a) The OH- concentration of 0.050 M ammonia, NH3, is 9.5x10^-4. Write the equilibrium equation for its ionization and the equilibrium constant expression; solve for the value of K.
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The equilibrium equation for the ionization of ammonia, NH3, can be written as:
NH3 + H2O ⇌ NH4+ + OH-
The equilibrium constant expression, K, for this reaction can be written as:
K = [NH4+][OH-] / [NH3][H2O]
Given that the OH- concentration of 0.050 M ammonia is 9.5x10^-4, we can substitute the values into the equilibrium constant expression to solve for K:
K = (9.5x10^-4) / (0.050)(1)
K = 1.9x10^-2
Therefore, the value of K for the ionization of ammonia is 1.9x10^-2.
To write the equilibrium equation for the ionization of ammonia, NH3, you need to understand that ammonia is a weak base that reacts with water to produce hydroxide ions, OH-. The equilibrium equation can be written as follows:
NH3 + H2O ⇌ NH4+ + OH-
Now, let's write the equilibrium constant expression, K, using the concentrations of the species involved:
K = [NH4+][OH-] / [NH3][H2O]
Given that the concentration of ammonia is 0.050 M and the concentration of hydroxide ions is 9.5 x 10^-4, and since water is in excess and its concentration remains nearly constant, you can assume that [H2O] does not change significantly. Thus, you can consider it to be constant and cancel it from the equilibrium constant expression. The equation simplifies to:
K = [NH4+][OH-] / [NH3]
To solve for the value of K, you need the concentration of NH4+, which can be determined by using the fact that NH4+ is in equilibrium with NH3 and H3O+ in an acid-base reaction. Since the concentration of OH- is given, you can use the Kw, the ion product of water, to calculate the concentration of H3O+. Kw is equal to 1.0 x 10^-14 at 25°C. Therefore:
Kw = [H3O+][OH-]
Given that [OH-] is 9.5 x 10^-4, you can rearrange the equation to solve for [H3O+].
[H3O+] = Kw / [OH-]
[H3O+] = (1.0 x 10^-14) / (9.5 x 10^-4)
[H3O+] = 1.05 x 10^-11
Since the concentration of NH3 is 0.050 M and the concentration of H3O+ is 1.05 x 10^-11, you can substitute these values into the equilibrium constant expression to solve for K:
K = [NH4+][OH-] / [NH3]
K = (x)(9.5 x 10^-4) / (0.050)
K = (x)(1.9 x 10^-2)
Substituting the value of [H3O+] for x and simplifying, you can solve for K:
K = (1.05 x 10^-11)(1.9 x 10^-2)
K ≈ 2.0 x 10^-12
Therefore, the value of K for the ionization of ammonia is approximately 2.0 x 10^-12.