Question

(a) How does the hydration of ions in aqueous solution affect their conductivity? Illustrate your answer with examples and diagrams. [10 marks]

(b) The conductivity of a 0.01 mol dm–3 solution of a monobasic organic acid in water is 5.07 × 10–2 S m–1. If the molar conductance at infinite dilution (Λ°) of aqueous sodium chloride, sodium formate and hydrochloric acid are 1.264 × 10–2, 1.046 × 10–2 and 4.261 × 10–2 respectively at 25°C determine the acid dissociation constant and the pKa for the acid.

Answer 1

answer (a):- Hydration of ion as a result of higher charge density decreases the mobility of ion because a hydrated ion has to drag along a shell of water as its move through the solution. As, Conductivity is proportional to mobility of ion & , the hydration of ion decreases its mobility. Hence, we can conclude that, Conductivity of ion in aqueous solution decreases with increase in hydration of ion.

For e.g. :- The ionic mobilities of Lithium and sodium ions become low due to hydration of ions as a result of higher charge density around the ion because of their smaller radii. As a result, conductivity of Li+ and Na+ ions decreases.  

Hydration of Na+ ion ( see fig.)

Exceptional case of H+ ion:-

However, conductivity of H+ ion, in spite of its small size and high charge density ( and it is also heavily hydrated. But its conductivity is very high rather than becoming low. This unexpected result can be explained by " Grotthus Type Mechanism" in which a proton moves rapidly from H3O+ to a H- bonded water molecule and is transferred further along a series of hydrogen bonded water m^{​​​​​​}olecules by a rearrangement of hydrogen bonds".

Answer (b):-   Concentration (c) = 0.01 mol dm^-3 = 10 mol m ^-3

Molar conductivity () = Conductivity / concentration = 5.07 × 10^-2 S m^-1/ 10 mol m^-3

  = 5.07 × 10^-3 S m² mol^-1

Molar conductivity at infinite dilution (₀) = _0HCOONa + _{0 HCl} - _{0 NaCl}

  _{0 HCOOH} = (1.046 + 4.261- 1.264) × 10^-2

_{0 ( HCOOH)} = 4.043 × 10^-2 S m² mol^-1

Degree of dissociation (a) =  / ₀

a = 5.07 × 10^-3/ 4.04 × 10^-2

a = 0.1254

Dissociation constant (ka) of formic acid = C a​​​​​​​²/ ( 1-a)

= 0.01 × (0.1254)²/0.99

ka = 1.589 × 10^-4 mol/ dm^-3

pka = - log₁₀ ka

pka = - log₁₀ 1.589 × 10^-4

   pka of formic acid = 3.78 (approx)