A 100. milliliter sample of 0.100-molar NH4Cl solution was added to 80.0 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?
a. pKb = -log Kb
b. Use the Henderson-Hasselbalch equation.
pH = pKa + log [(base)/(acid)]
You can find pKa for NH3 by pKa + pKb = pKw = 14. You have pKw here and pKb from part a so you can calculate pKa.
c. millimols NH3 = mL x M = 80 x 0.200 = 16
mmols NH4Cl = 100 x 0.1 = 10
mmols NaOH added = mg/molar mass = 200/40 = 5
The NaOH will react with the acid portion of the buffer (the NH4Cl) as
...............NH4^+ + OH^- ==> NH3 + H2O
I...............10............0..............16.................
add..........................5....................................
C..............-5...........-5.................+5
E................5............0................21
Plug the E line into the HH equation to solve for the new pH. I should point out here that you are plugging in values in millimoles. The HH equation is written in terms of concentration (molarity = ?) which is millimoles/mL. So you CAN calculate M of NH4^+ and NH3 (M NH4^+ = 5/180 = ? and M NH3 = 21/180 = ?) but you can see that the volume cancels and you're left with just millimols so including the volume is extra work you don't need to go through.
d. Looking at the H-H equation of pH = pKa + log [(base)/(acid)]
if (base) = (acid) which is what you would have if equal volumes of equal molarities were mixed, then (base)/(acid) = 1, log 1 = 0, and pH of the solution would be pKa
Post your work if you get stuck.