calculate the pH of a) 0.18 M boric acid
To calculate the pH of a boric acid solution, we need to consider the equilibrium reaction of boric acid (H3BO3) with water:
H3BO3 + H2O ⇌ H2BO3- + H3O+
The dissociation constant expression for boric acid is:
Ka = [H2BO3-][H3O+]/[H3BO3]
Since boric acid is a weak acid, it partially dissociates in water. However, the concentration of H2BO3- is approximately equal to the concentration of H3BO3 since boric acid is a monoprotic acid.
Therefore, we can simplify the expression as:
Ka ≈ [H3O+]^2 / [H3BO3]
Now, we can use the given concentration of boric acid, 0.18 M, to solve for the hydronium ion concentration ([H3O+]).
First, assume x represents the concentration of [H3O+] in M.
[H3BO3] ≈ x
[H2BO3-] ≈ x
[Ka] ≈ (x)^2 / [H3BO3] = (x)^2 / (0.18)
Since Ka for boric acid is 5.8 x 10^-10, we can set up the following equation:
5.8 x 10^-10 = (x)^2 / (0.18)
Rearranging the equation, we get:
(x)^2 = 5.8 x 10^-10 * 0.18
Solving for x:
(x)^2 = 1.044 x 10^-10
x ≈ 1.02 x 10^-5 M
Therefore, the concentration of hydronium ions [H3O+] in the boric acid solution is approximately 1.02 x 10^-5 M.
Finally, we can calculate the pH using the equation:
pH = -log[H3O+]
pH = -log(1.02 x 10^-5)
pH ≈ 4.99
Therefore, the pH of a 0.18 M boric acid solution is approximately 4.99.