Boric acid is a weak acid with an equilibruim constant(K)of 10-9.3. What is the pH balance of a 0.1 molar solution of boric acid?
..........H3BO3 ==> H^+ + H2BO3
I..........0.1M......0.......0
C...........-x........x.......x
E.........0.1-x........x......x
Ka = (H^+)H2BO3^-)/(H3BO3)
Subsitute from the ICE chart and solve for x = (H^+) and convert to pH.
To find the pH of a solution of boric acid, we need to first calculate the concentration of H+ ions in the solution.
Boric acid, H3BO3, can react with water in equilibrium to produce H+ ions and the borate ion, B(OH)4-. The equation for this reaction is:
H3BO3 + H2O ⇌ H+ + B(OH)4-
From the equilibrium constant, K, we can determine the concentration of H+ ions using the equation:
K = [H+][B(OH)4-] / [H3BO3][H2O]
Since the concentration of water remains relatively constant, it can be excluded from this equation. Therefore, the equation can be simplified to:
K ≈ [H+]/[H3BO3]
Given that the equilibrium constant, K, is 10^(-9.3), we can set up the equation as follows:
10^(-9.3) = [H+]/[H3BO3]
Since we have a 0.1 molar solution of boric acid, the concentration of H3BO3 is 0.1 M. Substituting this value into the equation, we have:
10^(-9.3) = [H+]/0.1
To find the concentration of H+, we can rearrange the equation:
[H+] = 10^(-9.3) × 0.1
Calculating this on a calculator, we find that [H+] is approximately 4.47 × 10^(-10) M.
The pH of a solution is defined as the negative logarithm (base 10) of the H+ concentration. Thus, to find the pH of the solution, we can use the formula:
pH = -log[H+]
Substituting the concentration of H+ into the equation, we have:
pH = -log(4.47 × 10^(-10))
Calculating this on a calculator, we find that the pH of the 0.1 molar solution of boric acid is approximately 9.35.