Base Dissociation Constant III

 Hello, we are here again to continue from where we left off yesterday. Yesterday, we solved a mathematical example that helped us consolidate on the ideas that we have discussed about the base dissociation constant Kb. Today I would solve another example and then give you one to try out on your own. You can drop your answer to the self test question as a comment under this post.


Example 2;

Calculate the concentration of the hydroxide ion in a 0.1 M solution of hydrazine if the pKa of the solution is 7.95.


Solution;

We are told to obtain the concentration of the hydroxide ion that is contained in a given concentration of hydrazine solution. We must first write down the reaction equation as follows;

N2H4(aq) + H2O(l) ->N2H5^+(aq) + OH^-(aq)


We can see that what we have here is the value of the pKa thus we must find a way to obtain the pKb because hydrazine is not an acid but a base(because it accepted a proton in this reaction).


Some days ago I introduced the idea of the ion product of water (Find the lesson on self ionization of water on the page timeline). I said that;

[H^+] [OH^-] = 1 * 10^-14

Then I said that [H^+] = [OH^-] = 1 * 10^-7 M (the definition of neutrality).

If I take negative logarithm of 1 * 10^-7 and 1 * 10^-14, I would have 7 and 14 respectively.

It then follows that;

pKa + pKb = 14 (for a neutral solution such as water)

Also;

Ka * Kb = 1 * 10^-14 


Now back to our question here, it is now clear that;

pKb = 14 – pKa

pKb = 14 – 7.95

pKb = 6.05

Kb = Antilog (-pKb)

Kb = Antilog (-6.05)

Kb = 8.9 * 10^-7


Then;

Kb = [N2H5^+] [OH^-]/[N2H4]

Given that the initial concentration of N2H4 is 0.1 M, we can now set up the ICE table as follows;


          N2H4(aq) + H2O(l) ->N2H5^+(aq) + OH^-(aq)

I         0.1                                    0                       0

C        -x                                      +x                    +x

E     0.1 – x                                 x                        x

Note: If your mobile device can not view the ICE table in portrait, you can flip it over to landscape or alternatively look for a device having a larger display such as a tablet or a Laptop PC.


Hence;

8.9 * 10^-7 = x^2/(0.1 – x)

8.9 * 10^-7(0.1 – x) = x^2

8.9 * 10^-8 - 8.9 * 10^-7x = x^2

x^2 + 8.9 * 10^-7x - 8.9 * 10^-8 = 0

We can now solve with an online quadratic equation solver

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

x = 0.0003 M (Recall that yesterday I said that we discard the negative result from the solution of the quadratic equation since concentration can not be negative).


I hope the working of the problem is clear? Reach out to me if there is still any confusion anywhere.


Self test question;

What is the hydroxide ion concentration produced when 0.05 M methanoic acid is protonated by a very strong acid? The pKa of methanoic is 3.75.


Drop your answers to the self test question under the comments section of this post before 7 AM WAT tomorrow July11, 2023. If you can show the working to help others learn, that would also be appreciated. See you tomorrow!


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