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The ka of hypochlorous acid (hclo) is 3.0 x 10-8 at . what is the % ionization of hypochlorous acid in a 0.015 m aqueous solution of hclo at ?

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Dejanras

Answer is: the % ionization of hypochlorous acid is 0.14.

Balanced chemical reaction (dissociation) of an aqueous solution of hypochlorous acid:

HClO(aq) ⇄ H⁺(aq) + ClO⁻(aq).

Ka = [H⁺] · [ClO⁻] / [HClO].

[H⁺] is equilibrium concentration of hydrogen cations or protons.

[ClO⁻] is equilibrium concentration of hypochlorite anions.

[HClO] is equilibrium concentration  of hypochlorous acid.

Ka is the acid dissociation constant.

Ka(HClO) = 3.0·10⁻⁸.

c(HClO) = 0.015 M.

Ka(HClO) = α² · c(HClO).

α = √(3.0·10⁻⁸ ÷ 0.015).

α = 0.0014 · 100% = 0.14%.

pstnonsonjoku

The percent ionization of the solution is 0.14%.

  • The equation of the reaction is;

        HClO(aq) ⇄      H^+(aq)   +    ClO^-(aq)

I        0.015                 0                    0

C       -x                       +x                   +x

E   0.015 - x                  x                      x

  • Now the Ka of the solution is obtained from;

Ka = [ H^+] [ClO^-]/[HClO]

3.0 x 10^-8= x^2/ 0.015 - x  

3.0 x 10^-8(0.015 - x) = x^2

4.5 x 10^-10 - 3.0 x 10^-8x = x^2

x^2 + 3.0 x 10^-8x - 4.5 x 10^-10 = 0

x = 0.000021 M

  • Percent ionization =  0.000021 M/0.015 M × 100/1

Percent ionization = 0.14%

Hence, the percent ionization of the solution is 0.14%.

Learn more about percent ionization: https://brainly.com/question/9173942

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