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

Answer:

The dissociation constant of phenol from given information is $9.34\times 10^{-11}$.

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:

$K_a=\frac{[C_6H_5O^-][H^+]}{[C_6H_5OOH]}$

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

$pH=-\log[x]$

$5.153=-\log[x]$

$x=7.03\times 10^{-6} M$

$K_a=\frac{x\times x}{(c-x)}=\frac{x^2}{(c-x)}=\frac{(7.03\times 10^{-6} M)^2}{ 0.529 M-7.03\times 10^{-6} M}$

$K_a=9.34\times 10^{-11}$

The dissociation constant of phenol from given information is $9.34\times 10^{-11}$.