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

Asked by Pragya Singh | 1 year ago |  122

#### 1 Answer

##### Solution :-

$$CH_3COOH ⇒ CH_3COO^{–} + H^{+}K_{a}$$

$$=\dfrac{[CH_{3}COO^{-}][H^{+}]}{[CH_{3}COOH]}$$

$$=\dfrac{[H^{+}]^{2}}{[CH_{3}COOH]}\\ \\ \Rightarrow [H^{+}]=\sqrt{K_{a}[CH_{3}COOH]}$$

$$=\sqrt{(1.74\times 10^{-5})(5\times 10^{-2})}$$

$$=9.33\times 10^{-4}M\\ \\$$

$$[CH_{3}COO^{-}] =[H^{+}]$$

$$=9.33\times 10^{-4}M\\ \\$$

$$pH=-log(9.33\times 1.0^{-4})$$

$$=4-0.9699=4-0.97=3.03$$

Answered by Abhisek | 1 year ago

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