Red Dye 2 Lab Report

Red Dye 1 Red dye 1 is most likely a polar covalent molecule since we have found that it dissolves into water but does not break apart into ions. Just because it dissolves in water does not make it solely polar covalent, but since it has a relatively low conductivity, 534 mS/cm in the 0-20000 range, then it most likely is polar covalent. This is further confirmed with the fact that it does not react with NaOH, NaCl, or AgNO3, all ionic substances. The red dye does not react with NaOH since there was no color change when added, and the conductivity simple rose as the NaOH was added, due to the fact the excess ions were being added to the solution and not reaction at all. There was no color change when NaCl or AgNO3 was added to red dye 1, …show more content…
Red dye 2 has a conductivity of 408 mS/cm in the 0-20000 range, which would suggest that it’s either not ion or has a low amount of free ions. There is a small amount of scattering seen, suggesting there may be a few, but minimal, particles in the solution. When it is reacted with NaOH, there is not color change and the conductivity graph is a positive, straight line which shows that there is no reaction and the conductivity is only changing due to excess NaOH. This may make it seen nonionic, but when it was reacted with NaCl and AgNO3 there were color changes. This shows that the substance is ionic, and that the products when it reacts with NaOH are soluble so it seems like there is no reaction. It is known that there are less free ions in this solution than in red dye 3 due to the drastic difference between charges (408 mS/cm vs 28965 mS.cm). Due to the fact that there was so odor to this substance, there can be no further predictions of molecular shape or make up. It may be helpful to continue reacting this substance with other ionic substances to narrow down the molecules by whether the products are soluble or not. Why does this not have an odor? Why is its conductivity so low if it is