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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: PleaseHelpMe1234 on April 13, 2018, 12:32:53 AM

Title: How much of each stock would you need for 100 ml of buffer?
Post by: PleaseHelpMe1234 on April 13, 2018, 12:32:53 AM: I don't really know how to do this question - please *delete me*

You are running a biochemical assay with an enzyme that needs to maintain a constant pH of 6.6 in order to operate effectively. Using starting stock solutions of 0.2M of NaH2PO4 and Na2HPO4, how much of each stock would you need for 100 ml of buffer at a final concentration of 0.01M and a pH of 6.6.

Hint: First calculate the dilution to determine total amount of phosphate stock required. Then use the Henderson-Hasselbalch equation to calculate the ratio of acid to base.
Title: Re: How much of each stock would you need for 100 ml of buffer?
Post by: Borek on April 13, 2018, 02:48:30 AM: Hints you were given are very good, hard tot ell which part makes it impossible for you to solve the problem.

Would you be able to solve the problem starting from 0.01 M solutions of both substances?
Title: Re: How much of each stock would you need for 100 ml of buffer?
Post by: Babcock_Hall on April 13, 2018, 11:12:08 AM: You need one more piece of information, namely the value of a particular constant. Do you know the constant to which I am referring, and do you know its value?
Title: Re: How much of each stock would you need for 100 ml of buffer?
Post by: PleaseHelpMe1234 on April 14, 2018, 03:10:38 AM: Quote from: Borek on April 13, 2018, 02:48:30 AM
Hints you were given are very good, hard tot ell which part makes it impossible for you to solve the problem.

Would you be able to solve the problem starting from 0.01 M solutions of both substances?

First I calculated the dilution to determine the total amount of phosphate stock required:

0.01 x 100 = 0.2V
V = 5 ml of phosphate stock

Secondly, I used Henderson HasselBalch's equations:

pH = 6.6 Pka = 7.2

6.6 = 7.2 + log([base]/[acid])
-0.6 = log([base]/[acid])
0.25 =([base]/[acid])

Therefore there ratio of base to acid is 0.25:1

Therefore, of the 5 ml of stock solution, there is 1 ml of base (NaHPO4), and 4 ml of acid (NaH2PO4).

Is this correct??
Title: Re: How much of each stock would you need for 100 ml of buffer?
Post by: PleaseHelpMe1234 on April 14, 2018, 03:12:02 AM: Quote from: Babcock_Hall on April 13, 2018, 11:12:08 AM
You need one more piece of information, namely the value of a particular constant. Do you know the constant to which I am referring, and do you know its value?

Do you mean the acid dissociation constant? And is it 7.2?
Title: Re: How much of each stock would you need for 100 ml of buffer?
Post by: Borek on April 14, 2018, 03:29:36 AM: Quote from: PleaseHelpMe1234 on April 14, 2018, 03:10:38 AM
Therefore, of the 5 ml of stock solution, there is 1 ml of base (NaHPO4), and 4 ml of acid (NaH2PO4).

Yes.
Title: Re: How much of each stock would you need for 100 ml of buffer?
Post by: Babcock_Hall on April 14, 2018, 10:21:37 AM: Quote from: PleaseHelpMe1234 on April 14, 2018, 03:12:02 AM
Quote from: Babcock_Hall on April 13, 2018, 11:12:08 AM
You need one more piece of information, namely the value of a particular constant. Do you know the constant to which I am referring, and do you know its value?

Do you mean the acid dissociation constant? And is it 7.2?
Yes, and yes, with one qualification. Your buffer is low enough in ionic strength that 7.20 is an appropriate value IMO. If you continue to more advanced chemistry courses, you may learn that the value of the ionic strength affects the apparent value of pK_a. However, it seems to me that it is unlikely that you need to worry about just now. I sometimes refer to things like this as second order effects.