# The pH of a neutralized solution

If pH is defined as the concentration of hydrogen ions in solution, then how can a ‘neutralized’ solution (defined as having an equal amount of hydrogen and hydroxide ions) have a pH other than 7?

Wikipedia writes: “In a reaction in water, neutralization results in there being no excess of hydrogen or hydroxide ions present in solution.” In my mind, an equivalent amount of hydrogen ions and hydroxide ions means a pH of 7 (in water), but Wikipedia carries on to write: “The pH of the neutralized solution depends on the acid strength of the reactants.”

This question doesn’t seem to answer my question.

• Are you sure is it completely neutralized? Because it's impossible – Simon-Nail-It Nov 21 '14 at 8:40
• When a solution is neutralized, the pH is equal to the point of equivalence, which is not 7 if either the acid or base is stronger (as with titration experiments). – ahorn Nov 21 '14 at 8:43
• Welcome to Chemistry.SE! To acquaint yourself with this page, take the tour and visit the help center. Furthermore this tutorial shows you how math and chemical formulae can be nicely formatted on this site. – Philipp Nov 21 '14 at 12:48
• The reason why I have never accepted an answer to this question is because, when I first read the answers, they were too advanced for me to understand, and subsequently my direction of study went strongly towards mathematics and economics. – ahorn May 17 '18 at 20:13

TL;DR
The neutral point is not the same as the equivalence point.
A neutral aqueous solution at room temperature, $25~^\circ\mathrm{C}$ and standard pressure $1~\mathrm{atm}$, has always $\ce{pH}=7$.

If you look for credible sources of definitions of neutral solutions on the internet, most likely you will find something along the lines of this:

Neutral solution
(Science) Has a $\ce{pH}$ level of $7$: a solution in which the concentration of hydrogen ions and hydroxide ions are equal (biology-online.org)

Or even worse:

Neutral Solution Definition:
An aqueous solution with a $\ce{pH}$ of $7.0$ $(\ce{[H^{+}]} = 1.0 \times 10^{-7}~\mathrm{M})$. (chemistry.about.com)

The second one especially is negligent of a lot of features.
Unfortunately the IUPAC does not provide an actual definition of a neutral solution. There are two possible definitions, which come to the same conclusion (compare the above), given the same external conditions.

1. The lazy definition:
The concentrations of hydronium and hydroxide ions are identical. $$c(\ce{OH-})= c(\ce{H3O+})$$
2. In terms of $\ce{pH}, \ce{pOH}, \mathrm{p}K_w$ one arrives at a more complete side.
Pure water is a neutral solution. It's autoprotolysis provides \begin{align} \ce{2H2O &~<=>~ OH- + H3O+}.\tag{1}\\ \end{align} Now we can formulate the equilibrium constant and furthermore assume that autoprotolysis is small compared to the overall concentration of water. \begin{align} K_c &= \frac{c(\ce{OH-})\cdot c(\ce{H3O+})}{c^2(\ce{H2O})}\\ K_w &= c(\ce{OH-})\cdot c(\ce{H3O+})\tag{2}\\ \end{align} Equation $(1)$ also provides the lazy definition. $$c(\ce{OH-})= c(\ce{H3O+})\tag3$$ We plug that into $(2)$ and we arrive at $$c(\ce{H3O+})=\sqrt{K_w}\tag{4}$$ Consider the actual definition of $$\ce{pH} = −\lg a(\ce{H+}) = −\lg\left( m(\ce{H+}) \gamma_m(\ce{H+}) / m^\circ\right)\tag{5}$$ rewritten using concentrations $$\ce{pH} = −\lg a(\ce{H+}) = −\lg\left( c(\ce{H+}) \gamma_c(\ce{H+}) / c^\circ\right)\tag{5a}$$ The identities used here are $a(\ce{H+})$ for the activity of a proton in aqueous solution, $\ce{H+ (aq)}$ and we will consider $\ce{H+ (aq) = H3O+}$ to be the same. Now we are going to assume, that the activity coefficient is $\gamma_c(\ce{H+})\approx 1$ for very diluted systems. Using the standard concentration $c^\circ$ we do not have to care about units. $$\ce{pH} = −\lg\left( \frac{c(\ce{H+})}{c^\circ}\right)\tag{5b}$$ Now we can substitute in $(4)$ and we will have a nice definition of the $\ce{pH}$ of a neutral solution. $$\ce{pH} = −\lg\left( \frac{\sqrt{K_w}}{c^\circ}\right)\tag{6}$$

The second formulation includes a very crucial point. The autoprotolysis $(1)$ is temperature, $T$, dependent, which is obvious following the definition of the standard equilibrium constant $$K^\circ = \exp\left\{\frac{-\Delta G^\circ}{\mathcal{R}T}\right\}\approx K_c.$$ Therefore also the ion product of water is temperature dependent, $K_w(T)$. Wikipedia actually has a couple of values for the $\mathrm{p}K_w$. Reaction $(1)$ is endothermic. A higher temperature means providing energy, means a higher $\mathrm{p}K_w$. For example, in the human body, $37~^\circ\mathrm{C}$, this value is slightly higher than for the one we usually assume as the "neutral point" $\mathrm{p}K_w(25~^\circ\mathrm{C})=14$.

Since $(6)$ is more correct in the following form, $$\ce{pH}(T) = −\lg\left( \frac{\sqrt{K_w(T)}}{c^\circ}\right)\tag{6a},$$ the neutral $\ce{pH}(T)$ therefore changes with temperature.

Another note is that it is also pressure dependent, but that might take it a little too far.

The equivalence point is usually defined for an arbitrary chemical reaction between an acid and a base, and it refers to the situation where there are stoichiometric quantities of acid and base are present. $$\ce{AH + B <=>[\ce{H2O}] A^- + HB+}$$ This means, that at the equivalence point the following statement holds: $$c_\text{initial}(\ce{HA}) = c_\text{final}(\ce{HB+}).$$ Therefore the $\ce{pH}$ at the equivalence point will only be governed by the reactions of $\ce{HB+}$ and might not be neutral.

The issue of solvent dependency is addressed by Fred Senese quite concise on antoine.frostburg.edu:

pH is often used to compare solution acidities. For example, a solution of pH 1 is said to be 10 times as acidic as a solution of pH 2, because the hydrogen ion concentration at pH 1 is ten times the hydrogen ion concentration at pH 2. This is correct as long as the solutions being compared both use the same solvent. You can't use pH to compare the acidities in different solvents because the neutral pH is different for each solvent. For example, the concentration of hydrogen ions in pure ethanol is about 1.58 × 10-10 M, so ethanol is neutral at pH 9.8. A solution with a pH of 8 would be considered acidic in ethanol, but basic in water!

• For me, the key to understanding how the pH at a point of equivalence is not necessarily 7 is as follows: a neutralization reaction produces a salt and water, but the salt is not necessarily neutral. – ahorn Nov 21 '14 at 12:40
• But I'm still not sure about whether a salt can have an excess of $\text{H}^+$ or $\text{OH}^-$ or not. – ahorn Nov 21 '14 at 12:43
• @ahorn A salt of any weak acid or base will be in equilibrium with the dissociation of water, thus producing protons and/or hydroxide ions. Therefore the pH will differ from the neutral point. – Martin - マーチン Nov 26 '14 at 4:31
• Just as a point of reference (though I think it should be clear to people by context), $\lg(x) \not= \log_2(x)$ here. This is a common notation in the US. – Tyberius Apr 24 '17 at 23:02
• @Tyberius I personally prefer having the charge where it should be, that's why I made the effort of writing $\ce{^{-}OH}$. I agree with the recent additions to mhchem, $\ce{H3+O}$ looks a bit silly. – Martin - マーチン Apr 25 '17 at 11:45

In addition to all other good answers, I'd like to actually show you why the "neutralized solution" is not equal to a neutral solution.

Whenever we want to describe a neutralisation reaction (that is nothing else than a titration) we first need to find out where protons come from and where they go. Let's examine a simple titration of a monoprotic acid with a monoprotic base:

$$c(\ce{H3O+})=c(\ce{OH-})+c(\ce{A-})-c(\ce{BH+})$$

Protons come from the autoprotolysis of water and the acid residue anions and they will react with the titration base. Extending the parts of this equation by their $\mathrm{pH}$, $c_0$ and $k_\mathrm a$ containing counterparts, we get:

$$c(\ce{H3O+})=\frac{k_\mathrm w}{c(\ce{H3O+})}+\frac{c_0(\ce{HA})~k_\mathrm a}{k_\mathrm a+c(\ce{H3O+})}-\frac{c_0(\ce{B})~k_\mathrm b~c(\ce{H3O+})}{k_\mathrm w+k_\mathrm b~c(\ce{H3O+})}$$

At the point of inflection, the equivalence point, the amount of both moles need to be equal and the total volume is the addition of previous acidic volume plus the added base volume $V_\mathrm{tot}=V_{\ce{A}}+V_{\ce{B}}$. The ratio of both concentrations $c(\ce{HA})$ and $c(\ce{B})$ is therefor equal to $1$. \begin{align} n_{\ce{A}}&=n_{\ce{B}}\\ \Leftrightarrow c_\mathrm A~V_\mathrm a&=c_\mathrm B~V_\mathrm b\\ \xrightarrow{V_\mathrm{tot}=V_\mathrm A+V_\mathrm B} \frac{c_B}{c_A}&=1=\tau~\text{or}~\gamma\end{align}

Knowing this, rearranging the second equation to $c_\mathrm A/c_\mathrm B$ gives us:

$$\tau=\frac{c_\mathrm B}{c_\mathrm A}=\left(1+\frac{k_\mathrm w}{k_\mathrm b~c(\ce{H3O+})}\right)\left(\frac{k_\mathrm w}{c(\ce{H3O+})}-c(\ce{H3O+})+\frac{k_\mathrm a}{c(\ce{H3O+})+k_\mathrm a}\right)$$

which we now somehow have to solve for the proton concentration after setting $\tau=1$. As this is rather annoying I did that for the titration of any more or less relevant acids for $\mathrm pK_\mathrm a$ values between $-2$ and $16$, with $\ce{NaOH}$ giving the following graph.

$\hskip1.2in$

You can see that strong acids will be neutralized a pH 7 and extremely weak acids at pH 14. In between there is a linear range in which one can calculate the pH at the equivalence point by the simple equation $$\mathrm pH_\mathrm{eq}=7+0.5~\mathrm pK_\mathrm a$$

This can be expanded for the neutralization of every monoprotic acid with every monoprotic base and again there is a simple equation for calculating the pH at the equivalence point.

$\hskip1.2in$

$$\mathrm pH_\mathrm {eq}=7+0.5~\mathrm pK_\mathrm a-0.5~\mathrm pK_\mathrm b$$

$$\ce{CH3COOH +NaOH->CH3COONa + H2O}$$
$$\ce{CH3COO- + H2O->CH3COOH + OH-}$$