Question

A 100 milliliter sample of 0.100-molar NH4Cl solution was added to 80 milliliters of a 0.200-molar solution of NH3. The value of Kb for ammonia is 1.79 x 10^-5.
(a) What is the value of pKb for ammonia? (b) What is the pH of the solution described in the question?
(c) If 0.200 grams of NaOH were added to the solution, what would be the new pH of the solution? (assume that the volume of the solution does not change.)
(d) If equal molar quantities of NH3 and NH4+ were mixed in solution, what would be the pH of the solution?

Answer 1

NH4^+ + OH^- ==> NH3 + H2O 

Set up an ICE chart. 
initial: 
NH4^+ = mL x M = 100 x 0.1 = 10 millimoles 
NH3 = mL x M = 80 x 0.20 = 16 millimoles. 

change: 
we add 0.2 g NaOH which is 0.2/40 = 0.005 moles or 5 millimoles. 
NH3 = +5 millimoles 
NH4^+ = -5 millimoles 

equilibrium: 
NH3 = 16 + 5 = 21 millimoles. 
NH4^+ = 10 - 5 = 5 millimoles. 

You may substitute millimoles in place of concn (since millimoles/mL = molarity and the mL (180 mL) appears in both numerator and denominator) OR you can divide millimoles/180 mL to arrive at concn for both base and acid and substitute those numbers. Plug those into the HH equation and solve for pH. 

For part d, just set up the HH equation and pH = pKa + log (base/acid). The question is asking you to calculate pH if base and acid were equal. So plug in the same number (any number you choose) for base and acid and calculate. Note that the log of 1 = 0