Creating a pH 6.7 Buffer Solution with Imidazole and C3H5N2Cl

  • Thread starter Thread starter ElectronicError
  • Start date Start date
  • Tags Tags
    Buffer
Click For Summary
SUMMARY

The discussion focuses on creating a pH 6.7 buffer solution using Imidazole (C3H4N2) as the weak base and C3H5N2Cl as the conjugate acid. The required buffer solution volume is 0.4 L, with a buffer capacity to absorb 20 mL of either 0.02 M HCl or 0.02 M NaOH without exceeding a pH change of +/- 0.1. The calculations involve using the Henderson-Hasselbalch equation and determining the necessary concentrations of the weak base and conjugate acid to achieve the desired pH.

PREREQUISITES
  • Understanding of the Henderson-Hasselbalch equation
  • Knowledge of buffer solutions and their components
  • Familiarity with molarity and dilution calculations
  • Basic acid-base chemistry principles
NEXT STEPS
  • Research the Henderson-Hasselbalch equation in detail
  • Learn about buffer capacity and its significance in solution preparation
  • Explore titration techniques for pH adjustment
  • Investigate the properties and applications of Imidazole and its derivatives
USEFUL FOR

Chemistry students, laboratory technicians, and anyone involved in buffer solution preparation and acid-base chemistry.

ElectronicError
Messages
7
Reaction score
0
I have worked through the problem and spent quite a bit of time on it but am not extremely confident in my solution. I would just like someone to check my work. Cheers.

Homework Statement



-Need 0.4 L of an aqueous buffer solution of pH=6.7
-Have access to 0.02 M HCl, 0.02 M NaOH, 0.01 M C3H4N2, and 0.01 M C3H5N2Cl
-Will use Imidazole (C3H4N2) as weak base. Will use C3H5N2Cl as conjugate acid.
-Given Kb for C3H4N2 = 9.0 x 10^-8
-The buffer must have the capacity to absorb 20 mL of either 0.02 M HCl or 0.02 M NaOH and undergo a pH change of no more than +/- 0.1.

Homework Equations



Henderson-Hasselbach

The Attempt at a Solution



(Ka)(Kb) = Kw = 1 x 10-14
Ka * 9.0*10^-8 = 1 * 10^-14
Ka = 1.1 x 10^-7

-log(1.1 x 10^-7) = pKa ~ pH = 6.95

pH = pKa + log([base]/[acid])

6.7 = 6.95 + log([base]/[acid])

[base] / [acid] = 10^-0.25 = 0.56 (reasonably close to 1)

Use 1.56 x 10^-5 M for [base] and 0.001 M for [acid]

Buffer concentration = 1.02 x 10^-3

#mol weak base = (buffer concentration) * (desired final volume) = (1.02 x 10^-3) * 0.4 L = 4.06 x 10^-4 mol

0.01 M = 4.06 x 10^-4 mol / ? L

4.06 x 10^-2 L = 40.6 mL C3H4N2

In a 400 mL beaker, add distilled water to 40.6 mL C3H4N2 until most of the desired final volume is reached. Add HCl or NaOH dropwise until the pH reaches 6.7. Add distilled water to the 400 mL line.
 
Physics news on Phys.org
Ummm..

:zzz:
 
This is my last, desperate call for help. Surely someone can lend a hand?
 
ElectronicError,

I have been trying to think about this one for you. Do you need to use all of those given solutions, or are you permitted to just use 2 or 3 of them?

My first attempt would be to begin with the imidazole hydrochloride and use it as an acid; find the amount of NaOH base titrant necessary to reach a pH of 6.7. Now, instead of performing that partial neutralization, think of the number of moles of imidazole which would be equivalent; and use correspondingly less amount of moles of the imidazole hydrochloride.

... I know that was not complete, but it might be a start. Take care in case I misjudged anything in that method.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 4 ·
Replies
4
Views
1K
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K