Base Dissociation Constant II

 Hello, welcome to today’s lesson. I want to delve more into the topic that we have been considering since yesterday. Let us recall that yesterday, we wrote the reaction; :B(aq) + H2O(l) -->BH^+(aq) + OH^-(aq). The specie BH^+ is formed by coordinate bonding between water and the base :B. Readers should review the post on July 6, 2023 because I explained the idea of coordinate bonding in that lesson using BF3 and NH3. 


Now consider the bond that is occurring in the BH^+. We have to note that the :B has a lone pair which could be shared with a proton from water. This would lead to the formation of a coordinate covalent bond between :B and H^+ and then we have BH^+(the positive charge just means that :B is bonded to more atoms than usual). Note that the formation of BH^+ implies that the electron pair no longer totally resides on the :B central atom (assuming that B is the central atom in a weak base such as NH3) and this creates a positive charge or electron deficient center in the product hence the positive charge on BH^+. We must note that :B could be a neutral molecule having a lone pair of electrons or an anion (a negative ion whose negative charge is caused by an excess electron pair).


We arrived yesterday at the equation;

Kb = [BH^+] [OH^-]/[:B]

In the image attached to the lesson of yesterday, we further saw that we could take a negative logarithm of Kb and that would give us a value that we call the pKb which is an easier number to deal with and also denotes the extent of dissociation of the base in solution.

Hence, pKb = - log(Kb)

For acids; pKa = -log(Ka)

We shall examine the pKb and pKa later in relation to another term that we shall call the pH.


Let’s consider a mathematical example involving the pKb.

Example 1;

What is the [OH^-] of a 0.5 M solution of ammonia if the pKb of ammonia is 4.74.


Solution;

Do we remember the ICE table? If you don’t, kindly review the lesson on June 20, 2023 found on the page timeline. Let us now set up the ICE table for the problem we have here.


              NH3(aq) + H2O(l)  <->NH4^+(aq) + OH^-(aq)

I             0.5                                   0                      0

C            -x                                    +x                   +x

E           0.5 – x                             x                         x

What we have is the pKb but we need the Kb to plug into the equilibrium equation.

Recall that pKb = -log Kb

Hence Kb= Antilog(-pKb)

Kb = 1.8 * 10^-5


Then we have;

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

Given that we have at equilibrium; 

Kb =1.8 * 10^-5 

[NH4^+] = [OH^-] = x

[NH3] = 0.5 – x

1.8 * 10^-5 = x^2/(0.5 – x)

1.8 * 10^-5 (0.5 – x) = x^2

9 * 10^-6 - 1.8 * 10^-5x = x^2

x^2 + 1.8 * 10^-5x - 9 * 10^-6  =0

Using Mathpapa online quadratic equation solver;

https://www.mathpapa.com/quadratic-formula/

x = 0.003 M (We discarded the negative result because concentration can not be negative)

Recall that; 

[NH4^+] = [OH^-] = x

Now x = 0.003 M

Hence; 

[OH^-] =  0.003 M

Hence, the concentration of the hydroxide ion is 0.003 M.


Tomorrow and next, we shall see more examples on mathematical problems that involve the Kb then we shall move on to discuss the percent dissociation(ionization) of a base. Hope the lessons are making sense? Get back to us if you are confused at any point.


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