# How do you find the pH of lime juice through titration?

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.

• 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. Feb 10, 2020 at 7:07
• 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. Feb 10, 2020 at 22:39
• Similarly, a citric acid solution with lower pH can lead to both higher or lower titration volume, depending on the overall solution composition. Feb 11, 2020 at 6:37
• 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. Sep 18, 2020 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.

• 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 $0.795$. Sep 24, 2020 at 17:04