Q: Question 25Starting from insoluble ZnO how would you obtain insoluble ZnCO3 by precipitation.

ZnO + 2HCl ⟶ ZnCl2 + H2OZnCl2 + Na2CO3 ⟶ ZnCO3 + 2NaClDissolve zinc oxide in dil. HCl. Add to it a saturated solution of Na2CO3. The precipitate formed by the interchange of radicals is filtered. It is dried to obtain the zinc carbonate.

Question 1

Question 5Name the ions formed when — HCl; HNO3; H2SO4; CH3COOH; NaOH and NH4OH ionise in aq. soln.

Accepted Answer

(i) When HCl is dissolved in water, it is ionised into hydrogen ion [or H3O+ ion] and chloride ion.HCl ⟶ H+ + Cl-The H+ cannot exist independently, therefore, it combines with water molecule to form hydronium ion (H3O+)H+ + H2O ⟶ H3O+(ii) When HNO3 is dissolved in water, it is ionised into hydronium ion and nitrate ion.HNO3 ⟶ H+ + NO3-H+ + H2O ⟶ H3O+(iii) When H2SO4 is dissolved in water, it is ionised into hydronium ion and sulphate ion.H2SO4 ⟶ 2H+ + SO42-H+ + H2O ⟶ H3O+(iii) When CH3COOH is dissolved in water, it is ionised into hydronium ion and acetate ion.CH3COOH ⟶ CH3COO- + H+H+ + H2O ⟶ H3O+(iii) When NaOH is dissolved in water, it is ionised into sodium ion and hydroxyl ion.NaOH ⟶ Na+ + OH-(iv) When NH4OH is dissolved in water, it is ionised into ammonium ion and hydroxyl ion.NH4OH ⟶ NH4+ + OH-

Question 2

Question 13Give equations for the decomposition of a metallic (i) chloride (ii) nitrate with conc. H2SO4.

Accepted Answer

(i) Decomposition of a metallic chloride —NaCl+H2SO4[conc.]→<200°CNaHSO4+HCl	ext{NaCl} + 	ext{H}_2	ext{SO}_4 [	ext{conc.}] \xrightarrow{\lt 200 \degree	ext{C}}	ext{NaHSO}_4 + 	ext{HCl}NaCl+H2​SO4​[conc.]<200°C​NaHSO4​+HCl(ii) Decomposition of a metallic nitrate with conc. H2SO4 —NaNO3+H2SO4[conc.]→<200°CNaHSO4+HNO3	ext{NaNO}_3 + 	ext{H}_2	ext{SO}_4 [	ext{conc.}] \xrightarrow{\lt 200 \degree	ext{C}}	ext{NaHSO}_4 + 	ext{HNO}_3NaNO3​+H2​SO4​[conc.]<200°C​NaHSO4​+HNO3​

Question 3

Question 16Define pH value. What would you say about the pH of a solution in which(i) H+ aq. ions = OH- ions(ii) evolves CO2 when heated with Na2CO3(iii) OH-ions > H+ aq. ions.

Accepted Answer

pH is defined as the negative logarithm [to the base 10] of the hydrogen ion concentration expressed in moles/litre. Thus, pH = -log10H+. It represents the strength of acids and alkalis, expressed in terms of hydrogen ion concentration [H+ aq.](i) When, H+ ions = OH- ions, the solution is neutral with pH = 7.(ii) When the solution evolves CO2 when heated with Na2CO3, it is acidic in nature with pH less than 7.(iii) when, OH-ions > H+ aq. ions, the solution is basic in nature with pH more than 7.

Question 4

Question 19State the colour change in a universal indicator e.g. pH paper on(i) slightly acidic soil(ii) slightly alkaline soil(iii) dairy milk(iv) human blood tested for medical diagnosis.

Accepted Answer

(i) The pH paper changes to yellow colour on a slightly acidic soil.(ii) The pH paper changes to blue colour on a slightly alkaline soil.(iii) The pH paper changes to green colour in a dairy milk.(iv) The pH paper changes to green colour in human blood tested for medical diagnosis.

Question 5

Question 20Define (i) salt (ii) normal salt (iii) acid salt – with relevant examples and equations.

Accepted Answer

(i) Salt — A salt is a compound formed by partial or complete replacement of the replaceable hydrogen ions of an acid by a metallic ion or ammonium ion [basic radical].NaOH + H2SO4 ⟶ NaHSO4 + H2O [Partial replacement]2NaOH + H2SO4 ⟶ Na2SO4 + 2H2O [Complete replacement](ii) Normal salt — The salt formed by complete replacement of the replaceable hydrogen ion of an acid molecule by a basic radical [metallic or ammonium ion].For example,2NaOH + H2SO4 ⟶ Na2SO4 + 2H2O2NaOH + H2SO3 ⟶ Na2SO3 + 2H2O[Both H ions in sulphuric and sulphurous acid are replaced by metallic radical — sodium. ](3) Acid salt — The salt formed by partial replacement of the replaceable hydrogen ion of an acid molecule by a basic radical [metallic or ammonium ion]. For example,NaOH + H2SO4 ⟶ NaHSO4 + H2ONaOH + H2SO3 ⟶ NaHSO3 + H2O[Only one H ion in sulphuric and sulphurous acid is replaced by metallic radical — sodium. ]

Question 6

Question 21State (i) the formation (ii) the components of – a basic salt.State which of following salts is an – acid, normal or basic salt.(i) bleaching powder(ii) potassium mercuric iodide(iii) sodium sulphite(iv) sodium hydrogen sulphite(v) sodium silver cyanide(vi) basic lead nitrate(vii) potassium zincate(viii) alum(ix) calcium bicarbonate(x) basic copper chloride(xi) trisodium phosphate.

Accepted Answer

(i) Formation of a basic salt — A basic salt is formed by partial replacement of hydroxyl radicals of a diacidic or triacidic base with an acid radical.(ii) Components of a basic salt — A basic salt contains a cation [metallic], a hydroxyl ion [of a base] and an anion [of an acid].For example — Basic copper nitrate Cu[OH]NO3 , basic copper chloride Cu[OH]ClAcid, normal or basic salt —(i) bleaching powder — Normal salt (Mixed salt)(ii) potassium mercuric iodide — Normal salt (Complex salt)(iii) sodium sulphite — Normal salt(iv) sodium hydrogen sulphite — Acid salt(v) sodium silver cyanide — Normal salt (Complex salt)(vi) basic lead nitrate — Basic salt(vii) potassium zincate — Normal salt(viii) alum — Normal salt (Double salt)(ix) calcium bicarbonate — Acid salt(x) basic copper chloride — Basic salt(xi) trisodium phosphate — Normal salt

Question 7

Question 22Name three (i) sulphates (ii) chlorides insoluble in water and two (i) oxides (ii) carbonates soluble in water.

Accepted Answer

Insoluble in Water(i) Three sulphates insoluble in water are:Lead sulphate (PbSO4)Calcium sulphate (CaSO4)Barium sulphate (BaSO4).(ii) Three chloride insoluble in water are:Silver chloride (AgCl)Lead chloride (PbCl2)Mercury chloride (Hg2Cl2).Soluble in Water(i) Two oxides soluble in water are:Sodium oxide (Na2O)Potassium oxide (K2O)(ii) Two carbonates soluble in water are:Sodium carbonate (Na2CO3)Ammonium carbonate [(Na4)2CO3]

Question 8

Question 23State the method only, generally used for the preparation of the following salts(i) Zn(NO3)2(ii) NH4Cl(iii) ZnSO4(iv) ZnS(v) CaCO3(vi) FeCl3(vii) PbCl2(viii) Pb(NO3)2

Accepted Answer

Method of preparation are as follows :(i) Zn(NO3)2 — By action of dilute acids on carbonate.(ii) NH4Cl — By neutralisation of an alkali (titration)(iii) ZnSO4 — By displacement of active metal and acid(iv) ZnS — Direct combination (Synthesis)(v) CaCO3 — By precipitation (double decomposition)(vi) FeCl3 — Direct combination (Synthesis)(vii) PbCl2 — Precipitation (double decomposition)(viii) Pb(NO3)2 — Action of dil. acid on carbonates and bicarbonates

Question 9

Question 24Give balanced equations for the preparation of the following salts –(a) (i) CuSO4(ii) NaHSO4(iii) Na2SO4(iv) FeSO4(v) BaSO4(vi) PbSO4 — using dil. H2SO4(b) (i) NaHSO4(ii) CuSO4 — using conc. H2SO4

Accepted Answer

(a) Using dil. H2SO4 :Preparation of CuSO4:CuO + H2SO4 ⟶ CuSO4 + H2OPreparation of NaHSO4:NaOH + H2SO4 ⟶ NaHSO4 + H2OPreparation of Na2SO4:Na2CO3 + H2SO4 ⟶ Na2SO4 + H2O + CO2Preparation of FeSO4:Fe + H2SO4 ⟶ FeSO4 + H2Preparation of BaSO4:BaCl2 + H2SO4 ⟶ BaSO4 + 2HClPreparation of PbSO4:Pb(NO3)2 + H2SO4 ⟶ PbSO4 + 2HNO3(b) Using conc. H2SO4 :Preparation of NaHSO4:NaCl+H2SO4[conc.]→<200°CNaHSO4+HCl	ext{NaCl} + 	ext{H}_2	ext{SO}_4 [	ext{conc.}] \xrightarrow{\lt 200 \degree	ext{C}}	ext{NaHSO}_4 + 	ext{HCl}NaCl+H2​SO4​[conc.]<200°C​NaHSO4​+HClPreparation of CuSO4:CuO + H2SO4 ⟶ CuSO4 + H2O

Question 10

Question 25Starting from insoluble ZnO how would you obtain insoluble ZnCO3 by precipitation.

Accepted Answer

ZnO + 2HCl ⟶ ZnCl2 + H2OZnCl2 + Na2CO3 ⟶ ZnCO3 + 2NaClDissolve zinc oxide in dil. HCl. Add to it a saturated solution of Na2CO3. The precipitate formed by the interchange of radicals is filtered. It is dried to obtain the zinc carbonate.

Question 11

Question 26Give balanced equations for the action of a dilute acid on(i) zinc carbonate,(ii) potassium bicarbonate for the preparation of the respective salt.

Accepted Answer

(i) ZnCO3 + 2HNO3 (dil) ⟶ Zn(NO3)2 + H2O + CO2(ii) 2KHCO3 + H2SO4 (dil) ⟶ K2SO4 + 2H2O + 2CO2

Question 12

Question 27Give balanced equations for the decomposition ofcalcium bicarbonate by dil. HClcalcium carbonate by dil. HNO3sodium sulphite by dil. H2SO4zinc sulphide by dil. H2SO4.

Accepted Answer

Decomposition of calcium bicarbonate by dil. HClCa(HCO3)2 + 2HCl ⟶ CaCl2 + 2H2O+ 2CO2Decomposition of calcium carbonate by dil. HNO3CaCO3 + 2HNO3 ⟶ Ca(NO3)2 + H2O+ CO2Decomposition of sodium sulphite by dil. H2SO4Na2SO3 + H2SO4 (dil.) ⟶ Na2SO4 + H2O + SO2Decomposition of zinc sulphide by dil. H2SO4ZnS + H2SO4 ⟶ ZnSO4 + H2S

Question 13

Question 28State what will be the effect of each of the following solution on blue litmus —(i) K2CO3 soln.(ii) KCl soln.(iii) NH4NO3 soln.

Accepted Answer

(i) As K2CO3 is a salt of a strong base (KOH) and weak acid (H2CO3), it hydrolyses in water to give alkaline solutions. They have pH more than 7, hence it will have no effect on blue litmus.(ii) KCl is a salt of a strong acid (HCl) and a strong base (KOH). Hence it's aqueous solution will be neutral in nature and will have no effect on blue litmus.(iii) NH4NO3 is a salt of a strong acid (HNO3) and weak base (NH4OH), it hydrolyses in water to give an acidic solution. They have pH less than 7, hence will turn blue litmus red.

Question 14

Question 1ACIDS — Ions present in acidsComplete and balance the equationsa. Definition — Dissolves in water yielding hydronium ions 1. Hydrochloric acidHCl ⇌ ............... + Cl-[H+ + H2O ⇌ ............... ]HCl + H2O ⇌ ............... + ...............2. Nitric acidHNO3 + H2O ⇌ ............... + ...............3. Sulphuric acidH2SO4 + H2O ⇌ ............... + ...............b. Classification     4. Monobasic acid [Basicity = 1 ]HCl [aq.] ⇌ ............... + ...............5. Dibasic acid [Basicity = 2]H2SO4 [aq.] ⇌ ............... + ...............6. Tribasic acid [Basicity = 3]H3PO4 [aq.] ⇌ ............... + ...............c. Preparation of acids     7. From non-metalsH2 + Cl2 ⟶ ...............8. From acidic oxidesCO2 + H2O ⟶ ...............   SO3 + H2O ⟶ ...............   P2O5 + H2O ⟶ ...............9. From normal saltsKNO3 + H2SO4 ⟶ ............... + ...............   NaCl+ H2SO4 ⟶ ............... + ...............10. By oxidation of non-metalsS + HNO3 ⟶ ............... + H2O ............... [g]d. Properties of acids     11. Neutralizes basesCuO + H2SO4 ⟶ ............... + ...............   NaOH + HCl ⟶ ............... + ...............Reaction with12. Chlorides and nitratesNaCl+H2SO4[conc.]→<200°C...............+...............	ext{NaCl} + 	ext{H}_2	ext{SO}_4 [	ext{conc.}] \xrightarrow{\lt 200 \degree	ext{C}} ............... + ...............NaCl+H2​SO4​[conc.]<200°C​...............+............... NaNO3+H2SO4[conc.]→<200°C...............+...............	ext{NaNO}_3 + 	ext{H}_2	ext{SO}_4 [	ext{conc.}] \xrightarrow{\lt 200 \degree	ext{C}} ............... + ...............NaNO3​+H2​SO4​[conc.]<200°C​...............+...............13. Carbonates and bicarbonatesNa2CO3 + H2SO4 ⟶ ............... + H2O + ............... [g]   NaHCO3 + H2SO4 ⟶ ............... + H2O + ............... [g]14. Sulphites and bisulphitesNa2SO3 + HCl ⟶ ............... + H2O + ............... [g]   NaHSO3 + HCl ⟶ ............... + H2O + ............... [g]15. Active metalsZn + HCl ⟶ ............... + ............... [g]

Accepted Answer

ACIDS — Ions present in acidsComplete and balance the equationsa. Definition — Dissolves in water yielding hydronium ions1. Hydrochloric acidHCl ⇌ H+ + Cl-[H+ + H2O ⇌ H3O+]HCl + H2O ⇌ H3O+ + Cl-2. Nitric acidHNO3 + H2O ⇌ H3O+ + NO3-3. Sulphuric acidH2SO4 + 2H2O ⇌ 2H3O+ + SO42-b. Classification     4. Monobasic acid [Basicity = 1 ]HCl [aq.] ⇌ H3O+ + Cl-5. Dibasic acid [Basicity = 2]H2SO4 [aq.] ⇌ 2H3O+ + SO42-6. Tribasic acid [Basicity = 3]H2PO4 [aq.] ⇌ 3H3O+ + PO43-c. Preparation of acids     7. From non-metalsH2 + Cl2 ⟶ 2HCl8. From acidic oxidesCO2 + H2O ⟶ H2CO3   SO3 + H2O ⟶H2SO4   P2O5 + 3H2O ⟶2H3PO49. From normal saltsKNO3 + H2SO4 ⟶KHSO4 + HNO3   NaCl+ H2SO4 ⟶NaHSO4 + HCl10. By oxidation of non-metalsS + 6HNO3 ⟶H2SO4 + 2H2O + 6NO2 [g]d. Properties of acids     11. Neutralizes baseCuO + H2SO4 ⟶ CuSO4 + H2O   NaOH + HCl ⟶ NaCl + H2OReaction with12. Chlorides and nitratesNaCl+H2SO4[conc.]→<200°CNaHSO4+HCl	ext{NaCl} + 	ext{H}_2	ext{SO}_4 [	ext{conc.}] \xrightarrow{\lt 200 \degree	ext{C}} \boldsymbol{NaHSO}_\bold{4} + \boldsymbol{HCl}NaCl+H2​SO4​[conc.]<200°C​NaHSO4​+HCl NaNO3+H2SO4[conc.]→<200°CNaHSO4+HNO3	ext{NaNO}_3 + 	ext{H}_2	ext{SO}_4 [	ext{conc.}] \xrightarrow{\lt 200 \degree	ext{C}} \boldsymbol{NaHSO}_\bold{4} + \boldsymbol{HNO}_\bold{3}NaNO3​+H2​SO4​[conc.]<200°C​NaHSO4​+HNO3​13. Carbonates and bicarbonatesNa2CO3 + H2SO4 ⟶ Na2SO4 + H2O + CO2 [g]   2NaHCO3 + H2SO4 ⟶ Na2SO4 + 2H2O + 2CO2 [g]14. Sulphites and bisulphitesNa2SO3 + 2HCl ⟶ 2NaCl + H2O + SO2 [g]   NaHSO3 + HCl ⟶ NaCl + H2O + SO2 [g]15. Active metalsZn + 2HCl ⟶ ZnCl2 + H2 [g]

Question 15

Question 2BASES — Ions present in bases     a. Alkali — dissociates yielding hydroxyl ions     16. Sodium hydroxideNaOH [aq.] ⇌ ............... + ...............17. Ammonium hydroxideNH4OH [aq.] ⇌ ............... + ...............b. Classification     18. Monoacidic base [Acidity = 1 ]KOH [aq.] ⇌ ............... + ...............19. Diacidic base [Acidity = 2]Cu(OH)2 [aq.] ⇌ ............... + ...............c. Preparation of bases     20. From metalsNa + O2 ⟶ ...............21. From metallic oxides and metalsK2O + H2O ⟶ ...............   K + H2O ⟶ ............... + ...............22.From saltsAlCl3 + NaOH ⟶ ............... + ............... ↓   FeSO4 + NaOH ⟶ ............... + ............... ↓23. By thermal decompositionZnCO3 ⟶ ............... + ............... [g]   Pb(NO3)2 ⟶ ............... + ...............[g] + .............. [g]c. Properties of bases     24. Neutralizes acidsPbO + HNO3 ⟶ ............... + ...............   Fe(OH)2 + HCl ⟶ ............... + ...............Reaction with25. Metallic salt solutionCuCl2 + NaOH ⟶ ............... + ............... ↓   FeCl3 + NaOH ⟶ ............... + ............... ↓26. Ammonium saltsNH4Cl + NaOH ⟶ ............... + ............... + ............... [g]   NH4Cl + Ca(OH)2 ⟶ ............... + ............... + ............... [g]

Accepted Answer

BASES — Ions present in bases     a. Alkali — dissociates yielding hydroxyl ions     16. Sodium hydroxideNaOH [aq.] ⇌ Na+ + OH-17. Ammonium hydroxideNH4OH [aq.] ⇌ NH4+ + OH-b. Classification     18. Monoacidic base [Acidity = 1 ]KOH [aq.] ⇌ K+ + OH-19. Diacidic base [Acidity = 2]Cu(OH)2 [aq.] ⇌ Cu2+ + 2OH-c. Preparation of bases     20. From metals4Na + O2 ⟶ 2Na2O21. From metallic oxides and metalsK2O + H2O ⟶ 2KOH   2K + 2H2O ⟶ 2KOH + H222. From saltsAlCl3 + 3NaOH ⟶ 3NaCl + Al(OH)3 ↓   FeSO4 + 2NaOH ⟶ Na2SO4 + Fe(OH)2 ↓23. By thermal decompositionZnCO3 ⟶ ZnO + CO2 [g]   2Pb(NO3)2 ⟶ 2PbO + 4NO2 [g] + O2 [g]c. Properties of bases     24. Neutralizes acidsPbO + 2HNO3 ⟶ Pb(NO3)2 + H2O   Fe(OH)2 + 2HCl ⟶ FeCl2 + 2H2OReaction with25. Metallic salt solutionCuCl2 + 2NaOH ⟶ 2NaCl + Cu(OH)2 ↓   FeCl3 + 3NaOH ⟶ 3NaCl + Fe(OH)3 ↓26. Ammonium saltsNH4Cl + NaOH ⟶ NaCl + H2O + NH3 [g]   2NH4Cl + Ca(OH)2 ⟶ CaCl2 + 2H2O + 2NH3 [g]

Question 16

Question 3SALTS — Ions present in salts   27a. Normal salt [formation]NaOH [excess] + H2SO4 ⟶ ............... + H2O27b. Acid salt [formation]NaOH [insufficient] + H2SO4 ⟶ ............... + H2OPreparation of salts — General methods   Direct combination — Synthesis   28. Iron and Chloride — salt iron [III] ChlorideFe + Cl2 ⟶ ...............29. Zinc and sulphur — salt zinc sulphideZn + S ⟶ ...............Displacement — Action of dilute acid on active metals   30. Iron — salt iron [II] sulphateFe + H2SO4 ⟶ ............... + ............... [g]31. Zinc — salt zinc sulphateZn + H2SO4 ⟶ ............... + ............... [g]32. Magnesium — salt magnesium chlorideMg + HCl ⟶ ............... + ............... [g]Precipitation — by double decomposition of two salt solutions   33. Lead nitrate and sodium chloridePb(NO3)2 + NaCl ⟶ ............... + ............... ↓34. Calcium chloride and sodium carbonateCaCl2 + Na2CO3 ⟶ ............... + ............... ↓Neutralization — Action of dilute acid on insoluble base   35. Oxide — salt copper sulphateCuO + H2SO4 ⟶ ............... + ...............36. Hydroxide — salt copper sulphateCu(OH)2 + H2SO4 ⟶ ............... + ...............Neutralization — [Titration] Action of dilute acid on an alkali   37. Hydroxide — salt sodium chlorideNaOH + HCl ⟶ ............... + ...............38. Hydroxide — salt ammonium chlorideNH4OH + HCl ⟶ ............... + ...............Action of dilute acid on carbonate & bicarbonate   39. Carbonate — salt lead nitratePbCO3 + HNO3 ⟶ ............... + ............... + ............... [g]40. Carbonate — salt copper chlorideCuCO3 + HCl ⟶ ............... + ............... + ............... [g]41. Bicarbonate — salt potassium sulphateKHCO3 + H2SO4 ⟶ ............... + ............... + ............... [g]

Accepted Answer

SALTS — Ions present in salts   27a. Normal salt [formation]2NaOH [excess] + H2SO4 ⟶ Na2SO4 + 2H2O27b. Acid salt [formation]NaOH [insufficient] + H2SO4 ⟶ NaHSO4 + H2OPreparation of salts — General methods   Direct combination — Synthesis   28. Iron and Chloride — salt iron [III] Chloride2Fe + 3Cl2 ⟶ 2FeCl329. Zinc and sulphur — salt zinc sulphideZn + S ⟶ ZnSDisplacement — Action of dilute acid on active metals   30. Iron — salt iron [II] sulphateFe + H2SO4 ⟶ FeSO4 + H2 [g]31. Zinc — salt zinc sulphateZn + H2SO4 ⟶ ZnSO4 + H2 [g]32. Magnesium — salt magnesium chlorideMg + 2HCl ⟶ MgCl2 + H2 [g]Precipitation — by double decomposition of two salt solutions   33. Lead nitrate and sodium chloridePb(NO3)2 + 2NaCl ⟶ 2NaNO3 + PbCl2 ↓34. Calcium chloride and sodium carbonateCaCl2 + Na2CO3 ⟶ 2NaCl + CaCO3 ↓Neutralization — Action of dilute acid on insoluble base   35. Oxide — salt copper sulphateCuO + H2SO4 ⟶ CuSO4 + H2O36. Hydroxide — salt copper sulphateCu(OH)2 + H2SO4 ⟶ CuSO4 + 2H2ONeutralization — [Titration] Action of dilute acid on an alkali   37. Hydroxide — salt sodium chlorideNaOH + HCl ⟶ NaCl + H2O38. Hydroxide — salt ammonium chlorideNH4OH + HCl ⟶ NH4Cl + H2OAction of dilute acid on carbonate & bicarbonate   39. Carbonate — salt lead nitratePbCO3 + 2HNO3 ⟶ Pb(NO3)2 + H2O + CO2 [g]40. Carbonate — salt copper chlorideCuCO3 + 2HCl ⟶ CuCl2 + H2O + CO2 [g]41. Bicarbonate — salt potassium sulphate2KHCO3 + H2SO4 ⟶ K2SO4 + 2H2O + 2CO2 [g]

Question 17

Question 7(a) Dry HCl gas does not change the colour of dry litmus paper. Why?(b) Is PbO2 a base or not? Comment.(c) Do basic solutions also have H+(aq.) ? Explain why are they basic by taking an example?

Accepted Answer

(a) The colour of litmus paper changes only in presence of ions like hydrogen [H+] or hydronium ions [H3O+]. HCl can produce these ions only in the form of aqueous solution. Hence, dry HCl gas does not change the colour of dry litmus paper.(b) A base reacts with an acid to form salt and water ONLY. The word ONLY is of importance in the definition of a base. Lead oxide reacts with hydrochloric acid to produce lead (II) chloride (a salt) and water, but the word "only" excludes it from the class of bases because chlorine is also produced.PbO2(s) + 4HCl(aq.) ⟶ PbCl2(aq.) + Cl2(g) + 2H2OThus, PbO2 is not a base.(c) Yes, basic solutions have H+ ions, but the concentration of OH- ions is more than the H+ ions which makes the solution basic in nature.Example — Sodium hydroxide (NaOH). It has more OH- ions than H+ ions, so it is basic in nature.

Question 18

Question 18You are supplied with five solutions : A, B, C, D and E with pH values as follows:A = 1.8, B = 7, C = 8.5, D = 13, and E = 5Classify these solutions as neutral, slightly or strongly acidic and slightly or strongly alkaline.Which solution would be most likely to liberate hydrogen with:(a) magnesium powder(b) powdered zinc metal.Give a word equation for each reaction.

Accepted Answer

A = Strongly acidicB = neutralC = slightly alkalineD = Strongly alkalineE = slightly acidic(a) Solution A (acidic solution)Solution A (acidic solution) + magnesium powder ⟶ Hydrogen + Mg salt(b) Solution A (acidic solution)Solution A (acidic solution) + powdered zinc metal ⟶ Hydrogen + Zn salt

Question 19

Question 4How would you prepare :(a) copper sulphate crystals from a mixture of charcoal and black copper oxide.(b) zinc sulphate crystals from zinc dust (powdered zinc and zinc oxide),(c) sodium hydrogen carbonate crystals.(d) calcium sulphate from calcium carbonate.

Accepted Answer

(a) Copper sulphate crystals from a mixture of charcoal and black copper oxideMethod of preparation: Neutralization of insoluble baseReaction:CuO + H2SO4 ⟶ CuSO4 + H2OCuSO4 + 5H2O ⟶ CuSO4.5H2O (blue vitriol)Procedure:Take dilute sulphuric acid in a beaker and heat it on wire gauze.Add black cupric oxide in small quantities at a time, with stirring till no more of it dissolves and the excess compound settles to the bottom.Filter it hot and collect the filtrate in a china dish. Evaporate the filtrate by heating to the point of crystallization and then allow it to cool andCollect the bright blue crystals of copper (II) sulphate pentahydrate CuSO4.5H2O (blue vitriol) and dry the crystals.(b) Zinc sulphate crystals from Zinc dustMethod of preparation: Displacement — By the action of dilute acid (dil. H2SO4) on an active metal (Zinc, Iron).Reaction:Zn(s) + H2SO4(aq) ⟶ ZnSO4 + H2↑ZnSO4 + 7H2O ⟶ ZnSO4.7H2O (white vitriol)Procedure:Take dilute sulphuric acid in a beaker and heat it on a wire gauze. Add some granulated zinc pieces with constant stirring.Effervescence take place because of the liberation of hydrogen gas.Add zinc till the Zinc settles at the base of the beaker.When effervescence stops, it indicates that all the acid has been used up.The excess of zinc is filtered off.Collect the solution in a china dish and evaporate the solution to get crystals. Filter, wash them with water and dry them between the folds of a filter paper. The white, needle-shaped crystals are of hydrated Zinc sulphate.(c) Sodium hydrogen carbonate crystals:Method of preparation: By passing carbon dioxide gas into a cold solution of sodium carbonate.Reaction:Na2CO3 + CO2 + H2O ⟶ 2NaHCO3Procedure:Dissolve 5 grams of anhydrous sodium carbonate in about 25 cm3 of distilled water in a flask.Cool the solution by keeping the flask in a freezing mixture.Pass carbon dioxide gas in the solution.Crystals of sodium bicarbonate will precipitate out after some time.Filter the crystals and dry it in the folds of filter paper.(d) calcium sulphate from calcium carbonate.Method of preparation: By Double Decomposition (precipitation)Reaction:CaCO3 + 2HNO3 ⟶ Ca(NO3)2 + H2O + CO2↑Ca(NO3)2 + Na2SO4 ⟶ CaSO4↓ + 2NaNO3Procedure:First insoluble Calcium Carbonate [CaCO3] is converted to soluble Calcium Nitrate [Ca(NO3)2] with the help of dilute nitric acid.The resulting solution is then treated with Sodium Sulphate [Na2SO4] and a precipitate of Calcium Sulphate is obtained.

Question 20

Question 6Name:(a) a chloride which is insoluble in cold water but dissolves in hot water,(b) a chloride which is insoluble(c) two sulphates which are insoluble(d) a basic salt,(e) an acidic salt,(f) a mixed salt,(g) a complex salt,(h) a double salt.

Accepted Answer

(a) a chloride which is insoluble in cold water but dissolves in hot water - Lead Chloride [PbCl2](b) a chloride which is insoluble - Silver chloride [AgCl](c) two sulphates which are insoluble - Barium sulphate [BaSO4] and Lead Sulphate [PbSO4](d) a basic salt - Basic lead chloride [Pb(OH)Cl](e) an acidic salt - Sodium hydrogen sulphate [NaHSO4](f) a mixed salt - Sodium potassium carbonate [NaKCO3](g) a complex salt - Sodium zincate [Na2ZnO2](h) a double salt - Dolomite [CaCO3.MgCO3]

Acids, Bases and Salts

Additional Questions

Equation Worksheet

Exercise 3A

Exercise 3B

Intext Questions 1

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