Anonymous
Anonymous asked in Science & MathematicsChemistry · 3 years ago

# The following equation shows the ionization of boric acid, H3BO3: H3BO3(aq) + H2O ⇌ H3O+(aq) + H2BO3– (aq); Ka = 5.4 x 10^–10?

(a) Calculate the pH of 0.100 M boric acid solution.

(b) Write an equation for the ionization of borate ion, H2BO3– , in aqueous solution (that is, its reaction with water) and calculate its Kb value, and determine the pH of solution of 0.10 M sodium borate, NaH2BO3(aq).

(c) A borate buffer solution contains 0.15 M boric acid and 0.050 M sodium borate. What is the pH of this solution?

(d) What is the molar ratio of borate ion, H2BO3–, to boric acid, [H2BO3–]/[H3BO3], in a buffer solution with pH = 8.70? If the concentration of boric acid in the buffer is 0.200 M, what should be the concentration of the borate ion?

(e) If the solution in (d) is reacted with 0.020 M NaOH, how would this change the ratio:

[H2BO3–]/[H3BO3], and what would be the pH of the resulting solution?

Relevance
• (a) Ka = [H3O+][H2BO3-]/[H3BO3] = 5.4X10^-10

Assume that [H3O+]=[H2BO3-] = x. Then [H3BO3] = 0.100 - x. Because Ka is a small number, x will be much smaller than 0.100, and so can be ignored. Then,

Ka = x^2 / 0.10 = 5.4X10^-10

x = 7.3X10^-6

pH = 5.13

(b) H2BO3- + H2O <--> H3BO3 + OH-

Kb = [H3BO3][OH-]/[H2BO3-]

For a weak base, the relationship between Ka of the weak acid and Kb of the conjugate base is given by:

Ka X Kb = Kw = 1.00X10^-14. So for H2BO3-,

Kb = 1X10^-14 / 5.4X10^-10 = 1.85X10^-5

(c) Either use the expression for Ka of boric acid given above, or use the Henderson-Hasselbalch equation pH = pKa + log [H2PO3-][H3BO3]. Using the equation for Ka:

Ka = 5.4X10^-10 = [H3O+](0.05)/(0.15)

[H3O+] = 1.62X10^-9

pH = 8.79

(d) Use this same equation and solve for that ratio, use it to express [H2BO3-] in terms of [H3BO3]. You also know that [H3BO3] + [H2BO3-] = 0.200. You should be able to solve for the actual concentrations.

(e) Adding a strong base to this buffer solution will convert H3BO3 into H2BO3- quantitatively. This increases the ratio [H2BO3–]/[H3BO3], and will raise the pH of the solution. Without knowing a volume of the buffer solution, you cannot calculate the actual pH of the resulting solution.