Question

determination of ph of acid solution like 0.1m hcl, 0.05m hcl, 0.025m hcl, and 0.0125m hcl by using quinhydrone and calomel electrode.

Answer 1

Answer on Question #52784 - Chemistry - Inorganic Chemistry

For calomel electrode:

$$pH = \frac{E - E_{kal}}{2.303 (RT / F)}$$

where E is the measured electromotoric force (in V), Ekal is the potential of the calomel electrode at given temperature (see Tab.2). R=8.314 Jkmol⁻¹ is the gas constant, F =96485 Cmol⁻¹ is the Faraday constant, T is the temperature.

For quinhydrone electrode:

$$\begin{array}{l} \mathbf{E} = \mathbf{E}_k + 000198 \text{ T. } \log [\mathrm{H}^+] \\ = \mathbf{E}_k - 000198 \text{ T. } p_{\mathrm{H}} \\ \text{or } \mathbf{E} = \mathbf{E}_k - 0577 p_{\mathrm{H}} \text{ at } 18^\circ \text{C.} \end{array}$$

So $pH = (E_k - E) / 0.0577$

where E is the measured electromotoric force (in V), $E_k$ is the potential of quinhydrone electrode and T is the temperature.

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