Question

In the laboratory, a general chemistry student measured the pH of a 0.529 M aqueous solution of phenol (a weak acid), C6H5OH to be 5.153. Use the information she obtained to determine the Ka for this acid.

Answer 1

The dissociation constant of phenol from given information is
`9.34* 10^(-11)`.

Step-by-step explanation:

The measured pH of the solution = 5.153

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

Initially c

At eq'm c-x x x

The expression of dissociation constant is given as:

`K_a=([C_6H_5O^-][H^+])/([C_6H_5OOH])`

Concentration of phenoxide ions and hydrogen ions are equal to x.

`pH=-\log[x]`

`5.153=-\log[x]`

`x=7.03* 10^(-6) M`

`K_a=(x* x)/((c-x))=(x^2)/((c-x))=((7.03* 10^(-6) M)^2)/( 0.529 M-7.03* 10^(-6) M)`

`K_a=9.34* 10^(-11)`

The dissociation constant of phenol from given information is
`9.34* 10^(-11)`.