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7.47     It has been found that the pH of a 0.01M solution of an organic acid is 4.15. Calculate the concentration of the anion, the ionization constant of the acid and its pKa.

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We have,
pH of organic acid is 4.15 and its concentration is 0.01M

Suppose the organic acid be HA. The dissociation of organic acid can be written as;

HA \rightleftharpoons H^+ + A^-

pH = 4.15
\\-\log[H^+] = 4.15\\ {[H^+]} = 7.08\times 10^{-5}(By taking antilog of -4.15)

Now,
K_a = [H^+] [A^-] / [HA]

[H^+] = [A^-] = 7.08\times10^{-5}

[HA]  = 0.01

Then,

\\K_a = (7.08\times10^{-5})^2 / 0.01\\ K_a = 5.01\times 10^{-7}

Thus 
pK_a=-\log ( 5.01\times 10^{-7}) = 6.30

Posted by

manish

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