Suppose we have 1 eq of acetic acid and we want neutralize it using sodium hydroxide. After adding 1 eq of sodium hydroxide 1 eq of sodium acetate should form and neutralization should be complete but because of hydrolysis little bit of acetic acid will form again so we would have to add more sodium hydroxide and this process would go on forever so how will neutralization ever be complete?

My guess is that after adding 1 eq of NaOH the conc. of acetic acid becomes so low that it is neglected and the neutralization is assumed to be over.

  • $\begingroup$ So, is the solution of sodium acetate incompletely neutralised sodium hydroxide or incompletely neutralised acetic acid ? Neutralisation involving weak acids/basis does not involve completeness but it does involve equilibrium. $\endgroup$ – Poutnik Apr 26 '19 at 3:58
  • $\begingroup$ @Poutnik ,@Sanom Dane I think he trying to tell , when dissociated h+ is consumed , more h+ ions are produces according to me chatliers principle. $\endgroup$ – Chemist Apr 26 '19 at 8:33
  • $\begingroup$ I wanted to say, neutralisation is over, if the solution is equivalent to solution of the equivalent amount of sodium acetate. $\endgroup$ – Poutnik Apr 26 '19 at 8:47
  • $\begingroup$ @Poutnik What I wanted to say is that as equivalent amount of sodium acetate will never be formed(exactly) neutralization will never be complete. $\endgroup$ – Sanom Dane Apr 26 '19 at 19:05
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    $\begingroup$ Also, the concentration of hydroxide ion at equivalence is the same as if sodium acetate were dissolved. So temptation to use extra hydroxide should be the same as to use the extra acid to get rid of the hydroxide. $$\ce{A- +H2O <=> HA + OH-}$$ $\endgroup$ – Poutnik Apr 26 '19 at 20:42

The $\ce{K_a}$ expression for acetic acid, in aqueous solution, is given by:

$\ce{K_a} =\dfrac{\ce{[CH3COO-][H+]}}{\ce{[CH3COOH]}}$

The expression can be rearranged to give

$\ce{[CH3COOH]} = \dfrac{\ce{[CH3COO-][H+]}}{\ce{K_a}}$

so no matter what the pH, all the acetic acid will never be reacted.

So in a titration let's assume that we start out with 0.1000 moles of whatever acid $\ce{[HA]_{initial}}$.

  • After base has added and $0.1000 \gg \ce{[HA]_{final}}$ and thus $0.0001 \gt \ce{[HA]_{final}}$, then the reaction is quantitatively complete.

    This is strictly a significant figures argument. You can't do a titration to 10 significant figures. 3 is decent, if you get 4 you're a magician.

  • When 0.1000 moles of base have been added to the solution the equivalent point has been reached.

    The key point is that the moles of base is equal to moles of acid, regardless of the pH.

  • When ever you get the "signal" that the titration is over you have reached the end point.

    The signal could be, for example, a color change of an indicator, the swing of the needle on a pH meter, or the formation of a precipitate.

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  • $\begingroup$ Thanks, will there ever be a case where [Salt]>>[Acid] is unachievable (maybe when initial conc. of acid is comparable to $K_h$) $\endgroup$ – Sanom Dane Apr 26 '19 at 19:10
  • $\begingroup$ I don't understand what you mean by $\ce{[salt] \gg [acid]}$. Could you explain? example? $\endgroup$ – MaxW Apr 26 '19 at 20:44
  • $\begingroup$ In your example if 0.1 mol of acid was taken then neutralization would be complete when [HA]<<0.1, also [BA(salt)]≈0.1? please correct me if I'm wrong, Thanks. $\endgroup$ – Sanom Dane Apr 27 '19 at 8:08
  • $\begingroup$ Not just 0.1 (1 significant figure), 0.1000 (4 significant figures). Let me clarify my answer about HA. $\endgroup$ – MaxW Apr 27 '19 at 14:39

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