Potassium Iodide Lab
We learning about how ions dissolve in water and also how dissolved ions of different compounds react with each other and finally examine the final product for the presence for ions.
Materials-
Sheet of Paper
Lead (II) Nitrate
Potassium Iodide
Deionized Water
Filter Paper
Charcoal
Matches
Equipment-
Clean dry spatula
Small Beakers
Glass stirring rod
Ring stand
Funnel
Tweezers
Bunsen Burner
Small Test Tubes
Safety-
During this lab we are dealing with some dangerous chemical including Lead Nitrate, Potassium Iodide. In this lab you will need to be wearing safety goggles at all times and closed toed shoes and for people that have long hair they need to have it tied back because we will be using a bunsen burner. Also when handling Lead it …show more content…
But when you have tap water there is ions from the sink that bond with the wrong elements and this would mess up the conductivity tests.
The Ions that were in the solid product would be our precipitate from out balanced equation. Since we found that Lead (II) Iodide was our precipitate from the solubility rules. These ions were found in the thick yellow solution when we combine both reactants and the precipitate would be found at the bottom. So in conclusion the ions in the solid product would be Pb-2 and I-1.
PbI2
The spectator ions would be found in the solution because they are not part of the solid product. To test to see if the spectator ions are present in the solution would be to conduct a conductivity test. This would work because the the ions conduct electricity and the solid product does not conduct electricity.
The reaction in this lab was a double replacement reaction. Pb(NO3)2(aq) + 2KI(aq) ---> 2KNO3(aq) + PbI2()
7. Net Ionic Equation: Pb+2(aq) + 2I-1(aq) …show more content…
In this lab we did complete this lab and accomplish our goal of combining dissolved ions of different compounds to form a precipitate. The data we collected shows that the two reactants we used Lead (II) Nitrate and Potassium Iodide were combined to form the solid product Lead (II) Iodide. The observation we made from this reactions shows how those to reactant reacted to form a thick yellow solid product which was Lead (II) Iodide.
In this lab there were many sources of error that could have affect our experiment. One of this possible errors could have been using a wet or a beaker with another substance inside of it that would have other compounds or residue from other past experiments. This source of error would cause the ions in the solution to dissolve with different compounds and the percent yield would be affected by that. Additionally, leaving some water left behind in the solid product which would leave ions in the solid product and this would ultimately mess up test 1 and test 2 by reacting the wrong
Materials-
Sheet of Paper
Lead (II) Nitrate
Potassium Iodide
Deionized Water
Filter Paper
Charcoal
Matches
Equipment-
Clean dry spatula
Small Beakers
Glass stirring rod
Ring stand
Funnel
Tweezers
Bunsen Burner
Small Test Tubes
Safety-
During this lab we are dealing with some dangerous chemical including Lead Nitrate, Potassium Iodide. In this lab you will need to be wearing safety goggles at all times and closed toed shoes and for people that have long hair they need to have it tied back because we will be using a bunsen burner. Also when handling Lead it …show more content…
But when you have tap water there is ions from the sink that bond with the wrong elements and this would mess up the conductivity tests.
The Ions that were in the solid product would be our precipitate from out balanced equation. Since we found that Lead (II) Iodide was our precipitate from the solubility rules. These ions were found in the thick yellow solution when we combine both reactants and the precipitate would be found at the bottom. So in conclusion the ions in the solid product would be Pb-2 and I-1.
PbI2
The spectator ions would be found in the solution because they are not part of the solid product. To test to see if the spectator ions are present in the solution would be to conduct a conductivity test. This would work because the the ions conduct electricity and the solid product does not conduct electricity.
The reaction in this lab was a double replacement reaction. Pb(NO3)2(aq) + 2KI(aq) ---> 2KNO3(aq) + PbI2()
7. Net Ionic Equation: Pb+2(aq) + 2I-1(aq) …show more content…
In this lab we did complete this lab and accomplish our goal of combining dissolved ions of different compounds to form a precipitate. The data we collected shows that the two reactants we used Lead (II) Nitrate and Potassium Iodide were combined to form the solid product Lead (II) Iodide. The observation we made from this reactions shows how those to reactant reacted to form a thick yellow solid product which was Lead (II) Iodide.
In this lab there were many sources of error that could have affect our experiment. One of this possible errors could have been using a wet or a beaker with another substance inside of it that would have other compounds or residue from other past experiments. This source of error would cause the ions in the solution to dissolve with different compounds and the percent yield would be affected by that. Additionally, leaving some water left behind in the solid product which would leave ions in the solid product and this would ultimately mess up test 1 and test 2 by reacting the wrong