The acid-dissociation constant at 25.0 °c for hypochlorous acid ( ) is 3.0 . at equilibrium, the molarity of in a 0.050 m solution of is11.48 11.78 1.7 10-12 ⋅ 2.22 6.0 10-3 ⋅ aqueous solutions of hno3 contain equal concentrations of h+ (aq) and oh- (aq) hno3 does not dissociate at all when it is dissolved in water hno3 produces a gaseous product when it is neutralized hno3 cannot be neutralized by a weak base hno3 dissociates completely to h+ (aq) and no3 - (aq) when it dissolves in water hclo ⋅ 10 −8 h3o

The molarity of H3O+ = 3.8*10^-5 M

The molarity of ClO- = 3.87*10^-5 M

The molarity of HClO = 0.05 - 3.87*10^-5M = 0.0499613 M

Explanation:

Step 1: The balanced equation

HClO (aq) + H2O(l) ⇄ CIO-(aq) + H3O+(aq)

Step 2: Given data

Temperature = 25.0 °C

Acid-dissociation constant of HClO= 3.0 *10^-8

Molarity of HClO = 0.050M

Step 3: Calculate molarity

Initial Molarity of HClO = 0.050 M

Since the mol ratio is 1:1; there will react x of HClO and there will produced x M of ClO- and H30+

At the equilibirium we have (0.050 - x)M of HClO and X of ClO- and H3O+

Ka = [ClO-][H3O+] /[HClO] = 3*10^-8

3*10^-8 = x*x / (0.05 -x)

x² = (0.05-x)*3*10^-8

x = √(3*10^-8*(0.05))

x = 3.87*10^-5

The molarity of H3O+ = 3.8*10^-5 M

The molarity of ClO- = 3.87*10^-5 M

The molarity of HClO = 0.05 - 3.87*10^-5M = 0.0499613 M

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