Ammonia, NH3, is a weak base with a Kb value of 1.8×10−5. What is the pH of a 0.205 M ammonia solution?

Question

Part B:

What is the percent ionization of ammonia at this concentration?

Answer 1

Hi, Elle.

The Kb is the equilibrium constant for the reaction of the base ammonia combining with water to produce ammonium, the conjugate acid, and a hydroxide anion (OH-).

Here is the reaction: NH3 + H2O --> NH4+ + OH-

The formula for any equilibrium constant is K= (product of products)/ (product of reactants) with units of concentration in molarity or pressure (for gasses only). Pure substances like H2O are not included in the formula.

The pH is directly related (as we'll see later) to the concentration of OH- which is one of the products.

K= (product of products)/ (product of reactants)

1.8×10^−5 = (OH-)(NH4+) / (NH3)

The given 0.205 M ammonia is the total concentration for both forms of that species, the acid and the base, NH3 and NH4+. However, for a weak base like ammonia, we can assume that the amount of NH3 that has been changed to NH4+ is so low (by a magnitude 5x less than 10, given by the kb) that doing the calculation wouldn't change the answer by very much at all. Going back to the chemical equation, we can see that for every 1 mol of NH3, 1 mol of NH4+ and 1 mol of OH- is made. In other words, the amount of NH4 and OH- will always be the same as each other here, so we can use one variable "x" to represent them both.

1.8×10^−5 = (OH-)(NH4+) / (NH3)

1.8×10^−5 = (x)(x) / (0.205 M)

1.8×10^−5 = x^2 / (0.205 M)

Solving for x, x = 0.00192 M where x = the concentration of OH-

To get from the concentration of the OH- to the pH we need to know that the "p" simply means "the negative log of the concentration" and that pOH + pH = 14 always.

pOH = the negative log of the concentration of the OH = - log (0.00192 M) = 2.72

pOH + pH = 14 so pH = 11.3

For part A, Final answer: pH = 11.3

The next part of the question asks for the percent ionization. Notice in the chemical equation that NH3 goes from having no charge to producing two charged species. The percent ionization is really just asking what percent of the NH3 becomes NH4+ and OH-. We already solved for the concentration of the OH- which was 0.00192 M. So 0.00192 M of the original 0.205 M was ionized. 0.00192 is 0.937 percent of 0.205 M.

For part B, Final answer: 0.937%