Question

What is the operational definition of $pH$and how is it used?

Short answer

The operational definition of pH corresponds to the direct calibration of the pH meter by the standard buffers. The electrode is immersed into the analyte solution that resulting in the development of a potential gradient across the membrane. The electrode potential depends on the activity of the ions. The pH is determined by measuring the electrode potential.

Step-by-step solution

Need for Operational definition of pH

It is necessary to define $pH$in operational terms i.e. by the way the measurement is made to assist easy duplication at various times at various laboratories throughout the world.

Determination of pH

As per the endorsements made by the National Institute of Standards and Technology (NIST), the International Union of Pure and Applied Chemistry (IUPAC), and similar national organizations, it is based on direct meter calibration with precisely established standard buffers, followed by potentiometric $pH$determination of unknown solutions.

Example

Consider the glass reference system in the below figure:

pH of standard buffer

When the electrodes are immersed in a standard buffer, its $pH$can be determined using the following equation:

${\mathrm{pH}}_{\mathrm{S}}=-\frac{E_{\mathrm{S}}-K}{0.0592}$

Where,

${\mathrm{pH}}_{\mathrm{S}}=$$pH$of standard buffer

$E_S=$cell potential when electrodes are immersed in standard buffer

and $K=E_j-E_{\mathrm{ref}}+L$,

with,

$E_j=$salt bridge

$E_{ref}=$reference electrode

$L=$standard electrode potential

Step 5: pH of unknown buffer

Similarly, when the electrodes are immersed in a standard buffer, its $pH$can be determined using the following equation:

${\mathrm{pH}}_{\mathrm{U}}=-\frac{\left(E_{\mathrm{U}}-K\right)}{0.0592}$

Where,

${\mathrm{pH}}_{\mathrm{U}}=pH$of unknown solution

$E_U=$cell potential when electrodes are immersed in a solution of unknown$pH$

Final equation of operational definition of pH

By subtracting the first equation from the second and solving for $p{H}_U$, we get,

${\mathrm{pH}}_{\mathrm{U}}={\mathrm{pH}}_{\mathrm{S}}-\frac{\left(E_{\mathrm{U}}-E_{\mathrm{S}}\right)}{0.0592}$

The above equation is accepted as the operational definition of$pH$.