Calculate the potential at equivalent point:

25 cm³ of a solution of HCl (0.1M) is being titrated potentiometrically against a std. (0.1 M) sol. of NaOH using a hydrogen electrode as the indicator electrode and saturated calomel electrode (S.C.E.) as the reference electrode. (a)  What would be the emf of the cell initially and after the addition of 20, 24.9, 25.10, 30.00 cm³ of NaOH solution?  (b) In addition calculate the potential at equivalent point and after the addition of 0.05 cm³ of NaOH.

Answer:

4. The galvanic cell can be represented as follows :

Pt, H₂ (1 atm.), H⁺ (c, unknown) ||  KCl sat. soln.,Hg = Cl₂(s), Hg.  The EMF of the cell is given by,

E_cell = E_SCE   - E_hydrogen

= 0.2422 - 0.0591log H⁺

= 0.2422 + 0.0591pH at 25°C

Initial pH of the titration solution viz., 0.1 M HCl = - log [H⁺]

= - log (0.1) = 0.1

so that E = 0.3013 V

Since the product of volume of the solution in cm³ and the concentration in mol dm-3  of a solute gives the amount of the solute in millimoles (mmol), hence, Amount of HCl initially present in the titration solution = 25 × 0.1 = 2.5 mmol

The amount of NaOH in 20 cm³ of 0.1 M solution added during titration = 20 × 0.1 = 2.0 mmol.

Amount of HCl left in the titration solution on adding 20 cm³ of NaOH = 2.5 - 2.0 = 0.5 mmol.

Total volume of titration solution = 25 + 20 = 45 cm³

Conc. of HCl or of H+ ions in the solution

No. mmol HCl remain after addition of NaO ₄/ Total volume of the solution

0.5/45 mol dm ^-3

pH of the titration solution = - log [H⁺]

= - log (0.5/45) = 1.95

The corresponding value of E is 0.2422 + 0.0591 × 1.95 = 0.3574 V. Proceeding as above the pH values of the titrant on the addition of 24.90, 25.10 and 30.00 cm³  of NaOH solution come out to 3.70, 10.30 and 10.96, respectively and hence the corresponding values of E are 0.4609, 0.8510 and 0.8900 V respectively.