Question

Boric acid frequently is used as an eyewash to treat eye infections. the ph of a 0.050m solution of borin acid is 5.28. what is the value of the boric acid ionization constant, ka

Answer 1

Based on  Henderson - Hasselbach equation, we have
pH = pka + log $\frac{[Acid]}{\text{[conjugate base]}}$
Given: pH = 5.28 and [Acid] = 0.05 m,
∴ pKa = 5.28 - log (0.05) = 6.58
∴ Ka = 2.624 X $10^{-7}$

Thus, ionization constant of boric acid is 2.624 X $10^{-7}$