Sorry I had to go to bed last night.
I'm going to represent the acid as BNH^+ and the base as BN
Let suppose we use ALL of the acid that is available so we will have 500 mL x 0.113M = 56.5 millimoles BNH^+ = acid.
How much NaOH must we add to neutralize a part of the acid to form BN in the right proportions.
If pKb = 6.24 then pKa = 7.76
millimiles NaOH added = 0.122M x mL
..........BNH^+ + OH^- ==> BN + H2O
I.........56.5.....0.........0....0
add..............0.122x.............
C.......-0122x...-0.122x....0.122x
E......56.5-0.122x..0......0.122x
Substitute all of that into the HH equation.
7.00 = 7.76 + log (0.122x/56.5-0.122x)
Solve for x and I obtained about 68 mL of the 0.122M base.
You can do part b by setting up an ICE chart as above, using 8.00 as the target pH and calculating mols NaOH that must be added.
Did you get the amine titration problem worked? If not I can help on that, too.
1. Suppose that you have 0.500 L of each of the following solutions, and an unlimited supply of water.
(Note: C9H7NHBr is a salt containing the ions C9H7NH+ and Br− and C9H7N is quinoline, an organic
base with pKb = 6.24 at 298 K. If you like, you may represent C9H7NH+ as HB+ and C9H7N as B.)
Data
0.113 mol L−1 C9H7NHBr (aq)
0.104 mol L−1 HBr(aq)
0.122 mol L−1 NaOH(aq)
(a) Provide simple instructions for preparing 1.00 L of a solution having pH = 7.00 at 298 K.
Your instructions should include the volumes of the solutions required.
(b) What is the buffer capacity of the resulting solution? (The buffer capacity is the
number of moles of NaOH that must be added to 1.0 L of solution to raise the pH by one unit.)
2 answers
I think you used the HH equation incorrectly. Shouldnt you have used log([acid]/[base])? It looks like you used them backwards.
1 answer