Question
In: Chemistry
What is the pH of a 0.11 M solution of C6H5OH (Ka = 1.3 x 10^-10)?
What is the pH of a 0.11 M solution of C6H5OH (Ka = 1.3 x 10^-10)?
Solutions
Expert Solution
C6H5OH ----------------------------> C6H5O- + H+
0.11 0 0 ----------------> initial
0.11-x x x -------------------> equilibrium
Ka = [C6H5O-][H+]/[C6H5OH]
Ka = x^2 / 0.11-x
1.3 x 10^-10 = x^2 / 0.11-x
x^2 + 1.3 x 10^-10 x -1.43 x 10^-11 = 0
by solving this
x = 3.78 x 10^-6
[H+] = x = 3.78 x 10^-6 M
pH = -log [H+] = -log (3.78 x 10^-6)
pH = 5.42 -----------------------> answer
queen_honey_blossom answered 2 years agoRelated Solutions
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