Question

# (a) Calculate the percent ionization of 0.00840 M hypochlorous acid (Ka = 3e-08). % ionization =...

(a) Calculate the percent ionization of 0.00840 M hypochlorous acid (Ka = 3e-08). % ionization = % (b) Calculate the percent ionization of 0.00840 M hypochlorous acid in a solution containing 0.0190 M sodium hypochlorite. % ionization = %

a) HClO + H2O <=> ClO- + H3O+

Ka= [ClO-][H3O+]/[HClO]

If [ClO-] = [H3O+] = x, then [HClO] = 0.00840 M - x

3*10-8 = x2/(0.00840 M - x)

x2 + 3*10-8 x - 2.52 x 10-10 = 0

Solving the quadratic equation we get x = [H3O+] = 1.586*10-5 M

The % of ionization is equal to ratio of ionized acid and the total acid concentration:

(1.586*10-5 M/0.00840)*100% = 0.1888 % = 0.19%

b) If a solution contains 0.0190 M sodium hypochlorite, [ClO-] = 0.0190 M

Ka= [ClO-][H3O+]/[HClO] => [H3O+] = Ka[HClO]/[ClO-]

We can approximate that [HClO] = 0.00840 M, bescause the ionization will be very suppressed in presence of sodium hypochlorite, therefore

[H3O+] = 3*10-8 * 0.00840 M/0.0190 M = 1.326*10-8M

The % of ionization will be:

(1.326*10-8M/0.00840)*100% = 1.58 x 10-4 %

#### Earn Coins

Coins can be redeemed for fabulous gifts.