# How to calculate the pH of a buffered solution with Henderson Hasselbalch?

Lets say you have a buffered solution containing $0.50~\mathrm{M}$ acetic acid ($K_a = 1.8 \cdot 10^{-5}$) and $0.50~\mathrm{M}$ sodium acetate, and you want to calculate the $\mathrm{p}\ce{H}$ of this solution.

This can be done in two ways: using the standard equilibrium chart (initial, change, equilibrium), or using the Henderson Hasselbalch equation.

My question is why do these two give the same result.

The Henderson Hasselbalch equation is derived starting with the equilibrium equation:

$K_a = \frac{\ce{[H+][CH3COO^{-}]}}{\ce{[CH3COOH]}}$ and then the $\ce{[H+]}$ is found, and then the $\log_{10}$ is taken. However, doesn't this assume that the system is already at equilibrium and that the concentrations of $\ce{[CH3COO^{-}]}$ and $\ce{[CH3COOH]}$ are the concentrations at equilibrium?

However, that is clearly not the case as shown by the ICE chart. However, both give the correct pH. Why is this?

The main equilibrium in such a solution is the following:

$\ce{CH3COOH +CH3COO^- <=>CH3COOH + CH3COO^-}$.

This should be your ICE chart's equation too. I suspect that the equation you have for your ICE table is the following:

$\ce{CH3COOH +H_2O <=>CH3COO^- + H_3^+O}$

You can certainly solve the second equation with an ICE table by plugging in the initial concentration of the conjugate base, $\ce{CH3COO^-}$.

However, doesn't this assume that the system is already at equilibrium and that the concentrations of $\ce{[CH3COO-]}$ and $\ce{[CH3COOH]}$ are the concentrations at equilibrium?

Yes. The concentrations (or more accurately, activities) used in the H-H equation are equilibrium concentrations.

In the case of a 0.5 M acetic acid/acetate ion buffer, we can, for most purposes, conclude that the initial molarity of the main system components - acetic acid and acetate ion will remain constant. This is because the main acid-base reaction in the system is the following:

$\ce{CH3COOH +CH3COO^- <=>CH3COOH + CH3COO^-}$.

As we can see, the above equation is symmetric. There is no net change in system composition resulting from the above equation and the given molarities.

We also have these equilibria:

$\ce{CH3COOH + H2O- <=>H_3^+O + CH3COO^-}$ $\ce{H2O + CH3COO^- <=>CH3COOH + HO^-}$

But both acetic acid and its conjugate base, acetate ion, are weak. So these won't do much to change the system composition. Plus these aren't the favored reaction in the system; that would be the one above (the one between the strongest acid and strongest base in the system).

If you wanted to solve this system analytically - i.e. you want an even more accurate measure of the system's pH, then you would have to employ the law of conservation of mass and invoke the concept of solution electro neutrality.

Mass conservation:

$1.0 M$ $\ce{=[CH3COOH] + [CH3COO^{-}]}$

Solution electroneutrality:

$\ce{[HO^-{}] + [CH3COO^{-}] = [H_3^+O]}$.

These two statements, coupled with $\ce{K_a(CH3COOH)}$ and $\ce{K_b(CH3COO^{-})}$, and $\ce{K_w}$, give you a system of equations in which you can solve for $\ce{[H_3^+O]}$.

• I think I got it. So this can also be used for the calculating the pH when titrating a weak acid with a strong base (or weak base with strong acid), right? Because once the reagent acid reacts with the original base, it will produce the conjugate acid and what is left is now a buffer, whose pH can be determined with this, right? Jun 20, 2014 at 23:03
• However, it cannot be employed for calculating pH when titrating a strong acid with a strong base because in that case the equilibrium concentrations will be completely different from the initial ones, right? Jun 20, 2014 at 23:04
• You can definitely use the HH equation to find the pH of a titration between a weak acid and a strong base. Just make sure you're using equilibrium concentrations; strong bases will completely abstract protons from the acid. Do not use initial concentrations when dealing with anything strong. Jun 20, 2014 at 23:17
• If you are titrating a strong acid with a strong base I would just take the easy way out by finding how much of the strong acid and strong base are consumed when they react with each other and then just take the negative log of the hydronium or hydroxide ion concentration. Note that this method only works for non-very dilute solutions. Jun 20, 2014 at 23:19
• @DissenterI try to solve this system analytically according to the following equation : $$\ [\ce{H+}]^3 +([\ce{CH3COONa}]_0 + 1)[\ce{H+}]^2 -(K_\mathrm{w}+[\ce{CH3COOH}]_0K_\mathrm{a})[\ce{H+}] -K_\mathrm{w}=0 \$$Is the equation correct?thank you. Oct 27, 2018 at 15:55