Ksp Value For His Experiment example essay topic

586 words
When ionic solids dissolve, they divide to give their positive and negative ions that make up the solids. These ions become hydrates and have the same relative proportions when in solution and when solid. The more the solid dissolves, the more the ion's concentration increases. This increase and build-up allows for the reverse reaction to occur. In this phase of the reaction the ions crystal lise out in order for the reaction to have a greater chance of occurring. Eventually the rate of dissolving will equal the rate of crystallization.

This is the state of saturation. This can be recognised by a constant colour or constant mass. The solubility product constant, Ksp is given in the following example: Ksp for Ag Cl is Ksp = [Ag] [Cl] Ksp for PbI 2 is Ksp = [Pb] [I] 2 This gives the relationship between the ions in the saturated solution and is the maximum concentration possible without creating precipitation. In this lab, solutions of lead nitrate and potassium iodide will be mixed at a number of dilutions. The reactions will then be observed to see at which point a precipitate no longer occurs. Ksp will then be stated as a range of values at room temperature, and the precipitate test tubes will be heated until the precipitate is dissolved so that Ksp may be observed and determined at different levels.

In this experiment various solutions of lead nitrate and potassium iodide were mixed at a number of different dilutions. Through the observation of the amount - or lack of precipitate formed in each dilution, the mathematical relationship between the ions in a saturated dilution may be determined. This relationship is known as the solubility product constant, or Ksp, and is defined as follows, 'The Ksp for an ionic solid is given by the product of the concentrations of the ions, each raised to the power of the coefficients in the dissolving reaction. ' ; (Heath Chemistry). The Ksp expression gives the maximum possible concentration of ions in a saturated solution without causing precipitation. Based on the equation for this experiment we can conclude that the ks p expression for this experiment is: Ksp = [Pb] [I] 2 The Ksp for lead iodide at 25 oC is 8.5 x 10^-9.

Through the substitution of the values for each test tube, the trial product, or KSPtrial was obtained. A precipitate then formed in those test tubes who's KSPtrial was greater than the Ksp, whereas if the KSPtrial was calculated as being less than the Ksp, no precipitate formed. Therefor the expression with the KSPtrial closest to the Ksp was the one used at room temperature. Through this information we were able to determine the Ksp value for his experiment, which ranges between 3.20 x 10^-8 and 1.35 x 10^-8, or between test tubes 4 and 5.

We may further determine that lead iodide is less soluble in a solution of. 1 M KI than in pure water. Adding KI would cause a shift in the equilibrium to the reactants, which would further result in the decrease of the lead's solubility. In order to dissolve the coating of lead iodide on the inside of a test tube, you could either decrease the amount of PB or of I in the solution. Thus causing a shift to the products, which would initiate the reaction to dissolve.