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Despite its formula, boric acid, H3BO3, is a monoprotic acid. A 25.00 mL sample of this...

Despite its formula, boric acid, H3BO3, is a monoprotic acid. A 25.00 mL sample of this acid required 15.66 mL of 0.1724 M NaOH to reach the equivalence point where the pH=11.13. Determine the original concentration of boric acid and the equilibrium constant, Ka.

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Answer #1

mmoles of NaOH = 15.66 x 0.1724 = 2.70

H3BO3 +   NaOH --------------------> NaH2BO3 + H2O

at equivalence point :

mmoles of acid = mmoles of base

M x 25 = 2.70

M = 0.10799

concentration of boric acid = 0.1080 M

H3BO3 +   NaOH --------------------> NaH2BO3 + H2O

at equivalence point salt only remains :

[salt] = 2.70 / (25 + 15.66) = 0.0664 M

it is salt of strong base and weak acid : so

pH = 7 + 1/2 (pKa + log C)

11.13 = 7 + 1/2 (pKa + log 0.0664)

pKa = 9.44

equilibrium constant, Ka = 3.63 x 10^-10

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