Chemistry 2000, Fall 96, Test 2 Solutions

1. If the osmotic pressure is rising, then the total concentration of solutes must be increasing. This suggests that the starch is breaking down into its constituent units.
2. 1. Solutes depress the freezing points of solvents. Since salt is very soluble in water, salt-water solutions can have very low freezing points.
   2. Ethanol is volatile. It might melt the ice on a road very quickly, but it would quickly evaporate and the ice would refreeze.

3. 

   In water, the sodium sulfite will dissociate into 1.98mol of and 3.96mol of ions. Using the molar density of water, we have

   

   The mole fraction of water is therefore

   

   Since only the water is volatile in this solution, the vapour pressure of the solution is entirely due to the water:

   

4. 1. 

      The concentration of hydroxide is therefore

      

      The activity of hydroxide is 1.3. From the equation, we have

      

   2. 

      Some of the acid is neutralized by the base, so after the addition, we have

      

      The total volume after the addition is 300mL so

      

5. 1. None of the protons of phosphoric acid have 's reasonably close to 4. The nearest is the first proton which has a 1.9 pH units removed from 4. Since effective buffering requires similar amounts of the acid and base form and since this is achieved when the pH is similar to the , acid phosphates are probably not the right materials from which to make a pH 4 buffer.
   2. The of the third proton is nearly 12 so a pH 12 buffer can be made from sodium hydrogen phosphate and sodium phosphate.
   3. The buffer will use the last acid equilibrium of the acid phosphates, namely

      

      

      We want a pH 12 buffer so . Therefore

      

      We also want . Putting the two equations together, we get

      

      We must dissolve 468g of sodium hydrogen phosphate and 279g of sodium phosphate in 10L of water.

6. 1. The reaction ratio is

      

      Since Q<K, the reaction will spontaneously occur as written, i.e. lead azide will be converted to lead and nitrogen. Furthermore, since this reaction can't materially affect the partial pressure of nitrogen in the atmosphere (the atmosphere is too big for that), the reaction will continue until the lead azide runs out. Thus, at equilibrium, only lead remains.

   2. Because the equilibrium constant is huge, the lead azide is quantitatively converted to lead and nitrogen. Initially

      

      Since 3 molecules are generated for every formula unit of lead azide, at equilibrium. The pressure due to this added nitrogen is

      

      There was already 1atm = 101325Pa of nitrogen in the box so