1. MES (2-(N-Morpholino)ethanesulfonic acid) is a common buffer used in biochemistry labs to stabilize proteins. The pKa for MES is 6.09.
a. What is the pH of 1 liter of a 200 mM MES aqueous solution? Clue: you will need to solve a quadratic equation. (2.5 points)
b. How much solid NaOH (in grams) needs to be added to the MES solution to make a buffer at pH 6.5? (1.5 points)
If we call MES just HS, then
...........HS ==> H^+ + S^-
initial....0.2.....0....0
change.....-x......x.....x
equil......0.2-x...x.....x
Substitute the equilibrium values into the Ka expression and solve for x, then convert to pH.
For the second part,
you know pH = pKa + log(base/acid)
(base) = (OH) that must be added
(acid) = 0.2-(OH)
I would plug that into the HH equation, solve for base and convert that to grams. I shall be happy to critique your work if you post it.
Part A...
Solving for x we will get 4.028x10^-4
Plugging it in into pH equation, we will get pH of 3.39
Part B
5.76g
Thanks Doc!
a. To find the pH of the MES aqueous solution, we need to use the Henderson-Hasselbalch equation:
pH = pKa + log([Salt]/[Acid])
Here, [Salt] represents the concentration of the conjugate base (NaMES) and [Acid] represents the concentration of the acid form (MES).
Given that the pKa of MES is 6.09, we can assume that in water (pure MES), the concentration of the conjugate base and the acid form are equal.
Let's denote the initial concentration as [MES]0. In a 200 mM MES aqueous solution, the concentration of MES will be 200 mM = 0.2 M.
Since the concentration of the conjugate base and the acid form will be the same in the initial solution, we can set it as [Salt]0 = [Acid]0 = 0.2 M.
Now, to solve the quadratic equation, we introduce the variable x. Let [Salt] be x (the concentration of the conjugate base NaMES formed) and [Acid] be (0.2 - x) (the concentration of MES that remains).
The equation becomes:
pH = 6.09 + log(x/(0.2 - x))
Now, let's rearrange this equation to solve for x:
10^(pH - 6.09) = x/(0.2 - x)
Multiply both sides by (0.2 - x):
(0.2 - x) * 10^(pH - 6.09) = x
Now, let's solve this quadratic equation to find the value of x.
b. To make a buffer at pH 6.5, we will need to add a calculated amount of NaOH to the MES solution to increase the concentration of the conjugate base (NaMES).
First, we need to calculate the concentration of the acid form (MES) that we want to achieve in the buffer. From the Henderson-Hasselbalch equation:
pH = pKa + log([Salt]/[Acid])
Rearranging this equation:
[Salt]/[Acid] = 10^(pH - pKa)
We know the desired pH (6.5) and the pKa (6.09) of MES. Let's calculate [Salt]/[Acid]:
[Salt]/[Acid] = 10^(6.5 - 6.09)
Now, multiply both sides by [Acid]:
[Salt] = [Acid] * 10^(pH - pKa)
We know the desired concentration of the acid form ([Acid]0 = 0.2 M) and the calculated ratio [Salt]/[Acid]. By using the above equation, we can find the concentration of the conjugate base ([Salt]):
[Salt] = [Acid]0 * [Salt]/[Acid]
Finally, multiply the concentration of the conjugate base ([Salt]) by the molar mass of NaOH to calculate the amount of solid NaOH in grams that needs to be added to the MES solution.