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%

explanation:

let the different atoms be x and y. each with four protons and same mass will be represented as follows.

mass number is number of protons + number of neutrons.

            and  

since, it is given that one has a negative charge while the other is neutral. therefore, there will be increase in number of protons and the atom will be represented as follows.

according to periodic table, mass number of beryllium is 9 and it has atomic number 4 so, the atoms will be and .  

the protons and neutrons will be located inside the nucleus whereas the electrons will be revolving around the nucleus.  

it would be a gas formation