For the reaction below, Kc = 1.10 × 10⁻⁸. What is the equilibrium concentration of OH⁻ if the reaction begins with 0.770 M HONH₂?

HONH₂ (aq) + H₂O (l) ⇌ HONH₃⁺ (aq) + OH⁻ (aq)

The equilibrium concentration of hydroxide ion is 0.00009 M.

The equation of the reaction is;

            HONH₂ (aq) + H₂O (l) ⇌ HONH₃⁺ (aq) + OH⁻ (aq)

I            0.770                                0                       0

C            -x                                     +x                     +x

E           0.770 - x                           x                        x

Note that water is present in large excess

Given that  Kc = 1.10 × 10⁻⁸

1.10 × 10⁻⁸ = [HONH₃⁺] [OH⁻]/[ HONH₂ ]

1.10 × 10⁻⁸ = [x] [x]/ 0.770 - x

1.10 × 10⁻⁸(0.770 - x) = x^2

8.27 × 10^-9 -  1.10 × 10^⁻8x = x^2

x^2 + 1.10 × 10^-8x - 8.27 × 10^-9  = 0

x=0.00009

Hence, the equilibrium concentration of hydroxide ion is 0.00009 M.

Learn more: link

 The Concentration of at equilibrium is 0.000092 M

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as

Initial conc.        0.770 M                           0  M        0 M

At eqm. conc.     (0.770-x) M                   (x) M   (x) M

The expression for equilibrium constant for this reaction will be,

Now put all the given values in this expression, we get :

By solving the term 'x', we get :

x = 0.000092 M

Concentration of at equilibrium= (x) M  = 0.000092 M