posted by Nicole on .
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?
I assume you can do (a).
(b) pH = pKa + log[(base)/(acid)]
Note this is pKa and NOT pKb.
(c)moles NaOH = grams/molar mass NaOH.
Exactly how much of this can you do? What do you not understand?
I don't understand C and D
The purpose of a buffer is to resist a change in pH. When an acid is added, one of the materials (the base--in this case NH3) neutralizes the acid forming MORE of the salt. So NH4^+ increases and NH3 decreases. When a base is added, it reacts with the acid (in this case NH4^+, thus decreasing NH4+ and increasing NH3.
can you help set up the equations i would need to use?
NH4^+ + OH^- ==> NH3 + H2O
Set up an ICE chart.
NH4^+ = mL x M = 100 x 0.1 = 10 millimoles
NH3 = mL x M = 80 x 0.20 = 16 millimoles.
we add 0.2 g NaOH which is 0.2/40 = 0.005 moles or 5 millimoles.
NH3 = +5 millimoles
NH4^+ = -5 millimoles
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.