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7.46     The ionization constant of acetic acid is 1.74 × 10-5. Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of acetate ion in the solution and its pH.

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It is given,
The ionisation constant of acetic acid is 1.74\times 10^{-5} and concentration is 0.05 M

The ionisation of acetic acid is;

CH_{3}COOH\rightleftharpoons CH_3COO^-+H^+
K_a = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]}= \frac{[H^+]^2}{[CH_3COOH]} 
\therefore [H^+] = \sqrt{(1.74\times10^{-5})(5\times 10^{-2})} = 9.33\times 10^{-4}=[CH_3COO^-] 

So, the pH of the solution = -\log(9.33\times 10^{-4})

We know that,
\alpha = \sqrt{\frac{K_a}{C}}
\alpha = \sqrt{\frac{1.74\times10^{-5}}{0.05}} =1.86\times 10^{-2}

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