Question

nicotinic acid (Ka =1.4 * 10 raise to power -5) is represented by the formula HNiC. calculate its percentage dissociation in a solution which contains 0.10 mole of nicotinic acid per 2 litre of solution ?

Answer 1

             H NiC     <=========> H+  +  NiC-
        (0.05 - x) M. ... .. . .. .. .. .  x M . . .  x M

Initial concentration of Nicotinic acid = 0.1/2 =0.05 Molar.
Let x moles/Litre of H NiC dissociate into x Molar of H⁺ and x molar of NiC⁻ ions.

        Ka = [ H+] * [ Ni C-] ÷ [ H Ni C]

=>   1.4 * 10⁻⁵ =  x * x / (0.05 - x).
=>   x² + 1.4 * 10⁻⁵ x - 7.0 * 10⁻⁷ = 0

We can solve this quadratic. Or  Follow a simplification method.  Actually Nicotinic acid is very weak.  It dissociates less than 5%.  So x is very small compared to 0.05 moles. Or dissociation constant α << 1.

So Let us take   0.05 - x ≈ 0.05, and find the answer approximately.

=>  x = √(0.05 * 1.4 * 10⁻⁵) Molar
      x =  0.0008366 Molar

% dissociation α = (x / 0.05) ×100 = 1.673 %

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     We can also take the initial concentration of Nicotinic acid as C (= 0.05) molar.  Then if α is the dissociation constant, α * C molar will be the concentrations of H⁺ & NiC⁻ ions.

    Further at the equilibrium,  C (1 - α) will be the concentration of the undissociated acid.  As usually α << 1, We could take it as C itself.

So   Kc = Ka = [ H⁺ ] [ Ni C⁻ ]  / [  H NiC ]
       1.5 * 10⁻⁵ = C α * C α / ( C (1-α) ) 
                        ≈ C α² 
=>       α = 1.673 %
 

Answer 2

Answer:

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