The Kb of ammonia is 1.8 x 10–5. What is the percent ionization of a solution of ammonia, 0.50 mol L–1 NH3, that contains 0.020 mol L–1 of ammonium nitrate, NH4NO3?
when i use the ICE table i don't end up with an unknown so im not sure what to do. Am i to try and make the equation with NH3 + NH4NO3 on the same side?
----NH3 + HOH => NH4OH(fm NH4NO3)
-----NH4OH <=> NH4^+ + OH^-
I:-->0.50M --------0.20M-----0
C:---(-x)---------- (+x)----(+x)
E:--(0.5-x)--------(0.2+x)---(x)
---->(~0.5)-------->(~0.2)---(x)
Kb=[NH4^+][OH^-]/[NH4OH]
1.8X10^-5= (0.2)(X)/(0.5)
Solve for 'x'
=>x=(0.50)(1.8x10^-5)/(0.2)
=>x=4.5x10^-5M (This is the fraction of (OH^-)ion delivered into solution on ionization of NH4OH)
%Ionization=[(4.5x10^-5)/(0.50)]x100%
= 0.009% Ionized
To calculate the percent ionization of a solution of ammonia, we need to consider the reaction between ammonia (NH3) and water (H2O) to form ammonium ions (NH4+) and hydroxide ions (OH-).
The chemical equation for this reaction can be represented as:
NH3 + H2O ⇌ NH4+ + OH-
First, let's find the initial concentration of ammonia (NH3). We are given that the solution contains 0.50 mol L–1 of NH3. Since we assume complete dissociation of NH4NO3, the ammonia concentration in the solution would also be 0.50 mol L–1.
Next, let's consider the reaction between NH3 and water. The Kb value given for ammonia is 1.8 x 10–5, which represents the equilibrium constant for the reaction:
Kb = ([NH4+][OH-]) / [NH3]
Since the concentration of OH- is determined by the concentration of NH4+ (from NH4NO3), and assuming that the reaction reaches equilibrium, we can substitute the concentrations into the equation as follows:
1.8 x 10–5 = ([NH4+] * [OH-]) / [NH3]
Now, we need to determine the change in concentration using the ICE (Initial, Change, Equilibrium) table.
Initial concentration:
[NH3] = 0.50 mol L–1
[NH4+] = 0.020 mol L–1 (from NH4NO3)
Change in concentration:
Let's assume x mol L–1 is the concentration of NH4+ and OH- at equilibrium. The change in concentration for NH3 would be (-x) mol L–1.
Equilibrium concentration:
[NH3] = (0.50 - x) mol L–1
[NH4+] = (0.020 + x) mol L–1
[OH-] = x mol L–1
Substituting the equilibrium concentrations into the Kb equation:
1.8 x 10–5 = [(0.020 + x) * (x)] / (0.50 - x)
Now, we need to solve this equation for x to find the concentration of NH4+ (and OH-) at equilibrium. We can rearrange the equation and solve for x using algebraic methods.
After finding the value of x, you can calculate the percent ionization using the equation:
% Ionization = (concentration of NH4+ at equilibrium / initial concentration of NH3) * 100
Remember to substitute the value of x into the equation when calculating the percent ionization.
Note: Calculating the exact value of x may require solving a quadratic equation, as the equation is not linear. However, if the value of x is small compared to the initial concentration of NH3 (0.50 mol L–1), you can make the approximation (0.50 - x) ≈ 0.50 and ignore the x term in the denominator when calculating the percent ionization.