# Chemistry

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If you start with 200 ml of 0.1M HEPES buffer (pH 7.5) and add 1.25 ml of 0.5M HCl, what will the final pH be? (The pKa of HEPES is 7.5)

• Chemistry -

I'm a little confused by the problem. USUALLY a statement that the buffer is 0.1M HEPES means that the acid form + basic form = 0.1M; however, with no pH given there is no way to calculate the ratio of the acid form to the base form. Therefore, I must assume that this is 0.1M in the base form (since we're added HCl to it). 200 mL x 0.1M = 20 mmoles.
1.25 x 0.5M = 0.625 mmols HCl added.
.........base + HCl ==> baseH^+ + Cl^-
I.........20.....0........0
C........-0.625..-0.625...+0.625
E.........19.375....0......0.625

pH = pKa + log base/acid
Substitute and solve for pH.

• Chemistry -

Thank you DrBob222. This is what I had attempted on my own, but I get a pH of 9, which seems wrong. If the pH of my buffer is 7.5, shouldn't the pH slightly decrease, given that I am adding a strong acid?

• Chemistry -

You are absolutely right and that bothers me too. I worked it and came out with about 9.0 but that can't be right. The pH should go down. That must mean that the assumption I made is not valid.Do you know how the HEPES buffer was prepared. I have found several recipes on the web that use the acid (HEPES is an acid) and treated it with KOH or NaOH to adjust to pH 7.5. If that is how the HEPES system is used, that means the base form is 0.05 and the acid form is 0.05 and if you put those values into this problem the pH does decrease with the addition of the HCl (from 7.5 to 7.44).HOLD ON. I just re-read the problem and is DOES, indeed, give a pH for the solution.I should have read the problem more carefully. It says pH = 7.5 which allows us to calculate the ratio of base/acid and at pH = 7.5 the ratio is 1 which means base = 0.05M and acid = 0.05M which is 10 mmols for a 200 mL solutioin.
..............B + HCl ==> acid
I............10.0...0......10.0
C.........-0.675..-0.675...0.675
E..........9.325.....0......10.675
Then pH = 7.5 + (9.325/10.675) = about 7.4

• Chemistry -

Thank you so much!

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