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A 0.02M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine.

7.62     A 0.02M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine.

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M manish

Given,
  p^H  =  3.44

We know that

p^H=-\log [H^+]
By taking antilog on both sides we get,
[H^+] = antilog (- 3.44)

\therefore [H^+] = 3.63\times 10^{-4}

pyridinium hydrochloride completely ionised.

Then K_h =  (conc. of products)/ (conc, of reactants)
                 =\frac{ ( 3.63\times10^{-2} )^2 }{0.02}        (? Concentration is 0.02M)

\Rightarrow K_h = 6.58 \times 10^-6

Now,
K_h = K_w / K_a

\Rightarrow Ka = K_w / K_h


   = 10^{-14} / 6.58 \times 10^{-6} = 1.51 \times 10^{-9}(approx)

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