Experiment 4: Developmental Of An Equation
Experiment 4: Developmental of an Equation
Purpose
While performing this lab, the student will learn how to determine the formula of the ionic solid produced from the reaction. Also, the student will learn to write the chemical equation corresponding to the reaction.
Procedure
Please refer to General Chemistry Laboratory Experiments, Volume 1, 2011, by Robert Tatz and Judith Casey, Exp 4, pages 32-33, for proper procedure.
Data
See attached sheet.
Report Questions 1. The charge on an iodide ion is -1. It is determined by the formula for potassium iodide in which neither element has subscripts. This signifies that they have equal amount charges and cancel each other out. The charge of the lead ion is +2, as shown on the periodic …show more content…
table and exemplified in the formula PbI2. The formula for lead sulfide would be PbS as the charges cancel each other again. 2. The charge on the chromium ion in the first compound would be +3. The charge on chromium in the second compound would be -2. The two compounds would be named chromium (III) phosphate and chromium (II) phosphate, respectively. 4.
The ion ratio is 3:2. Because of this, none of the test tubes would have either ion in excess.8 drops of sodium phosphate should be added to 12 drops of tin chloride to have neither ion in excess. a. 3SnCl2 + 2Na3PO4 →Sn3(PO4)2 + 6NaCl 7. Yes, if the concentration of potassium iodide was doubled, the correct ratio of ions could still be determined using the same procedure. Test tube three would have no ions in excess due to the new 1:2 ratio.
Results & Conclusion
This experiment was run so that the students could learn to find the formula of ionic solids formed in reactions and to write the chemical equation that corresponds to it. In order to do this, students must first run qualitative tests for the ions in the solution. While testing for anions, iodide and nitrate react with H2O2. Iodide formed separate layers and nitrate had no observable change. While testing for cations, lead and potassium reacted with H2S. Lead turned grey then a dark precipitate settled at the bottom. Potassium remains clear even after heating. Next, lead nitrate was added to potassium iodide. During the test for anions (again with H2O2), the solution turned purple. This signifies the presence of iodine. During the test for cations (again with H2S), the solution turned grey and a grey precipitate formed after heating. This signifies the presence of lead. Lastly, numerous tests were run to determine the exact ratio of ions in the precipitate. Lead nitrate and potassium iodide were
reacted with water at varying ratios. Afterwards, the 5 tubes were divided up into 10 more tubes. 5 reacted with KI and 5 reacted with Pb(NO3)2. The results showed that test tube four had the correct ratio of lead to iodide (1:2). This is determined by the lack of formation of precipitate and the clear solution.
Purpose
While performing this lab, the student will learn how to determine the formula of the ionic solid produced from the reaction. Also, the student will learn to write the chemical equation corresponding to the reaction.
Procedure
Please refer to General Chemistry Laboratory Experiments, Volume 1, 2011, by Robert Tatz and Judith Casey, Exp 4, pages 32-33, for proper procedure.
Data
See attached sheet.
Report Questions 1. The charge on an iodide ion is -1. It is determined by the formula for potassium iodide in which neither element has subscripts. This signifies that they have equal amount charges and cancel each other out. The charge of the lead ion is +2, as shown on the periodic …show more content…
table and exemplified in the formula PbI2. The formula for lead sulfide would be PbS as the charges cancel each other again. 2. The charge on the chromium ion in the first compound would be +3. The charge on chromium in the second compound would be -2. The two compounds would be named chromium (III) phosphate and chromium (II) phosphate, respectively. 4.
The ion ratio is 3:2. Because of this, none of the test tubes would have either ion in excess.8 drops of sodium phosphate should be added to 12 drops of tin chloride to have neither ion in excess. a. 3SnCl2 + 2Na3PO4 →Sn3(PO4)2 + 6NaCl 7. Yes, if the concentration of potassium iodide was doubled, the correct ratio of ions could still be determined using the same procedure. Test tube three would have no ions in excess due to the new 1:2 ratio.
Results & Conclusion
This experiment was run so that the students could learn to find the formula of ionic solids formed in reactions and to write the chemical equation that corresponds to it. In order to do this, students must first run qualitative tests for the ions in the solution. While testing for anions, iodide and nitrate react with H2O2. Iodide formed separate layers and nitrate had no observable change. While testing for cations, lead and potassium reacted with H2S. Lead turned grey then a dark precipitate settled at the bottom. Potassium remains clear even after heating. Next, lead nitrate was added to potassium iodide. During the test for anions (again with H2O2), the solution turned purple. This signifies the presence of iodine. During the test for cations (again with H2S), the solution turned grey and a grey precipitate formed after heating. This signifies the presence of lead. Lastly, numerous tests were run to determine the exact ratio of ions in the precipitate. Lead nitrate and potassium iodide were
reacted with water at varying ratios. Afterwards, the 5 tubes were divided up into 10 more tubes. 5 reacted with KI and 5 reacted with Pb(NO3)2. The results showed that test tube four had the correct ratio of lead to iodide (1:2). This is determined by the lack of formation of precipitate and the clear solution.