If you make up a solution of 100mL of 0.1 M Hepes in the basic form, what will be the pH? pKa = 7.55

I suppose the problem is asking for the anion. If HB is the acid, then the base is B^- and that hydrolyzes in water solution this way.

.............B^- + HOH ==> HB + OH^-
I..........0.1M............0....0
C...........-x.............x....x
E.........0.1-x............x....x

K for the base is not listed but it can be calculated as Kbase = Kw/Kacid = 1E-14/2.82E-8
Then Kbase = (HB)(OH^-)/(HB)
2.81E-8 = (x)(x)/(0.1-x)
Solve for x = OH^- and convert to pH.

Dr. Bob...I have seen you trying to post answers to several questions, but in reality all you are doing is confusing everyone more. Please be more straight forward with your answers. In your other posts, I have seen that several people can not follow what you are saying.

10.275

To determine the pH of a solution of 0.1 M Hepes in the basic form, we need to consider the acid-base properties of Hepes and calculate the concentration of its conjugate acid and base forms.

Hepes (4-(2-hydroxyethyl)-1-piperazineethanesulfonic acid) is a zwitterionic buffer used to maintain a stable pH in biological experiments. It can exist in both acidic and basic forms, depending on the pH of the solution.

The pKa value of Hepes is 7.55. The pKa represents the pH at which half of the molecules will exist in the acidic form (the conjugate acid, H2Hepes) and the other half will be in the basic form (the conjugate base, Hepes-).

In this case, we have 100 mL of 0.1 M Hepes in the basic form. Since Hepes is a weak acid, we can assume that the concentration of the conjugate base (Hepes-) is equal to the molarity of the solution (0.1 M). We can also assume that the concentration of the conjugate acid (H2Hepes) is negligible since we have an excess of the basic form.

To calculate the pH, we need to determine the ratio of the conjugate base (Hepes-) to the conjugate acid (H2Hepes). Using the Henderson-Hasselbalch equation:

pH = pKa + log10([Conjugate Base]/[Conjugate Acid])

Since we have an excess of the basic form, we can approximate the concentration of the conjugate acid (H2Hepes) to be very small. This means that [Conjugate Base] ≈ 0.1 M and [Conjugate Acid] ≈ 0. Therefore, we can simplify the equation as:

pH ≈ pKa + log10(0.1/0)

Since the logarithm of 0 is undefined, we need to use a small positive number (approaching zero) instead of zero. Let's choose 0.000001 for convenience:

pH ≈ 7.55 + log10(0.1/0.000001)

Calculating this becomes:

pH ≈ 7.55 + log10(100,000)

Using a common logarithm calculator, we find:

pH ≈ 7.55 + 5

pH ≈ 12.55

Therefore, the pH of the solution of 100 mL of 0.1 M Hepes in the basic form is approximately 12.55.