By the Henderson–Hasselbalch equation, log (salt)/(acid) term is a ratio of the concentrations. So, how can dilution change the pH of the acid as the concentrations of both salt and acid would change with dilution and the term would remain the same?

  • $\begingroup$ I don’t see how you arrived at that maths … $\endgroup$
    – Jan
    May 31, 2016 at 21:49

1 Answer 1


Consider a strong acid such as $\ce{HCl}$ which — within experimental error — can be considered fully dissociated in aquaeous solution. This can be expressed as:

$$c(\ce{H3O+}) = c_0(\ce{HCl})$$

Assume you have a concentration $c = 1~\mathrm{\frac{mol}{l}}$. pH is defined as:

$$\mathrm{pH} = - \lg [\ce{H3O+}]$$

Thus our solution has a pH of:

$$\mathrm{pH} = - \lg [\ce{H3O+}] = - \lg 1 = - 0 = 0$$

(Remember that to calculate pH we need dimensionless concentration. In this post, $c$ shall always contain a dimension while square brackets shall not.)

Take $1~\mathrm{l}$ of this solution and add $1~\mathrm{l}$ of water. Our new solution has a concentration of $c = \frac{1~\mathrm{mol}}{2~\mathrm{l}} = 0.5~\mathrm{\frac{mol}{l}}$. Our pH is:

$$\mathrm{pH} = - \lg [\ce{H3O+}] = - \lg 0.5 = - (-0.301) = 0.301$$

The solution’s pH value obviously changed with concentration change.

Now, let’s do the same thing with a weak acid such as acetic acid ($\mathrm{p}K_\mathrm{a} = 4.76$). But first, I need to do a bit of maths. Remember that we cannot use the Henderson–Hasselbalch equation, because it assumes a buffered system. A simple acid is unbuffered. Instead, I will start at the definition of the acid constant:

$$K_\mathrm{a} = \frac{[\ce{H3O+}][\ce{OAc-}]}{[\ce{HOAc}]}$$

And since chemistry dictates that $[\ce{H3O+}] = [\ce{A-}]$.

$$K_\mathrm{a} = \frac{[\ce{H3O+}]^2}{[\ce{HOAc}]}$$

Because $c_0(\ce{HOAc}) = c(\ce{HOAc}) + c(\ce{H3O+})$, we can transform to:

$$K_\mathrm{a} = \frac{[\ce{H3O+}]^2}{[\ce{HOAc}]_0 - [\ce{H3O+}]}$$

Which finally gives us an equation with only one unknown. Solving this:

$$K_\mathrm{a} = \frac{[\ce{H3O+}]^2}{[\ce{HOAc}]_0 - [\ce{H3O+}]}\\ K_\mathrm{a} \left ( [\ce{HOAc}]_0 - [\ce{H3O+}] \right ) = [\ce{H3O+}]^2\\ [\ce{H3O+}]^2 + K_\mathrm{a} [\ce{H3O+}] - K_\mathrm{a} [\ce{HOAc}]_0 = 0$$

Since the determinant is positive, we can immediately write:

$$[\ce{H3O+}]_{1/2} = \frac{- K_\mathrm{a} \pm \sqrt{K_\mathrm{a}^2 + 4 K_\mathrm{a} [\ce{HOAc}]_0}}{2}$$

Now plug the values $c_{0,1}(\ce{HOAc}) = 1~\mathrm{\frac{mol}{l}}$ and $c_{0,2}(\ce{HOAc}) = 0.5~\mathrm{\frac{mol}{l}}$ as above. You will need to add the value of the square root or you will get a negative concentration. This is what you should be getting (assuming OpenOffice isn’t buggy):

$$\begin{array}{ccc} \hline \text{expression} & c_0 = 1~\mathrm{\frac{mol}{l}} & c_0 = 0.5~\mathrm{\frac{mol}{l}}\\ \hline K_\mathrm{a}^2 + 4 K_\mathrm{a} [\ce{HOAc}]_0 & 6.95 \cdot 10^{-5} & 3.48 \cdot 10^{-5}\\ \sqrt{K_\mathrm{a}^2 + 4 K_\mathrm{a} [\ce{HOAc}]_0} & 8.34 \cdot 10^{-3} & 5.90 \cdot 10^{-3}\\ - K_\mathrm{a} \pm \sqrt{K_\mathrm{a}^2 + 4 K_\mathrm{a} [\ce{HOAc}]_0} & 8.32 \cdot 10^{-3} & 5.88 \cdot 10^{-3}\\ [\ce{H3O+}] & 4.16 \cdot 10^{-3} & 2.94 \cdot 10^{-3}\\ -\lg [\ce{H3O+}] & 2.38 & 2.53\\ \hline \end{array}$$

Obviously, once again, the pH went up when diluting the solution.

The Henderson–Hasselbalch equation, as I skimmed past earlier, is meant for buffer systems. The concentrations $[\ce{A-}]$ and $[\ce{HA}]$ are those concentrations you added to the buffer system. Since we did not add any acetate ions (other than acetic acid), the logarithmic argument would become zero causing an undefined error. We cannot use Henderson–Hasselbalch here.

However, if you have a solution containing $1~\mathrm{\frac{mol}{l}}$ acetic acid and $1~\mathrm{\frac{mol}{l}}$ sodium acetate, the logarithm’s argument will be $1$, therefore the logarithm will disappear and the pH value will be $4.76$ — exactly the $\mathrm{p}K_\mathrm{a}$ value of acetic acid. This buffer system is concentration independent. Diluted acids to whom the conjugate base has not been added as a salt are not!

  • 2
    $\begingroup$ Just what I was looking for! Diluting a buffer wont change its pH but diluting an acid would. Thanks a lot :) $\endgroup$
    – Polisetty
    Jun 1, 2016 at 19:27
  • $\begingroup$ @A.K. A logarithm needs a base. $\log$ by itself is meaningless. The answer often uses $\log_{10}$ for which the shorthand notation $\lg$ exists. $\endgroup$
    – Jan
    Oct 16, 2018 at 13:34

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