Question

I MUCEDURES SHOP log indoor Your team will examine the impact of changing concentration and temperature on the position of equilibrium (Part I). Compare your observations with different teams and look for patterns in results to identify factors that shift the position of equilibrium toward reactants or products. In Part II, test your understanding as you make predictions and explain any shift in the equilibrium position of a new equilibrium system. Part Observations of Equilibria You and your teammate(s) will observe and interpret the impact of changing conditions (concentration of reactants or products, addition of different reagents, and temperature change) on several chemical equilibrium systems and collect data. When you are done, share your results with the other teams. In case of discrepancies in observations, repeat the procedure. Part IA-Saturated Sodium Chloride - moto bomb A saturated solution of sodium chloride, NaCl, is 5.4 M. What conditions will make the precipitation of NaCl (S) as complete as possible? The reaction is: duto Na+ C Na" (aq) + CI+ (aq) NC NaCl (s) button to be 1900 mundo coloresso White Shop Smoobrodo muito bodo to mozog 1. Transfer about 3 mL of saturated NaCl to each of six separate labeled (1-6) test tubes in a test tube rack. Save tube 1 for reference. 2. Take your test tube rack to the fume hood. CAREFULLY add a few drops of concentrated hydrochloric acid, HCl, to the contents of tube 2. Caution: Keep HCl and HNO3 off your skin and clothing. Wash any contact points with copious amounts of water. Add concentrated HCl one drop at a time until you have added about 5-6 drops or a change is observed. Compare the contents of this tube with your reference tube. Record the total number of drops added and your observations. Since no additional Nal(aq) ions were added, how do you account for the observations? 3. CAREFULLY add concentrated nitric acid, HNO3, instead of hydrochloric acid, to tube 3. Ilse the pipet provided in the bottle of acid to add the same number of drops of HCl (step 2) Compare the contents of this tube with the contents of the second tube. Record your observations. reactants and products from steps 1-3. Which ion is responsible for the observed 4. Compare the reactants and products from steps 1_3 change in tube 2? How do you know?

Answer 1

The concentration of the saturated solution of NaCl is given as 5.4 M. This means addition of Na+ and Cl- ions up to 5.4 M does not cause any precipitation. The addition of the ions beyond 5.4 M makes it to get precipitated. The solubility product can be given as,

ksp = [Na⁺][Cl^-] = 5.4 M.

If  [Na⁺][Cl^-] > 5.4 M, Precipitation occurs.

[Na⁺][Cl^-] < 5.4 M, No precipitation.

Answer to the question number 4:

Addition of Con. HCl: (test tube 2)

The addition of Con. HCl will increase the concentration of one of the reactant i.e. Cl^- ions. Now the concentration of Cl^- ion is greater than 5.4 M. The equilibrium gets disturbed and now the reaction quotient is greater than ksp. To overcome this situation more NaCl gets precipitated (i.e. more NaCl crystals will get formed).  

Addition of HNO₃ followed by addition of HCl: (test tube 3)

Addition of HNO₃ will not make any change in the reaction as it does not possess any ions in common. But it will increase the volume of water in the solution which makes the precipitate formed to get dissolved. On further addition of HCl increases the Cl^- ion concentration and makes the solution to get precipitate. Therefore, the final inference is the test tube remains same as appeared before with increase in solution volume.