I am doing an experiment on the effect of time on the $\mathrm{pH}$ of Lime juice. I was wondering how you would calculate the $\mathrm{pH}$ through titrations using $\pu{1 Molar}$ $\ce{NaOH}$. I have been racking my brain on this for a few hours and searching online.

Sorry if this has already been answered. Thank you for your time.

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    $\begingroup$ There is no direct relation between pH and titration result, unless there is known there is negligible amount of citric acid salts, of other acids and other pH buffering systems. $\endgroup$ – Poutnik Feb 10 '20 at 7:07
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    $\begingroup$ To expand on @Poutnik comment, pH and titration (measuring number of moles) are two different concepts, measuring two different chemical properties. For example, 1 molar HCl is far more acid (lower pH) than 5 molar citric acid, but it would take more NaOH to titrate the citric acid. $\endgroup$ – DrMoishe Pippik Feb 10 '20 at 22:39
  • $\begingroup$ Similarly, a citric acid solution with lower pH can lead to both higher or lower titration volume, depending on the overall solution composition. $\endgroup$ – Poutnik Feb 11 '20 at 6:37
  • $\begingroup$ pH paper would be the cheapest and easiest way to estimate the pH of the lime juice. Even if titration could measure pH, it's annoying to do. $\endgroup$ – user137 Sep 18 '20 at 7:49

As said, titration and the concept of $\mathrm{pH}$ is very different. The process of titration is used to figure out an estimate of how much of acid $\ce{A}$ (of come concentration) will neutralize base $\ce{B}$ in an acid-base reaction (of come concentration) (also the reverse). The acid and base reaction is solely dependent on the coefficient of a chemical equation regardless of the $\mathrm{pH}$.

e.g., One mole of aqueous $\ce{HCl}$ will react always with one mole aqueous of $\ce{NaOH}$.

The $\mathrm{pH}$ in this case only indicates the "acidity" of either of the reactants based on the amount of $\ce{H3O+/OH-}$ ions that dissociated form them:

$$\mathrm{pH} = - \log[\ce{H3O+}]$$

You can of course figure out the concentration of a base/acid in titration if you know the basicity of one of the solution and the mole of acid/base in the other. You can do this by:

Let's say you want to know the concentration of a solution of ethanoic acid ($\ce{CH3COOH}$), we will have to prepare a suitable base to use as a titrant. We mix a universal indicator with the solution of $\ce{CH3COOH}$ and add it to a flask. Dissolving one mole of $\ce{NaOH}$ into x volume of water and put it in the buret. Run the buret until the resultant solution is neutral.

If the original volume of the $\ce{CH3COOH}$ solution is y and the amount of $\ce{NaOH}$ solution deposited is k.

Concentration of original acid is given by: $$[\text{acid}] = {\frac{k}{x*\text{(original basicity of acid)}*y} }$$

as in \begin{align} \\ [\text{acid}] &= {\frac{(\text{mole of}\, \ce{NaOH}\,\text{used})\cdot(\text{volume of}\, \ce{NaOH}\,\text{used})}{(\text{volume of}\, \ce{NaOH})\cdot(\text{original basicity of acid})\cdot(\text{volume of acid})} } \end{align}

However, we still wouldn't be able to determine the $\mathrm{pH}$ of the original acid as we do not know how many of the acid in the original solution actually ionized: $$[\text{acid}]*(\text{basicity}) \neq [\ce{H3O+}]$$

Thus we do not know the $[\ce{H3O+}]$ and the $\mathrm{pH}$ cannot be determined.

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    $\begingroup$ A further difficulty is the fact that the formula pH = - log [H+] is an approximation which is only valid at rather low concentrations (<< $0.1$ M). At higher concentrations, the concentration [H+] should be replaced by the activity $ a(\ce{H^+})$, which can be rather different from [H+]. For example, a solution HCl $1$ M has an activity $0.795$. $\endgroup$ – Maurice Sep 24 '20 at 17:04

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