1.

Calculate the pH of 0.1M CH_3COOH solutoion. Dissociation constant of acetic acid is 1.8 times 10^-5.

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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