# What volume of water will dilute an acidic solution back to neutral?

Now we are talking about simple dilution, not acid-base neutralization, so we expect asymptotic behavior that never quite gets us back to pH=7; that is, an infinite amount of water would be required. However, most pH probes read to two decimal places, so diluting to reach pH=6.995 is a sufficient end condition. So, given an initial pH of $pH_0$ and an initial solution volume $V_0$, what additional volume $V$ of pure water would be required to dilute the pH to 7.00 (6.995)?

I've wondered this for a while, and today I attempted to solve it. I may have found the solution, but I am skeptical of its implications. If I have 1 L of a solution at pH=2, my derived formula indicates it would require additional volume of about 8.6 million liters of water for the solution to read neutral. Is this reasonable?

My derived formula:

$$V=\Big(\frac{1}{1-10^{-0.005}}\Big)\Big(10^{6.995-pH_0}-1\Big)V_0 \\ \approx \Big(86.36\cdot10^{7-pH_0}-87.36\Big)V_0$$

Derivation:

Moles of $H^+$ from original soln: $n_0 = [H^+]_0V_0$

from added water: $n = [10^{-7}]V$ (via autodissociation)

Total moles: $n'=n_0+n$

Total volume: $V'=V_0+V$

New molarity after addition of $V H_2O$:

$$[H^+]'=\frac{n'}{V'}=\frac{[H^+]_0V_0+[10^{-7}]V}{V_0+V}$$

$$\Rightarrow V=\Big(\frac{[H^+]_0-[H^+]'}{[H^+]'-[10^{-7}]}\Big)V_0$$

(Note that $V \to \infty$ as $[H^+]' \to 10^{-7}M$, an infinite volume is require to exactly reach neutral pH).

Letting $[H^+]'=10^{-6.995}M$ (paragraph 1), $[H^+]_0=10^{-pH_0}$, and multiplying top and bottom by $10^{6.995}$, obtain:

$$V=\Big(\frac{10^{6.995-pH_0}-1}{1-10^{-0.005}}\Big)V_0$$

Question: Is this derivation valid? Are there assumptions I should or should not have made? Would it truly take 8.6 million times the original volume to dilute a pH=2 solution back to neutral pH=7.00?

• Since water establishes an equilibrium with $\ce{OH^-}$ as well as with $\ce{H^+}$, I don't think you can leave that out when considering what happens in the system as you add water. – J. Ari Apr 5 '17 at 20:04

Like J. Ari said in the comments, there is an equilibrium involving $\ce{H+}$ and $\ce{OH-}$, and it will be shifted by the dissolved acid (that is, the concentrations will shift so the equilibrium constant may stay the same). Therefore you can't assume that water will always contain precisely $10^{-7}$M protons from autodissociation (plus maybe some extra from acid). It doesn't work that way.
$$\ce{[H+]\cdot[OH-]=10^{-14}}\\ \ce{[H+]-[OH-]=}c(\rm{HCl})$$ (since the acid is strong, and hence fully ionized even in not-so-diluted solutions) $$\ce{[H+]=10^{-6.995}}\\ \ce{[OH-]=10^{-7.005}}\\ c\ce{(HCl)=10^{-6.995}-10^{-7.005}=10^{-6.995}(1-10^{-0.01})}\\ V_{total}=V_0\cdot{c_0\over c}\\ V_{added}=V_0\cdot\left({c_0\over c}-1\right) =V_0\cdot\left({10^{6.995-pH_0}\over1-10^{-0.01}}-1\right)$$ So it is only 4.3 million times the original volume, which is about 2 times less than your estimate.