# 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. – Poutnik Feb 10 '20 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. – DrMoishe Pippik Feb 10 '20 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. – Poutnik Feb 11 '20 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. – 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.

• 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 $0.795$. – Maurice Sep 24 '20 at 17:04