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.
NH3 + HOH ==> NH4^+ + OH^-
Kb = (NH4^+)(OH^-)/(NH3)
Set up an ICE chart and substitute into the Ka expression. Solve for Kb.
sum ting wong
The equation for the ionization of ammonia, NH3, in water can be written as:
NH3 + H2O ⇌ NH4+ + OH-
The equilibrium constant expression, K, for this reaction is given by the ratio of the concentration of the products (NH4+ and OH-) to the concentration of the reactants (NH3 and H2O), with each concentration raised to the power of its stoichiometric coefficient:
K = [NH4+][OH-] / [NH3][H2O]
However, since water is in large excess, its concentration can be considered approximately constant. Therefore, we can simplify the expression by removing the concentration of water:
K = [NH4+][OH-] / [NH3]
Given that the OH- concentration is 9.5x10^-4 M and the NH3 concentration is 0.050 M, we can substitute these values into the equilibrium constant expression to solve for K:
K = (9.5x10^-4)(0.050) / (0.050)
K = 9.5x10^-4
So, the equilibrium constant, K, for the ionization of ammonia is 9.5x10^-4.
To write the equilibrium equation for the ionization of ammonia (NH3) and determine the equilibrium constant expression, we need to understand the reaction that takes place when NH3 dissolves in water.
When NH3 dissolves in water, it acts as a weak base and undergoes a reaction to produce hydroxide ions (OH-) and ammonium ions (NH4+). This can be represented by the following equation:
NH3 + H2O ⇌ NH4+ + OH-
The equilibrium constant expression, K, is a measure of the extent to which the reaction proceeds at equilibrium. It can be written using the concentrations of the species involved in the equation. In this case, it can be expressed as:
K = [NH4+][OH-] / [NH3][H2O]
Now, we are given that the OH- concentration of 0.050 M NH3 is 9.5x10^-4. We can use this information to determine the value of K.
Substituting the given value into the equation:
9.5x10^-4 = [NH4+][OH-] / (0.050)[H2O]
Since we are dealing with a dilute solution, we can assume that the concentration of water (H2O) remains constant and can be treated as a constant value. Therefore, we can rewrite the equation as:
9.5x10^-4 = [NH4+][OH-] / K'
Here, K' represents the new equilibrium constant, combining the constant concentration of water.
To solve for K', we need to rearrange the equation and isolate K':
K' = [NH4+][OH-] / 9.5x10^-4
Since the concentration of water (H2O) does not affect the value of K', it is essentially constant. Therefore, we can consider K' to be equal to K. Hence, the value of the equilibrium constant, K, is:
K = [NH4+][OH-] / 9.5x10^-4