{ZnCl_2}\) forms; (c) no gas; (d) \(\displaystyle \mathrm{H_2}\) evolves, forming \(\displaystyle \mathrm{Na_2ZnO_2}\).

  2. Exercise 44

    For making cake, baking powder is taken. If at home your mother uses baking soda instead of baking powder in cake,
    (a)
    how will it affect the taste of the cake and why?
    (b)
    how can baking soda be converted into baking powder?
    (c)
    what is the role of tartaric acid added to baking soda?

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    NCERT’s answer
    (a)
    Baking soda is sodium hydrogencarbonate. On heating, it is converted into sodium carbonate which is bitter to taste
    \[2 \mathrm{NaHCO}_3 \xrightarrow{\text { Heat }} \mathrm{Na}_2 \mathrm{CO}_3+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2 \]
    (b)
    Baking soda can be converted into baking powder by the addition of appropriate amount of tartaric acid to it.
    (c)
    The role of tartaric acid is to neutralise sodium carbonate and cake will not taste bitter.
    (a)
    Baking soda alone leaves sodium carbonate on heating, so the cake tastes bitter.
    \[\mathrm{2NaHCO_3(s) \xrightarrow{\Delta} Na_2CO_3(s) + H_2O(l) + CO_2(g)} \]
    (b)
    Mix baking soda with a mild edible acid, tartaric acid (plus a starch filler), to get baking powder.
    (c)
    Tartaric acid neutralises the sodium carbonate, removing the bitter taste, and reacts with the soda to release \(\displaystyle \mathrm{CO_2}\), which makes the cake rise.
    Answer: (a) bitter, from \(\displaystyle \mathrm{Na_2CO_3}\); (b) add tartaric acid (and starch); (c) it neutralises the \(\displaystyle \mathrm{Na_2CO_3}\).

  3. Exercise 45

    A metal carbonate X on reacting with an acid gives a gas which when passed through a solution Y gives the carbonate back. On the other hand, a gas G that is obtained at anode during electrolysis of brine is passed on dry Y, it gives a compound Z, used for disinfecting drinking water. Identity X, Y, G and Z.

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    NCERT’s answer
    The gas evolved at anode during electrolysis of brine is chlorine (G) When chlorine gas is passed through dry \(\displaystyle \mathrm{Ca}(\mathrm{OH})_2\) (Y) produces bleaching powder (Z) used for disinfecting drinking water. \[2 \underset{\text{Slaked lime}}{\mathrm{Ca}(\mathrm{OH})_2}+2 \mathrm{Cl}_2 \rightarrow \underset{\text{Bleaching powder}}{\mathrm{Ca}(\mathrm{ClO})_2}+\mathrm{CaCl}_2+2 \mathrm{H}_2 \mathrm{O} \] Since Y and Z are calcium salts, therefore X is also a calcium salt and is calcium carbonate. \[\begin{aligned} & \mathrm{CaCO}_3+2 \mathrm{HCl} \rightarrow \mathrm{CaCl}_2+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O} \\ & \mathrm{Ca}(\mathrm{OH})_2+\mathrm{CO}_2 \rightarrow \mathrm{CaCO}_3+\mathrm{H}_2 \mathrm{O} \end{aligned} \]
    Brine electrolysis gives chlorine at the anode, so \(\displaystyle \mathrm{G = Cl_2}\). \[\mathrm{2NaCl(aq) + 2H_2O(l) \xrightarrow{electrolysis} 2NaOH(aq) + Cl_2(g) + H_2(g)} \] Chlorine on dry Y gives the water disinfectant, bleaching powder, so Y is slaked lime. \[\mathrm{Ca(OH)_2(s) + Cl_2(g) \rightarrow CaOCl_2(s) + H_2O(l)} \] The gas from X and an acid is \(\displaystyle \mathrm{CO_2}\); through a solution of Y (lime water) it gives the carbonate back, so X is calcium carbonate. \[\mathrm{CaCO_3(s) + 2HCl(aq) \rightarrow CaCl_2(aq) + H_2O(l) + CO_2(g)} \] \[\mathrm{CO_2(g) + Ca(OH)_2(aq) \rightarrow CaCO_3(s) + H_2O(l)} \] Answer: \(\displaystyle \mathrm{X = CaCO_3}\), \(\displaystyle \mathrm{Y = Ca(OH)_2}\) (slaked lime), \(\displaystyle \mathrm{G = Cl_2}\), \(\displaystyle \mathrm{Z = CaOCl_2}\) (bleaching powder).

  4. Exercise 46

    A dry pellet of a common base B, when kept in open absorbs moisture and turns sticky. The compound is also a by-product of chloralkali process. Identify B. What type of reaction occurs when B is treated with an acidic oxide? Write a balanced chemical equation for one such solution.

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    NCERT’s answer
    Sodium hydroxide (NaOH) is a commonly used base and is hygroscopic, that is, it absorbs moisture from the atmosphere and becomes sticky. The acidic oxides react with base to give salt and water. The reaction between NaOH and \(\displaystyle \mathrm{CO}_2\) can be given as \[2 \mathrm{NaOH}+\mathrm{CO}_2 \rightarrow \mathrm{Na}_2 \mathrm{CO}_3+\mathrm{H}_2 \mathrm{O} \]
    B is deliquescent and forms in the chlor-alkali process, so B is sodium hydroxide, \(\displaystyle \mathrm{NaOH}\). With an acidic oxide such as \(\displaystyle \mathrm{CO_2}\), \(\displaystyle \mathrm{NaOH}\) undergoes a neutralisation reaction, giving a salt and water. \[\mathrm{2NaOH(aq) + CO_2(g) \rightarrow Na_2CO_3(aq) + H_2O(l)} \] Answer: \(\displaystyle \mathrm{B = NaOH}\); neutralisation (salt + water), e.g. \(\displaystyle \mathrm{2NaOH + CO_2 \rightarrow Na_2CO_3 + H_2O}\).

  5. Exercise 47

    A sulphate salt of Group 2\displaystyle 2 element of the Periodic Table is a white, soft substance, which can be moulded into different shapes by making its dough. When this compound is left in open for some time, it becomes a solid mass and cannot be used for moulding purposes. Identify the sulphate salt and why does it show such a behaviour? Give the reaction involved.

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    NCERT’s answer
    The substance which is used for making different shapes is Plaster of Paris. Its chemical name is calcium sulphate hemihydrate \(\displaystyle \left(\mathrm{CaSO}_4 . \frac{1}{2} \mathrm{H}_2 \mathrm{O}\right)\). The two formula unit of \(\displaystyle \mathrm{CaSO}_4\) share one molecule of water. As a result, it is soft. When it is left open for some time, it absorbs moisture from the atmosphere and forms gypsum, which is a hard solid mass. \[\underset{\substack{\text{Plaster of Paris} \\ \text{(Soft)} \\ \text{(Sulphate salt)}}}{\mathrm{CaSO}_4 . \frac{1}{2} \mathrm{H}_2 \mathrm{O}}+1 \frac{1}{2} \mathrm{H}_2 \mathrm{O} \rightarrow \underset{\substack{\text{Gypsum} \\ \text{(Hard mass)}}}{\mathrm{CaSO}_4 . 2 \mathrm{H}_2 \mathrm{O}} \]
    Plaster of Paris, \(\displaystyle \mathrm{CaSO_4\cdot\tfrac{1}{2}H_2O}\), absorbs water from moist air and sets to gypsum, a rigid solid that cannot be remoulded. \[\mathrm{CaSO_4\cdot\tfrac{1}{2}H_2O(s) + \tfrac{3}{2}H_2O(l) \longrightarrow CaSO_4\cdot 2H_2O(s)} \] Answer: Plaster of Paris, \(\displaystyle \mathrm{CaSO_4\cdot\tfrac{1}{2}H_2O}\); it absorbs moisture from the air and sets into hard gypsum, \(\displaystyle \mathrm{CaSO_4\cdot 2H_2O}\).

  6. Exercise 48

    Identify the compound X on the basis of the reactions given below. Also, write the name and chemical formulae of A, B and C. NCERT_Question_Class10_Science_Exemplar_Ch2_Q48

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    NCERT’s answer
    X— NaOH (Sodium hydroxide) A— \(\displaystyle \mathrm{Na}_2 \mathrm{ZnO}_2\) (Sodium zincate) B— NaCl (Sodium chloride) C— \(\displaystyle \mathrm{CH}_3 \mathrm{COONa}\) (Sodium acetate)
    Salt + water with both acids marks a base; \(\displaystyle \mathrm{H_2}\) with \(\displaystyle \mathrm{Zn}\) marks a strong alkali, so X is sodium hydroxide. \[\mathrm{2NaOH(aq) + Zn(s) \xrightarrow{\Delta} Na_2ZnO_2(aq) + H_2(g)} \] \[\mathrm{NaOH(aq) + HCl(aq) \rightarrow NaCl(aq) + H_2O(l)} \] \[\mathrm{NaOH(aq) + CH_3COOH(aq) \rightarrow CH_3COONa(aq) + H_2O(l)} \] Answer: \(\displaystyle \mathrm{X = NaOH}\); \(\displaystyle \mathrm{A = Na_2ZnO_2}\) (sodium zincate), \(\displaystyle \mathrm{B = NaCl}\) (sodium chloride), \(\displaystyle \mathrm{C = CH_3COONa}\) (sodium acetate).