I learned in class that the equivalence point in an acid-base titration is reached when the solution contains an equal amount of substance of $\ce{OH-}$ and $\ce{H+}$ ions. However, in a weak acid and strong alkali titration, the pH at the equivalence point (when I added the right amount of acid and alkali to make it a neutral solution) is 9. Also, in the titration of a strong acid and weak alkali, the pH of the solution when I have added the correct amounts is 5.

My question:

If I add 1 mole of ethanoic acid to 1 mole of sodium hydroxide, I should get a neutral solution and the pH should be 7. However, why is the pH 9? This does not make sense to me as a pH 9 suggest that the concentration of $\ce{OH-}$ is higher than the concentration of $\ce{H+}$. But how is that, I reacted the correct amounts?

  • 1
    $\begingroup$ when you add 1 mole of sodium hydroxyde to 1 mole of ethanoic (=acetic) acid, you will get 1 mole of sodium acetate (try to write down the neutralization reaction). Acetate is the conjugated base of a weak acid and is therefore strong enough to significantly alter the pH: that's why the final pH is 9. $\endgroup$
    – mannaia
    Jan 25, 2014 at 15:19
  • $\begingroup$ @mannaia: could you please elaborate "strong enough to significantly alter pH" $\endgroup$
    – Eliza
    Jan 25, 2014 at 16:44
  • $\begingroup$ Please note that the proper term for "(number of) moles" is amount of substance. The former would be the same as referring to the mass as "(number of) kilograms". $\endgroup$ Jan 29, 2018 at 15:32

4 Answers 4


Two kinds of reactions have to be considered:

  1. Neutralisation of the weak acid with the strong base
  2. Secondary reactions from the products

The neutralisation is fairly straightforward: $$\ce{HA + NaOH -> NaA + H2O} $$

However, and here comes number two from the list: It doesn't just stop there because we still have unreacted acid in the solution and lots of water. So what will happen? The conjugated base of the acid can also react (as a base): $$\ce{A- + H2O <=> OH- + HA}\; ,$$

thus raising the pH of the solution. This depends on the equilibrium constant $\mathrm{K_b}$ of the conjugated base.

So although you have reacted exactly equal amounts of a (strong) base with the (weak) acid, you also have to consider the (weak) conjugate base of the acid, which also swims around in your titration soup.

  • $\begingroup$ What do you mean with "we still have unreacted acid in the solution" ? $\endgroup$
    – mannaia
    Jan 25, 2014 at 17:36
  • $\begingroup$ It's a phrase used to make you aware of all the other variables that might have something to do with how the solution reacts. $\endgroup$
    – tschoppi
    Jan 25, 2014 at 17:39
  • 1
    $\begingroup$ I think it's rather confusing. I would simply say that, at the equivalent point, the acid has been fully neutralised by the base (no unreacted acid present anymore) $\endgroup$
    – mannaia
    Jan 25, 2014 at 17:42
  • $\begingroup$ @tschoppi: in case of strong acid and strong alkali titration, aren't there conjugate bases and conjugate acids... won't there be a secondary reaction?Also, in the case of strong acid and weak base, will a weak conjugated acid form? $\endgroup$
    – Eliza
    Jan 25, 2014 at 17:42
  • $\begingroup$ @mannaia The term "neutralized" is slightly ambiguous in this context, as you don't necessarily have pH=7. $\endgroup$
    – tschoppi
    Jan 25, 2014 at 17:46

Let's consider your example where you are titrating acetic acid with sodium hydroxide. You are correct in saying that when equal amount of substance of acid and base is added, the acid will be completely neutralized. The reaction is: $$\ce{CH3COOH + NaOH -> CH3COONa + H2O}$$ Hence, at the equivalence point, the conical flask will only contain $\ce{CH3COONa}$ (which is a salt) and $\ce{H2O}$. However, acids and bases are not the only substances which have a pH different from 7. Some salts are also able to be acidic or basic. In the case of $\ce{CH3COONa}$, it is a basic salt. The reason why is because $\ce{CH3COONa}$ actually consists of $\ce{CH3COO-}$ (a base) and $\ce{Na+}$ (an acid). Since $\ce{CH3COO-}$ is the conjugate base of a weak acid, it is strong enough to be able to react with water to produce $\ce{OH-}$ ions: $$\ce{CH3COO- + H2O <=> CH3COOH + OH-}$$ Meanwhile, since $\ce{Na+}$ is the conjugate acid of a strong base, it won't be strong enough to react with water.

Therefore, at the equivalence point, even though there is no acid or base present, $\ce{[OH-] > [H3O+]}$, hence the pH will be $>7$.

  • $\begingroup$ Why is the conjugate base/acid of a strong acid/base weak? $\endgroup$
    – Tyto alba
    Apr 15, 2017 at 20:06
  • $\begingroup$ Lets consider HCl which is a strong acid. When HCl ionises in water, their is the following equilibrium: $\ce{HCl + H2O <=> Cl- + H3O+}$. Since HCl is a strong acid, the equilibrium lies heavily to the right. Therefore, it follows that the equilibrium for the reverse reaction (where Cl- is acting as a base) must lie heavily to the left. Hence Cl- is a weak base. This can be applied to any conjugate acid/base pair, so we can get the general statement the conjugate base of a strong acid is weak and vice versa. $\endgroup$
    – Nanoputian
    Apr 16, 2017 at 2:15

My answer to your question in the headline is: Not always.

  1. Neutralisation and neutral $\mathrm{pH}$ are often different things.

  2. Neutral $\mathrm{pH}$ is when $\ce{[H3O+]} = \ce{[OH-]} = 7$

  3. Neutralisation of a weak acid by a strong base means that you completely have converted the acid to its corresponding base. The $\mathrm{pH}$ at the equivalence point will depend on the acid used.

  4. Neutralisation of a weak base by a strong acid means that you completely have converted the base to its corresponding acid. The pH at the equivalence point will depend on the base used.

I will give you three examples:

  1. Suppose you titrate a strong base with a strong acid: $$\ce{NaOH + HCl = NaCl + H2O}$$

    The equivalence point is reached when all $\ce{NaOH}$ has been converted to sodium chloride. The result, at the equivalence point, will be the same as dissolving sodium chloride in water at the same concentration as you have at the equivalence point. We expect the pH of the solution at the equivalence point to be neutral, i.e. $\mathrm{pH} = 7$.

    Why do we expect that? Well, neither $\ce{Na+}$ nor $\ce{Cl-}$ will react with water in any protolysis reaction. [We do not consider any ionic strength effects on the $\mathrm{pH}$ in this reasoning.]

  2. Suppose you titrate a weak base with a strong acid: $$\ce{NH3 + HCl = NH4+ + Cl-}$$

    The equivalence point is reached when all ammonia has been converted to ammonium ions. The result, at the equivalence point, will be the same as dissolving ammonium chloride in water at the same concentration as you have at the equivalence point. Now, suppose the concentration of the ammonium and chloride ions were $\pu{0.01 M}$ at the equivalence point. The $\mathrm{pH}$ would then be about $5.6$.

  3. Suppose you titrate a weak acid with a strong base: $$\ce{NH4+ + OH- = NH3 + H2O}$$

    The equivalence point is reached when all ammonium ions have been converted to ammonia. The result, at the equivalence point, will be the same as dissolving ammonia in water at the same concentration as you have at the equivalence point. Now, suppose the concentration of the ammonia was $\pu{0.01 M}$ at the equivalence point. The $\mathrm{pH}$ would then be about $10.5$.

From this we learn, neutralisation can give $\mathrm{pH} = 7,$ but not always.


You have confused neutrality point, inflection point and equivalence point. The former is where there are equal numbers of protonated and deprotonated solvent species ($\mathrm{pH}~7$ in water at $\pu{25 ^\circ C}$). The inflection point is where the second derivative of the titration curve (amount of base added plotted against $\mathrm{pH}$) is a maximum (maximum buffering) and the third the point where the amount of analyte and titrant are equivalent. If you take acetic acid and start adding base to it even a small increment of added base will produce a large change in $\mathrm{pH}$. The base isn't reacting with the $\ce{HOAc}$, it is reacting with the solvent. The happens until the inflection point $\mathrm{pH}~4.76$ for acetic acid, is approached in which region the base is reacting with the $\ce{HOAc}$:
$$\ce{HOAc + OH- -> OAc- + H2O}$$

and a small change in $\mathrm{pH}$ requires much more base. Eventually, (about $\mathrm{pH}~4.76,$ that is two $\mathrm{pH}$ units above the inflection point) nearly all the $\ce{HOAc}$ is reacted and additional base just goes to deprotonate water, increasing the $\ce{OH-}$ ion concentration and $\mathrm{pH}$ at a rate much faster than near the inflection point. Thus the fact that you wind up at a $\mathrm{pH}$ near $9$ is simply a consequence of the fact that you are well past the inflection point of your acid which is, because the inflection point is at $4.76$, considered a weak acid. Once all the acid has been deprotonated, additional additions of base go to deprotonate water and the pH rises.

The pH at the equivalence point is the pH at which the solution is neutral. Finding this pH requires that we compute the charge of each species in the solution and totalling them up as a function of pH. We then seek the pH value that zeroes the sum. The species are $\ce{Na+}$, $\ce{OH-}$, $\ce{AcOH}$ and $\ce{AcO-}$. The charge on a liter of water is $Q_w(pH) = 10^{-pH} - 10^{(pH - pK_w)}$ which is the sum of the hydronium and hydroxyl ion charges. For the acetate its a bit trickier. We start with the Henderson Hasselbach equation writing it in the form $r_j = 10^{(pH - pK_j)}$ This is the ratio of the concentration (activity, really) of species which have lost j protons to the that of the species which has lost $j-1$. In the case of acetic acid there is only one proton to lose and so there is only one $r$ but in the case of polyprotic acids there is, of course, 1 $r$ for each proton. Now it shouldn't take you long to figure out that if the ratios are $r_1, r_2, r_3...$ the fraction of undissociated acid molecules is $f_0 = 1/(1 + r_1 + r_1r_2 + r_1r_2r_3)$ As r_1 is the ratio of singly deprotonated to undissociated its clear the the fraction of singly deprotonated is $f1 = r_1f_0$. At this point we have all we need to calculated the charge. Assuming 1 mole $\ce{NaOH}$ and 1 mole $\ce{AcOH}$ in 1 L of water the total charge would be $+1 + 1*Qw(pH) -1*f_1(pH)$ So the question now turns to how we find the value of pH which brings this expression to 0. Clearly the fastest way to a solution is to have and Excel spreadsheet compute it and either grope for the answer (pH 9 gives + $4.75·10^-5$ and pH 9.5 gives $4.75·10^-5$ so clearly the answer lies between those 2 pH values. This suggests a bisecting root finder as a path to a solution and that works very well. Excel has the Solver which does the iterations for you and takes you directly to the solution $pH = 9.38$. Note that the solution for 1 mole each in 10 liters of solution is $pH = 9.38$ which makes it clear that the water is playing a significant role here. Not all the $\ce{OH-}$ are going to deprotonate acid. Many are going to shift the waters charge balance.

  • 2
    $\begingroup$ "are equal numbers of solvent and protonated solvent species" That cannot be right, i.e. it would mean that that in pure water half of all water molecules would be dissociated, which doesn't really fit into $K_\mathrm{w}(\pu{25 ^\circ C}) = 10^{-14} = c[\ce{OH- (aq)}]\cdot c[\ce{H3O+ (aq)}].$ Also, I feel like that the part you are omitting ("but I won't go into that here."), would actually answer the question of why the pH is different from 7 at the equivalence point. $\endgroup$ Jan 29, 2018 at 16:05
  • 3
    $\begingroup$ I have updated your post with chemistry markup. If you want to know more, please have a look here and here. We prefer to not use MathJax in the title field, see here for details. $\endgroup$ Jan 29, 2018 at 16:05
  • $\begingroup$ A more productive comment would be "you left out a key word". I don't need to be schooled in the fact that this leads to an absurd situation, which, of course, it does. I can easily add in the how - to - determine - pH part if anyone wants me to. $\endgroup$ Jan 29, 2018 at 19:42

Not the answer you're looking for? Browse other questions tagged or ask your own question.