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We know that for strong acids with a $1:1$ mole ratio between all particles involved are suitable for use with the aforementioned formula for titration problems. I want to know the limits on this formula, that is:

Consider if we had $$\ce{H2SO4(aq) + 2KOH(aq) -> 2H2O(l) + K2SO4(aq)}$$

Since the moles of $\ce{H+}$ = moles of $\ce{~^{-}OH}$ in order to neutralize, how do we apply $M_aV_a=M_bV_b$ to this situation, if at all possible?

Secondly, does this formula work if we had weak acids? I would think not, because with weak acids we don't have the benefit of complete dissociation and thus not a $1:1 = \text{acid}:\ce{H+}$ ion ratio nor a $1:1 = \text{base}:\ce{~^{-}OH}$ ratio.

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Regarding the ionization extent of sulfuric acid solutions:

$\text{1)}$ The first ionization is complete at most molarities (until you start reaching extremely high molarities - i.e. $\text{18 M}$). So sulfuric acid is a strong acid.

$\text{2)}$ The second ionization (i.e. the ionization of the hydrogen sulfate ion, $\ce{HOSO3^-}$, can be complete at low starting concentrations of hydrogen sulfate ion. This is because $K_a(\ce{HOSO3^-}) = 1.2 \times 10^{-2}$ is rather sizable!

So we can generalize these two cases of sulfuric acid solutions:

For very dilute sulfuric acid solutions:

$[\ce{H3O^+}]=2M_i - [\ce{HOSO3^-}] = [\ce{HOSO3^-}]+ 2\times[\ce{SO4^{2-}}]$, in which we generally take $[\ce{HOSO_3^-}]$ as negligible (i.e. = $\epsilon \to 0$).

For non-dilute solutions of sulfuric acid, we only have these relationships:

$$[\ce{H3O^+}]=M_i +[\ce{SO4^{2-}}] = [\ce{HOSO3^-}]+2[\ce{SO4^{2-}}]$$

As we can see, the first relationship (one for very dilute solutions) follows from the second because in a very dilute solution, $[\ce{HOSO3^-}]\approx 0$ - i.e. it has fully ionized into $[\ce{SO4^{2-}}]$.

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  • $\begingroup$ Ionisation is adding or taking an electron. What you are describing is dissociation as in heterolytic bond cleavage resulting in ions. For your second equation it is better to write $$\ce{[H3+O]} = 2 M_{i} - \ce{[HOSO_{3}^{-}]}.$$ $\endgroup$ – Martin - マーチン♦ May 10 '14 at 9:59
  • $\begingroup$ Yes, ionization can also be used in that context - i.e. ionization of vaporized sodium in the Born-Haber cycle. But textbooks also use ionization to refer to the loss of protons by acids. $\endgroup$ – Dissenter May 11 '14 at 3:43
  • $\begingroup$ I agree on your edit for the second equation; that would be a more accurate way of getting at the hydronium ion concentration. $\endgroup$ – Dissenter May 11 '14 at 3:43
  • $\begingroup$ Unfortunately you are correct, as ionisation provides the base definition for dissociation, too. Well, well, well, we are all here to learn. $\endgroup$ – Martin - マーチン♦ May 12 '14 at 12:50
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Actually, you can. $\ce{H2SO4}$ liberates $\ce{2 H+}$ ions which is neutralized by the $\ce{OH^-}$ ions. The $K_\mathrm{a}$ for water is $10^{-14}$, this means that the concentration of those ions are very low compared to the $\ce{K^+}$ and the $\ce{SO4^{2-}}$ ions. You just need to use the stoichiometry of the equation properly to apply it to $M_aV_a=M_bV_b$.

You can use weak acids and bases if you allow them time to react. Most of them will react in less than a second, but I seen a few that take several seconds. In fact, the best way to determine the concentration of sodium hydroxide solution (since it can absorb carbon dioxide from the air) is to weigh dried potassium hydrogen phthalate (a weak acid) to determine the concentration.

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  • $\begingroup$ If you apply this to a weak acid (equal amount of substance, mol), you will not get a neutral solution. Since at $\ce{pH} = 7~(25^\circ{}C)$ not all of the acid/base is dissociated. $\endgroup$ – Martin - マーチン♦ Apr 10 '14 at 6:47
  • $\begingroup$ Manual titrations usually use pH indicators. Of the dozen listed on wikipedia, one has a pH change near 7. Most titrations using NaOH use phenolphthalain, which changes near 8. At this pH, the amount of undisassociated acid (from a weak acid) is insignificant. en.wikipedia.org/wiki/Ph_indicator $\endgroup$ – LDC3 Apr 10 '14 at 13:23
  • $\begingroup$ They use phenophtalein, because it changes exactly (at least for $\ce{H3C\bond{-}COOH}$) at that point where $\ce{n_0(HA)=n(OH^{-})}$. The solution is not neutral. If it is neutral at pH 7, then for most weak acids like $\ce{HSO4-}$ there is a significant amount undissociated. This is why you can use the as buffers. $\endgroup$ – Martin - マーチン♦ Apr 11 '14 at 1:28
  • $\begingroup$ @Martin $H_2SO_4$ is a strong acid, both protons. Look it up. $\endgroup$ – LDC3 Apr 11 '14 at 1:52
  • $\begingroup$ $\text{p}K_a(H2SO4)=-3$, $\text{p}K_a(H3O+)=-1.7$, $\text{p}K_a(HSO4-)=1.9$ Hence the second deprotonation is incomplete, especially with excess hydronium. $\endgroup$ – Martin - マーチン♦ Apr 11 '14 at 7:05

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