This lab experiment explored how the concentration and conductivity of a sodium chloride solution changed as sodium chloride was added incrementally. The solution's conductivity increased with each addition as the number of ions increased in the fixed volume of water. Eventually, adding more sodium chloride no longer increased the conductivity, indicating the solution had reached saturation. The saturated solution was in dynamic equilibrium, with ions continuously dissolving from and crystallizing onto the solid.

1. Lab 3 Solubility

4. Graph to compare conductivity and concentration conductivity concentration Increasing concentration — Increasing conductivity Conductivity no longer increases

8. Sample Data

9. EOS The attractions of water dipoles for ions pulls the ions out of the crystalline lattice and into aqueous solution

11. Dynamic Nature of Equilibrium When a system reaches equilibrium , the forward and reverse reactions continue to occur … but at equal rates. We are usually concerned with the situation after equilibrium is reached. After equilibrium the concentrations of reactants and products remain constant.

12. Dynamic Equilibrium Illustrated NaCl dissolves and recrystallizes continuously. After a time, the solution contains radioactive Na + … NaCl containing radioactive Na + is added to a saturated NaCl solution. … and the added salt now contains some stable Na + .

21. Again, a molar solution The volume is exactly 1.0 L Note that not all of the initial 1.0 L of water was used to prepare the solution

24. Mole Fraction and Mole Percent The mole fraction ( x i ) of a solution component i is the fraction of all the molecules in the solution that are molecules of i EOS The mole percent of a solution component is its mole fraction multiplied by 100 Sum of x i must equal 1

26. Dilution The desired molarity solutions are often prepared from concentrated stock solutions (routinely used solutions prepared in concentrated form) by adding water. This process is called dilution. Moles of solute before dilution = moles of solute after dilution M 1 V 1 = M 2 V 2

27. Figure 4.11 (a) A Measuring Pipet (b) A Volumetric (transfer) Pipet

28. Example: What volume of 16 M sulfuric acid must be used to prepare 1.5 L of 0.10 M H 2 SO 4 solution? V 1 = Volume before dilution = ? M 1 = Concentration before dilution = 16 M V 2 = Volume after dilution = 1.5 L M 2 = Concentration after dilution = 0.10 M M 1 V 1 = M 2 V 2 V 1 =M 2 V 2 / M 1 = (0.10 M x 1.5 L)/ 16 M = 0.0094 L = 9.4 mL