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Questions 1. Write balanced molecular and ionic equations for each successful precipitation reaction that you observed. a. b. Were the solubility rules completely adequate? Explain. 2. Describe how red litmus paper is affected by acids and bases. a. b. Describe how blue litmus paper is affected by acids and bases. Which litmus paper would you use to test for an acid? Why? c. d. Which litmus paper would you use to test for a base? Why? Write balanced molecular and ionic equations for each reaction of a solid with HCl. 3. a. Write balanced molecular and ionic equations for each neutralization b. reaction in which heat was evolved. . Results 1. Testing the solubility rules 1gtcoti3 b sscisds NaBr Na SO NaOH Solurde No Heat No reaction Do hat Mscluble Nohat Na,CO, Solubla No het Na Yation No feaden Solubeo No heat Scubla NH NO Selulole o heat Solble Seluble Cder dhomes ho Ba(NOs)2 | Arecipitte teodhte peaph SoctePeptate sobble no eat Seluble no readken white Calor dhanc Chanze no hedt no faschon cordy soluble no heast AgNO, nsole No heut Precipbte WD veachion duale No renctesn te wte cole Pb(NOs)2 Caor chonge Precipitate cavdy Seluble Ni(NOs)2 Preciptere no heat C+ alers 33 eoiblo os PIYO Precipitote disappented 2. Looking at dacids and bases HCI+ precipitate From Ni (NO2)s + NaOH gas ewles HCI+CaCO, afu Blue Litmus Paper Red Litmus Paper le idolo Color changed to Pink No reaction HС Color changed to pnk Do veaction HС-Но, No reuction No veachn NHS vityo coler changed Onid to dark blue Color changal NaOH Heat Evolved? No heat HCI NH evolues eligt НС-Н,02 + NaОН

Answer 1

1a) Molecular equation:  Ba(NO3)2 (aq) + Na2SO4 (aq) → BaSO4 (s) + 2NaNO3 (aq)

Ionic equation: Ba²⁺ + SO₄^2- → BaSO4 (s)

Molecular equation: Ba(NO3)₂(aq) + 2NaOH(aq) → Ba(OH)₂(s) + 2NaNO3(aq)

Ionic equation: Ba²⁺(aq) + 2OH^-(aq) → Ba(OH)₂(s)

Molecular equation: Ba(NO3)2(aq) + Na2CO3(aq) --> BaCO3(s) + 2NaNO3(aq)

Ionic equation: Ba²⁺(aq) + CO3^2-(aq) → BaCO3(s)

Molecular equation: 2 AgNO3(aq) + Na2SO4(aq) → Ag2SO4(s) + 2 NaNO3(aq)

Ionic equation: 2Ag⁺(aq) + SO₄^2-(aq) → AgSO₄(s)

Molecular equation: AgNO3(aq) + NaOH(aq)→ AgOH(s) + NaNO3(aq)

Ag⁺(aq) + OH^-(aq) → AgOH(s)

Molecular equation: 2AgNO3(aq) + Na2CO3(aq) → Ag2CO3(s) + 2NaNO3(aq)

2Ag+ + CO3^2- → Ag2CO3(s)

Molecular equation: Na₂SO₄ (aq) + Pb(NO3)2(aq) → 2NaNO₃ (aq) + PbSO₄ (s)

Pb²⁺ (aq) + SO₄^2-(aq) → PbSO₄(s)

Molecular equation: 2NaOH (aq) + Pb(NO3)2(aq) → 2NaNO₃ (aq) + Pb(OH)2(s)

Pb²⁺ (aq) + 2OH^-(aq) → Pb(OH)2(s)

Molecular equation: Na2CO3 (aq) + Pb(NO3)2(aq) → 2NaNO₃ (aq) + PbCO3(s)

Pb²⁺ (aq) + CO3^2-(aq) → PbCO3(s)

Molecular equation: Ni(NO3)2(aq) + Na2SO4(aq) → NiSO4(aq) + 2NaNO3(aq)

Ni⁺²(aq) + SO4^-2(aq) → NiSO4 (s)

Molecular equation: Ni(NO3)2(aq) + 2 NaOH(aq) → Ni(OH)2(s) + 2 NaNO3(aq)

Ni⁺²(aq) + 2HO^-(aq) → Ni(OH)2 (s)

Molecular equation: Ni(NO3)2 (aq) + Na2CO3 (aq) → NiCO3 (s) + 2NaNO3 (aq)

Ni⁺²(aq) + CO3^-2(aq) → NiCO3 (s)

1b) Compounds with group 1A ions, nitrates and sulphates are soluble. Hydroxide ions are soluble with group 1A ions, ammonium ion, Ca2+, Sr2+, Ba2+. Carbonates and phosphates are also soluble with group 1A ions and ammonium ion.

2a) Red litmus paper turns blue when comes in contact with base while it remains red with acid or neutral solution. When base solution comes in contact with red litmus paper, hydrogen ion reacts with base to form conjugate base which has blue color, due to presence of weak diprotic acid in red litmus paper.

2b) Blue litmus paper turns red when comes in contact with acid whereas it remains blue with base and neutral solution. The pigment in blue litmus paper reacts with hydrogen ions wich tunes the bond to give color with longer wavelength i.e. red color.

2c) For acid blue litmus paper is used because change in color of pH below 4.5 turns red.

2d) For base red litmus paper is used because change in color of pH above 8.3 turns blue.

3a) CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + CO₂(g) + H₂O(l)           

Ionic equation: CaCO₃(s) + 2H+(aq) → Ca2+(aq) + CO₂(g) + H₂O(l)   

3b) In Both reaction heat is not evolved HCl + NH3 and NaOH + HC2H3O2