Calculate the pH of a solution (to 2 decimal places) which is 0.106 M in phenol, Ka = 1.0 x 10-10.
Let's let HPh equal phenol with H representing the hydrogen that ionizes and Ph representing the remainder of phenol.
HPh ==> H^+ + Ph^-
Write the Ka expression.
Ka = 1 x 10^-10 = [(H^+)(Ph^-)]/(HPh)
Before the ionization (HPh) = 0.106 M.
Before the ionization (H^+)=(Ph^-)=0
After ionization, (H^+)= x
(Ph^-) = x
(HPh) = 0.106 - x
Plug these variables into the Ka expression and solve for x = (H^+).
Then convert (H^+) to pH with pH = - log(H^+). Post your work if you get stuck.
pH= -log(1.0x10^-10) + log x/0.106-x.
how??
To calculate the pH of a solution, we need to determine the concentration of H+ ions in the solution.
In this case, we are given the concentration of phenol (C6H5OH) in the solution, which is 0.106 M. Phenol is a weak acid, and we are given its Ka value, which is 1.0 x 10^-10.
Phenol can undergo the following equilibrium reaction with water:
C6H5OH (aq) + H2O (l) ⇌ C6H5O- (aq) + H3O+ (aq)
According to the Ka expression, Ka = [C6H5O-][H3O+]/[C6H5OH].
To solve for [H3O+], we can assume that the [C6H5O-] concentration is negligible compared to the [C6H5OH] concentration since phenol is a weak acid. Therefore, we can simplify the expression as:
Ka = [H3O+]^2 / [C6H5OH]
Solving for [H3O+], we get:
[H3O+]^2 = Ka * [C6H5OH] = (1.0 x 10^-10)(0.106)
[H3O+] = √(1.0 x 10^-10)(0.106)
[H3O+] ≈ 3.26 x 10^-6 M
Since pH is defined as -log[H3O+], we can calculate the pH as:
pH = -log(3.26 x 10^-6)
pH ≈ 5.49
Therefore, the pH of the solution is approximately 5.49.