Chem 145 Lab 4

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Jan 9, 2024

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Name: Elias Douramanis Partner: Austin Cincotta Date: 10/7/2023 EXPERIMENT 4 TITLE: Qualitative Identification of Ionic Solutions OBJECTIVES: 1. To develop a method for the separation and identification of a group of ionic solutions using ionic reactions 2. To learn how to predict possible ionic reactions using solubility rules PROCEDURE REFERENCE: SUNY Broome Community College Department of Chemistry MODIFICATIONS: None
DATA/RESULTS: Tray 2: Lette r Compound J 2M HNO3 G 1M HNO3 I NH3 K H2SO4 M NaOH E CaCl2 L NaBr F NaNO3 H Pb(NO3) N AgNO3 Acids/Bases Salts:
CALCULATIONS: H 2 SO 4 + Pb(NO 3 ) 2 = HNO 3 + PbSO 4 H2SO4 + Pb(NO3)2 = 2HNO3 + PbSO4 2H + ( aq ) + SO 4 2- ( aq ) + Pb 2+ ( aq ) + 2NO 3 - ( aq ) = 2H + ( aq ) + 2NO 3 - ( aq ) + PbSO 4 ( s ) SO 4 2- ( aq ) + Pb 2+ ( aq ) = PbSO 4 ( s ) Pb(NO 3 ) 2 + NaBr = PbBr 2 + NaNO 3 Pb(NO 3 ) 2 + 2NaBr = PbBr 2 + 2NaNO 3 Pb 2+ ( aq ) + 2NO 3 - ( aq ) + 2Na + ( aq ) + 2Br - ( aq ) = PbBr 2 ( s ) + 2Na + ( aq ) + 2NO 3 - ( aq ) Pb 2+ ( aq ) + 2Br - ( aq ) = PbBr 2 ( s ) Pb(NO 3 ) 2 + CaCl 2 → PbCl 2 + Ca(NO 3 ) 2
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Pb 2+ ( aq ) + 2NO 3 - ( aq ) + Ca 2+ ( aq ) + 2Cl - ( aq ) = PbCl 2 ( s ) + Ca 2+ ( aq ) + 2NO 3 - ( aq ) Pb 2+ ( aq ) + 2Cl - ( aq ) = PbCl 2 ( s ) AgNO 3 + CaCl 2 = AgCl + Ca(NO 3 ) 2 2AgNO 3 ( aq ) + CaCl 2 ( aq ) = 2AgCl( s ) + Ca(NO 3 ) 2 ( aq ) 2Ag + ( aq ) + 2NO 3 - ( aq ) + Ca 2+ ( aq ) + 2Cl - ( aq ) = 2AgCl( s ) + Ca 2+ ( aq ) + 2NO 3 - ( aq ) 2Ag + ( aq ) + 2Cl - ( aq ) = 2AgCl( s ) CaCl2 + NaOH = NaCl + Ca(OH) 2 CaCl2 ( aq ) + 2NaOH( aq ) = 2NaCl( aq ) + Ca(OH) 2 ( s ) Ca 2+ ( aq ) + 2Cl - ( aq ) + 2Na + ( aq ) + 2OH - ( aq ) = 2Na + ( aq ) + 2Cl - ( aq ) + Ca(OH) 2 ( s ) Ca 2+ ( aq ) + 2OH - ( aq ) = Ca(OH) 2 ( s ) Pb( NO 3 ) 2 + NaOH = Pb(OH) 2 + NaNO 3 Pb(NO 3 ) 2 ( aq ) + 2NaOH( aq ) = Pb(OH) 2 ( s ) + 2NaNO 3 ( aq ) Pb 2+ ( aq ) + 2NO 3 - ( aq ) + 2Na + ( aq ) + 2OH - ( aq ) = Pb(OH) 2 ( s ) + 2Na + ( aq ) + 2NO 3 - ( aq ) Pb 2+ ( aq ) + 2OH - ( aq ) = Pb(OH) 2 ( s ) HNO 3 + NaOH → NaNO 3 + H 2 O
HNO 3 ( aq ) + NaOH( aq ) = NaNO 3 ( aq ) + H 2 O( l ) H + ( aq ) + NO 3 - ( aq ) + Na + ( aq ) + OH - ( aq ) = Na + ( aq ) + NO 3 - ( aq ) + H 2 O( l ) H + ( aq ) + OH - ( aq ) = H 2 O( l ) RESULTS AND CONCLUSION: This experiment illustrated the method of identifying unknown compounds by employing solubility rules, pH levels, and a flow chart. The use of a litmus test is crucial for distinguishing acids from bases, and understanding pH levels is essential for differentiating between acids and bases. The flow chart is useful in recognizing an unknown compound by applying knowledge of solubility rules, acid-base properties, and specific conditions. According to the solubility rule, a substance will form a precipitate when its concentration in a solution surpasses its solubility limit. The combination of two ionic compounds containing elements that are not soluble can lead to precipitation. Consequently, integrating the solubility rule with the flow chart facilitates the identification of an unknown compound. QUESTIONS: 1. How can you distinguish between an acid and a base (liquids)? The litmus test is used to tell if something is an acid or a base by checking its pH level. If you put blue litmus paper in acid, it turns red. If you put red litmus paper in a base, it turns blue. 2. How can you distinguish between two bases of different concentrations (liquids)? Use litmus paper, and see the shade of blue it produces. 3. How can you distinguish between solid sodium nitrate and solid lead (II) nitrate? By adding CaCl2 to each and seeing what compound PPT.
Solid lead (II) nitrate is insoluble and sodium nitrate is soluble. 4. How can you distinguish between solid sodium hydroxide and solid aluminum hydroxide? By adding NH4OH, Aluminum hydroxide would produce PPT and sodium hydroxide would not produce PPT.
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