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Calculate the pH of 0.1M CH_3COOH solutoion. Dissociation constant of acetic acid is 1.8 times 10^-5. |
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Answer» Solution :`pH=-log[H^+]` For weak ACIDS, `[H^+]=sqrt(K_s times C)` `=sqrt(1.8 times 10^-5 times 0.1)` `=1.34 times 10^-3 M pH=-log(1.34 times 10^-3)` `=3- log1.34` `=3-0.1271` `=2.8729 APPROX 2.87` |
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