Theoretical vs. Experimental Results (or why the result of an equilibrium calculation might not agree with an experimental result) Thermodynamics and equilibrium chemistry are powerful theoretical tools for modeling a chemical reaction. Using these theories we can predict a reaction’s favorability, predict the reaction’s composition at equilibrium, and, when it is at equilibrium, predict how the reaction will respond to a change in conditions. Such models assume what we might call “ideal conditions;” when conditions are not ideal, our experimental results may not agree with our theoretical predictions. In this exercise we seek to define what we mean by ideal conditions. Theoretical Predictions Consider the equilibrium reaction and the equilibrium constant for the solubility of AgBr \[\ce{AgBr}(s) ⇌ \ce{Ag+}(aq) + \ce{Br-}(aq) \hspace{50px} K_\textrm{sp} = \ce{[Ag+][Br- ]} = 5.0×10^{–13}\nonumber\] and answer the following questions: What are the concentrations of Ag + and Br – in a saturated solution of AgBr (s) ? Will the concentration of Br – increase, decrease, or remain the same if you add AgNO 3 (s) to this saturated solution? Why? Will the concentration of Ag + increase, decrease, or remain the same if you add KBr (s) to this saturated solution? Why? Will the concentration of Br – increase, decrease, or remain the same if you add KNO 3 (s) to this saturated solution? Why? Will the concentration of Ag + increase, decrease, or remain the same if you add KNO 3 (s) to this saturated solution? Why? Review your answers with at least two classmates and, if your answers are not the same, work to resolve your differences. Experimental Results The table below shows experimental results for [Ag + ] and [Br – ] in saturated solutions of AgBr (s) that contain different concentrations of KNO 3 , a strong electrolyte that dissolves completely to form equal concentrations of K + and NO 3 – . [KNO3], M [Ag+], M [Br–], M 0 7.1×10–7 7.1×10–7 1×10–5 7.1×10–7 7.1×10–7 1×10–4 7.2×10–7 7.2×10–7 1×10–3 7.3×10–7 7.3×10–7 1×10–2 7.9×10–7 7.9×10–7 1×10–1 9.5×10–7 9.5×10–7 To what extent do these experimental results agree with and to what extent do they disagree with our theoretical predictions? What is the effect on the molar solubility of AgBr of increasing the concentration of KNO 3 ? What is the effect on the K sp for AgBr of increasing the concentration of KNO 3 ? Review your answers with at least two classmates and, if your answers are not the same, work to resolve your differences. Building a New Theoretical Model AgBr is an ionic compound held together by the electrostatic attraction between the Ag + ions and the Br – ions. What factors affect the strength of an ionic bond? Given the strong attraction between ions of opposite charge, what might prevent all the Ag + ions and Br – ions from recombining to form AgBr (s) ? Explain your answer in terms of the factors that affect the formation of an ionic bond. [ Hint: the solvent, water is a polar molecule. ] What happens at the level of individual ions when you add an inert, strong electrolyte, such as KNO 3 , to a saturated solution of AgBr (s) . Sketch a picture that shows a Ag + ion in an solution that contains NO 3 – ions. Repeat for Br – ions in an solution that contains K + ions. Explain why the molar solubility of AgBr (s) is greater in solutions with higher concentrations of KNO 3 . Making a New Theoretical Prediction Do you expect that AgBr is more soluble, less soluble, or equally soluble in 0.1 M Mg(NO 3 ) 2 than in 0.1 M KNO 3 ? Explain your reasoning. Review your answers with at least two classmates and, if your answers are not the same, work to resolve your differences. Experimental Results The table below shows K sp values for AgBr (s) in deionized water and in several different solutions of strong electrolytes. Solution [electrolyte], M [Ag+] = [Br–] Ksp Deionized water 0 7.1×10–7 5.0×10–13 KNO3 1×10–5 7.1×10–7 5.0×10–13 NaN 1×10–4 7.2×10–7 5.2×10–13 NaN 1×10–3 7.3×10–7 5.3×10–13 NaN 1×10–2 7.9×10–7 6.2×10–13 NaN 1×10–1 9.5×10–7 9.0×10–13 Mg(NO3)2 1×10–5 7.1×10–7 5.0×10–13 NaN 1×10–4 7.2×10–7 5.2×10–13 NaN 1×10–3 7.6×10–7 5.8×10–13 NaN 1×10–2 8.7×10–7 7.6×10–13 NaN 1×10–1 1.2×10–6 1.4×10–12 Do these experimental results agree with your theoretical predictions? If your answer is no, then return to the previous page and reconsider your answers to the questions found there. Do not proceed until you understand why AgBr (s) is more soluble in a solution of Mg(NO 3 ) 2 than in an equimolar solution of KNO 3 . Given what you have learned in completing this exercise, explain why a calculation for the pH of a solution of 0.10 M sodium acetate is not likely to give the same result as an experimental determination of the solution’s pH. Why is it okay to makes simplifying assumptions when solving equilibrium problems? Review your answers with at least two classmates and, if your answers are not the same, work to resolve your differences. So, Where Do Equilibrium Constants Come From? The equilibrium constants in your textbook are for ideal solutions; that is, for solutions in which there are no ions and, therefore, no interactions between ions. Interestingly, for a solubility reaction this constraint is impossible since the equilibrium reaction itself involves ions. The effect of an ion on an equilibrium constant involves both the ion’s concentration and its charge. A common method to account for both an ion’s concentration and its charge is ionic strength, μ, which we define as \[\mu=\dfrac{1}{2}\sum_{i=1}^{n}(c_iz_i^2)\nonumber\] where n is the number of different types of ions, c i is the concentration of an ion, and z i is that ion’s charge. The unit for ionic strength is molarity. Calculate the ionic strength for each of the solutions in the table on page 4. Be sure to include the concentrations of Ag + and Br – , as well as the concentrations of the ions from the strong electrolyte. Feel free to divide the work with other classmates. Solution [electrolyte], M [Ag+] = [Br–] Ksp μ, M Deionized water 0 7.1×10–7 5.0×10–13 NaN KNO3 1×10–5 7.1×10–7 5.0×10–13 NaN NaN 1×10–4 7.2×10–7 5.2×10–13 NaN NaN 1×10–3 7.3×10–7 5.3×10–13 NaN NaN 1×10–2 7.9×10–7 6.2×10–13 NaN NaN 1×10–1 9.5×10–7 9.0×10–13 NaN Mg(NO3)2 1×10–5 7.1×10–7 5.0×10–13 NaN NaN 1×10–4 7.2×10–7 5.2×10–13 NaN NaN 1×10–3 7.6×10–7 5.8×10–13 NaN NaN 1×10–2 8.7×10–7 7.6×10–13 NaN NaN 1×10–1 1.2×10–7 1.4×10–12 NaN For a relatively small range of ionic strengths, a plot of K sp (on the y -axis) as a function of \(\sqrt{μ}\) is a straight-line. Extrapolating this line to an ionic strength of zero gives the thermodynamic equilibrium constant. Construct such a plot using the data in the table above. Fit a straight-line to the data and report the thermodynamic K sp for AgBr (s) . The Moral of the Story A thermodynamic K sp applies only to a solution with an ionic strength of zero. Calculations that use a thermodynamic equilibrium constant will give a predicted result different from an experimental result, and this difference is greater for larger ionic strengths. The relationship between K sp and \(\sqrt{μ}\), however, suggests that appropriate corrections are possible; however, that is a topic for another day and another course.