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In this exam question comes to the electrical conductivity of a weak electrolyte . Formic
acid is a weak acid with a K( S)- value of 1.7E -4 mol / L . In water it dissociated
incompletey into protons and formate ions. 1 mol of formic acid is diluted with water
to a volume of 1 liter. Determine the degree of dissociation (alpha) of the acid and its
electric conductivity . According to Ostwald's dilution law (alpha) ² / (1- alpha) times
c ° = K (S) Substituting the numerical values , and solve the quadratic equation for alpha,
we obtain a degree of dissociation of 0.013 , ie 1.3 %. By adding the limiting conductivities
of protons and formate ions , we obtain the limiting equivalent conductivity lambda ( infinity)
of formic acid to 404.4 Scm ² / mol. Because of the its incomplete dissociation, 1M formic
acid has an equivalent conductivity corresponding to only 1.3 % of this value , ie ,:24 Scm
² / mol. By multiplying this value (which is also the molar equivalent conductivity
) by the concentration ( in mol / cm ³), we end up with a specific conductivity of
5,24 mS / cm .