 Henderson - Hasselbalch equation

(1) HA + H2O D H3O+ +A- KA = [A-][H3O+] / [HA]

(2) A- + H2O D OH- + HA KB = Kw/KA

(3) 2H2O D OH- + H3O+ KW = [OH-] [H3O+]

if at time t = 0 we add an amount of an acid ChHA and another amount of the acids conjugate base CA- to water and then allow it to equilibrium…

Then from (1) we know that [HA] decreases by [H3O+]

And from (2) [HA] increases by an amount of [OH-]

Also from (2) [A-] decreases by [OH-]

And from (1) [A-] increases by [H3O+]

Thus at equilibrium;

(3) [A-] = CA- - [OH-] + [H3O+]

(4) [HA] = ChHA - [H3O+] + [OH-]

Now if KA < 10-3 (HA is a weak acid) then we can assume that [H3O+] and [OH-] are nearly zero thus

(5) [A-] = CA-

(6) [HA] = ChHA

Solving for KA in line (1) and substituting (5) and (6) we get

[H3O+] = KA CHA/CA-, Taking the negative Log of both sides gives

-log [H3O+] = - log KA – log (CHA/CA-)

pH = pKA + log (CA-/CHA) (Henderson – Hasselbalch Equation)

What we have just done is derived an equation that allows us to predict the pH of an aqueous mixture of a weak acid and the salt of its conjugate base based on assumptions that are fairly accurate if the acid is a weak acid such as acetic acid (vinegar). This type of mixture is known as a buffer. A buffer is a system that tries to maintain a certain pH even if you add more acid or more base.

1. HA + H2O D H3O+ +A- KA = [A-][H3O+] / [HA]

(2) A- + H2O D OH- + HA KB = Kw/KA

As we add more protons making it more acidic the hydroxide in (2) (OH-) bonds with them producing water and the pH is maintained and the equilibrium in (2) shifts to the right. If we add more hydroxide and increase the pH, the protons in (1) (H3O+) bonds with them and makes water thus maintaining the pH and shifts that equilibrium to the left.

Keep in mind that CA- and CHA are the amounts of acid and conjugate base that we add to the solution so if you were making a buffer then you would measure these quantities out.

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