Question

If you have 120. mL of a 0.100 M HEPES buffer at pH 7.55 and you add 2.00 mL of 1.00 M HCl, what will be the new pH? (The pK_a of HEPES is 7.55.)

pH =

Answer 1

Since,

pH = pKa + log { [base] / [acid] }

7.55 = 7.55 + 0

So, log { [base] / [acid] } = 0

log (1) = 0

So, [base] = [acid]

Hence, in the pH 7.55 solution, the concentrations of the acid and base forms of HEPES are equal.

Now, the sum of acid and base concentrations is 0.1 M, each form is at a concentration of 0.050 M.

Since you have 0.12 L, the moles of each form in the original solution are

Moles of acid (or base) = 0.050 M x 0.12 L = 0.006 moles

Now, you are adding 2.0 mL of 1.00 M HCl,

So, moles of HCl = 0.002 L x 1.0 M = 0.002 moles

This will stoichiometrically react with the basic form of HEPES and convert it to the acidic form. So, after the addition you have:

Moles of acid = 0.006 moles + 0.002 moles = 0.008 moles
Moles of base = 0.006 moles - 0.002 moles = 0.004 moles

Now total volume = 120 mL + 2 mL = 122 mL = 0.122 L

So,

[acid] = 0.008 moles / 0.122 L = 0.066
[base] = 0.004 moles / 0.122 L = 0.033

Now,

pH = pKa + log { [base] / [acid] }

     = 7.55 + log (0.033/0.066)

     = 7.55 – 0.30

     = 7.25