1.

A0.1M sodium acetate solution was prepared. The K_(h)=5.6xx10^(-10)

Answer»

The degree of hydrolysis is `7.48xx10^(-5)`
The `[OH^(-)]` concentration is `7.48xx10^(-3)M`
The `[OH^(-)]` concentration is `7.48xx10^(-6)M`
The PH is approximately `8.88`

Solution :`underset(0.1(1-H))(CH_(3)COO^(-))+H_(2)OhArrunderset(0.1h)(CH_(3)COOH)+underset(0.1h)(OH^(+))`
`K_(h)=((0.1h)(0.1h))/(0.1(1-h))=0.1h^(2)`
`implies5.6xx10^(-10)=0.1h^(2)impliesh=7.48xx10^(-5)`
`[thereforeltltltlt1]`
`[OH^(-)]=ch=7.48xx10^(-5)xx10^(-1)=7.48xx10^(-6)implies`
`[H^(+)]=(K_(w))/([OH^(-)])=(10^(-14))/(7.48xx10^(-6))=1.33xx10^(-9)`implies pH=8.8` approx
Hence choices (a), (c) and (d) are correct while (b) is wrong.


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