Ammonia, NH3, is a weak base with a Kb value of 3.95×10−5. What is the percent ionization of 0.085 M ammonia?

% = 2.11%

Explanation:

In order to do this, you need to do the acid base equilibrium reaction. In this case, the ionization of NH3 in water. The reaction taking place is the following:

NH3 + H2O <> NH4+ + OH-

Now, the percent ionization is a relation between the initial moles and the final moles. You could also take the concentrations. In this problem, we have the concentrations, therefore, we'll work with that.

The expression to calculate the percent ionization of ammonia would be:

% = [NH4+]/[NH3] * 100 (1)

In other words, we need the initial and final concentration of ammonia. We have the innitial, to get the final concentration, we do an ICE chart of the reaction and then, solve for the concentration.

The ICE chart is the following:

         NH3 + H2O <> NH4+ + OH-      Kb = 3.95x10⁻⁵

i)        0.085                           0           0

c)          -x                               +x         +x

e)     0.085-x                          x            x

Writting the expression of equilibrium we have:

Kb = [NH4+][OH-] / [NH3]

Replacing the above values:

3.95x10⁻⁵ = x² / 0.085-x

Kb is a very small value, so the difference of 0.085-x can be depreciated so:

3.95x10⁻⁵ = x²/0.085

x² = 3.95x10⁻⁵ * 0.085

x = √3.35x10⁻⁶

x = 0.0018 M

This result means that this is the final concentration of NH4+ and OH-, therefore the percent ionization is:

% = 0.0018 / 0.085 * 100

% = 2.11%

i believe it is a.) released into

explanation:

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