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

Given:

Concentration of boric acid, H3BO3 = 0.050 M

pH = 5.28

To determine:

The Ka of H3BO3

Explanation

The dissociation of boric acid is represented as

H3BO3 ↔ H2BO3- + H+

Ka = [H2BO3-][H+]/[H3BO3]

pH = -log [H+]

[H+] = 10^-pH = 10^-5.28 = 5.25*10⁻⁶ M

At equilibrium we have:

[H2BO3-] = [H+] = 5.25*10⁻⁶M

[H3BO3]= 0.050 - 5.25*10⁻⁶≅ 0.050 M

Ka = [5.25*10⁻⁶]²/[0.050] = 5.51 * 10⁻⁴

Ans: Ka of H3BO3 = 5.51 * 10⁻⁴

periodic trends is an atom that gets larger as the number of electronic shells increase, the radius of atoms increases as you go down a certain group in the periodic table of elements. in general, the size of an atom will decrease as you move from left to right of a certain period.