Question

In the laboratory a student measures the percent ionization of a 0.529 M solution of phenol (a weak acid) , C6H5OH, to be 1.33×10-3 %. Calculate value of Ka from this experimental data.

Answer 1

The value of dissociation constant from this experimental data fro phenol is
`9.3576* 10^(-11)`.

Step-by-step explanation:

`C_6H_5OH\rightleftharpoons C_6H_5O^-+H^+`

Initially c 0 0

At eq'm
`(c-c\alpha )`
`c\alpha`
`c\alpha`

Concentration of phenol , initially = c = 0.529 M

Degree of dissociation =
`\alpha =1.33* 10^(-3) \%=1.33* 10^(-5)`

An expression for dissociation constant is given as:

`K_a=([C_6H_5O^-][H^+])/([C_6H_5OH])`

`K_a=(c\alpha * c \alpha )/((c-c\alpha ))=(c(\alpha )^2)/((1-\alpha ))`

`K_a=(0.529 M* (1.33* 10^(-5))^2)/((1-1.33* 10^(-5)))`

`K_a=9.3576* 10^(-11)`

The value of dissociation constant from this experimental data fro phenol is
`9.3576* 10^(-11)`.