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.)
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.
I think you used the HH equation incorrectly. Shouldnt you have used log([acid]/[base])? It looks like you used them backwards.
To prepare a solution with a pH of 7.00 at 298 K using the given solutions, follow these steps:
(a)
1. Calculate the amounts of C9H7NHBr and HBr needed to form the buffering system. The buffer system should consist of the weak base C9H7NH+ and its conjugate acid HB+, which is formed by the dissociation of C9H7NHBr.
- Given that the concentration of C9H7NHBr is 0.113 mol/L, we need to find the volume required to obtain the appropriate amount.
- Using the formula: Concentration (C) = moles (n) / volume (V), we can rearrange it to solve for volume V. Rearranging the formula, V = n / C, we have:
V = 0.113 mol / 0.113 mol/L
V = 1 L
Therefore, we need to use 1.00 L of the 0.113 mol/L C9H7NHBr solution.
- Since the concentration of HBr is 0.104 mol/L, we need to use the same volume of HBr solution as the volume of C9H7NHBr solution, which is 1.00 L.
2. Mix the C9H7NHBr and HBr solutions together to form the buffering system. The HBr will provide the Br− ions necessary to convert C9H7NH+ into the weak acid HB+ according to the reaction: C9H7NHBr + H2O ⇌ HB+ + Br− + H3O+.
- Combine the 1.00 L of the 0.113 mol/L C9H7NHBr solution with 1.00 L of the 0.104 mol/L HBr solution. This will give you a total volume of 2.00 L of the buffering system.
3. Adjust the pH of the solution to 7.00 by using NaOH. Given that the concentration of NaOH is 0.122 mol/L, we need to find the volume required to raise the pH to 7.00.
- The buffer system will initially have a pH close to the pKb of C9H7NH+ (6.24). To increase the pH to 7.00, we need to add a strong base (NaOH) to neutralize the acid present in the solution. As NaOH is added, it will react with HB+ and form more C9H7NH+ and H2O, shifting the equilibrium to the left.
- To calculate the buffer capacity, we need to find the moles of NaOH needed to raise the pH by one unit. The Henderson-Hasselbalch equation gives the relationship between the pH and the ratio of the concentration of the weak acid to that of its conjugate base.
- pH = pKb + log([HB+]/[B])
- pH = 7.00, pKb = 6.24
- Rearranging the equation, [HB+]/[B] = antilog(pH - pKb)
- Substituting the values, [HB+]/[B] = 10^(7.00 - 6.24)
- [HB+]/[B] = 5.6234 (approximately)
Since the ratio [HB+]/[B] is 5.6234, it means that for every mole of HB+ (from HBr), we need 5.6234 moles of NaOH to raise the pH by one unit.
- Therefore, we need to add 5.6234 moles of NaOH to 1.00 L of the buffer solution.
(b) The buffer capacity is the number of moles of NaOH that must be added to 1.0 L of the solution to raise the pH by one unit. From the previous calculations, we found that for every mole of HB+ (from HBr), we need 5.6234 moles of NaOH to raise the pH by one unit.
Thus, the buffer capacity of the resulting solution is 5.6234 moles of NaOH.