Calculate the hydroxide ion conc. and the % ionization @20oC in a 1.25 M NH3 solution. The Kb for NH3 = 1.70x10-5 @20oC.
To solve this problem, we'll use the concept of the ionization constant (Kb) for NH3. The Kb is defined as the equilibrium constant for the reaction of NH3 with water to form the hydroxide ion (OH-) and its conjugate acid (NH4+).
The balanced equation for this reaction is:
NH3 + H2O ⇌ NH4+ + OH-
To calculate the hydroxide ion concentration, we'll assume that the amount of NH4+ produced is negligible compared to the amount of NH3 initially present. This means that we can assume that x moles of NH3 react with water to produce x moles of OH-. Therefore, the hydroxide ion concentration is equal to x.
Now, let's set up an equilibrium expression for the reaction:
Kb = [NH4+][OH-] / [NH3]
Using the initial concentration of NH3 (1.25 M), we can assume that the initial concentration of NH4+ and OH- is zero. Thus, the equilibrium expression simplifies to:
Kb = x^2 / (1.25 - x)
The percent ionization (% ionization) is given by:
% ionization = (concentration of OH-) / (initial concentration of NH3) x 100
To solve for x and the hydroxide ion concentration, we'll use the quadratic equation:
x^2 / (1.25 - x) = Kb
1.70x10^-5 = x^2 / (1.25 - x)
Now, we'll solve this equation to find the value of x and then calculate the hydroxide ion concentration and the percent ionization.
Let's solve this equation: