Creating a pH 6.7 Buffer Solution with Imidazole and C3H5N2Cl

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To create a pH 6.7 buffer solution using Imidazole (C3H4N2) as the weak base and C3H5N2Cl as the conjugate acid, the required concentrations and volumes were calculated. The Henderson-Hasselbalch equation was applied, leading to a ratio of base to acid of approximately 0.56. The calculations suggested using 40.6 mL of Imidazole in a total volume of 400 mL, adjusting the pH with HCl or NaOH as needed. A suggestion was made to consider starting with Imidazole hydrochloride and titrating with NaOH to achieve the desired pH. The discussion emphasizes the importance of verifying calculations and methods for accuracy in buffer preparation.
ElectronicError
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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.
 
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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.
 

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