# Calculate the pH and boiling point of the following solution

Submitted by rocktown1990 on Wed, 04/18/2012 - 23:47

Calculate the pH and boiling point of a solution that has a mole fraction of 0.01786 for aluminium
nitrate in water. The density of this solution is 1.055 g/mL.

thanks

Assuming that the aluminum nitrate is the only solute, the mole fraction of the water must be

1.000 -  0.01785  =  .982 mol water

The mass of .982mol of water is .964 g  or 9.64 x 10-2kg

The molality of the solution in terms of  aluminum nitrate would be .01785 mol / 9.64 x 10-2 kg = 1.85x10-2 molal

However, since 1 mole of aluminum nitrate dissociates into 4 moles of ions, the molality in terms of solute particles will be 4 x 1.85 x 10-2   or  7.40 x 10-2 molal

Then we use the molality to calculate the change in freezing point

change in fp  =  Kf   x molality  =  1.86 C/molal  x  7.40 x 10-2 molal =  1.38 x 10-1 C

To find the pH we need to know the volume of the solution.  To find the volume of the solution we need to find the total mass of the solution.  So, we need to convert 0.01785 mol into grams.

mass of Al(NO3)3  =  moles x molar mass =  0.01785 mol  x  213g = 3.80 g

Thus the total mass of the solution would .964g  +  3.40g  =  4.77g

Given the density value we can convert to volume    4.77g  / 1.055 g/mL  =    4.52 mL

Thus the molarity of the solution would be .01785 mol / .00452 L  =  3.95 M

Since 1 Al+3 ion is produced for every formula unit the molariity of the Al+3 is the same as the molarity of the aluminum nitrate

Ksp = [Al3+][OH-]^3 = 4.6 x 10^-33  =  3.95  x3  = 4.6 x 10-33

x = 1.05 x 10-11  =  [OH-]        pOH =  10.98      pH = 3.02