Mathematical Problems on Acid Dissociation Constant Part 2

 Hello fam, today is yet another day! As I informed you yesterday, today we are going to take an example in which the Ka is given and we would need to obtain the equilibrium concentration of either of the ions H3O^+ and A-. 


I hope all these are making sense? If you are confused, feel free to reach out to me via the comments section, Whatsapp or messenger. We also take phone calls and text messages on our dedicated phone number displayed on the page. 


Now lets look at this example.

Question;

Vinegar is a dilute solution of acetic acid(HC2H3O2). If the concentration of HC2H3O2 in a vinegar solution is 0.210 M, calculate the concentration of the C2H3O2^- ion present.

(Ka for acetic acid = 1.75 * 10^-5)


Solution;

As we saw yesterday, we would have to set up the ICE table in order to help us to navigate the solution to the problem. If we set up the ICE table, then we would have that;


              HC2H3O2(aq) + H2O(l) <---> C2H3O2^-(aq) + H3O^+(aq)

I            0.210                                          0                            0

C          -x                                                +x                        +x

E         0.210 – x                                     x                            x

(Note that 0 + x = x. This explains the equilibrium concentration of the acetate ion and the hydronium ion).

Recall that; 

Ka = [C2H3O2^-] [H3O^+]/[HC2H3O2]

The Ka is given in the question as 1.75 * 10^-5 and [C2H3O2^-] = [H3O^+] =x

We then have that;

1.75 * 10^-5 = x * x/0.210 – x

1.75 * 10^-5 = x^2/0.210 – x

1.75 * 10^-5(0.210 – x) = x^2

3.7 * 10^-6 - 1.75 * 10^-5x  = x^2

x^2 + 1.75 * 10^-5x - 3.7 * 10^-6 = 0

This could be finished off using any free online quadratic equation calculator(I particularly love to use Mathpapa quadratic calculator)

x = 1.9 * 10^-3

Thus it follows that; [C2H3O2^-] = [H3O^+] = 1.9 * 10^-3 M.

The concentration of the acetate (C2H3O2^-) ion in the solution is  1.9 * 10^-3 M.


Now try this on your own;

Calculate the hydronuim ion concentration when 0.250M of acetic acid is dissolved in water. (Ka = 1.8 * 10^-5)


You can drop your answer in the comments section. Comments on this post would close at 5 AM (WAT) tomorrow 22nd June 2023. I look forward to your active participation in this exercise. See you tomorrow!


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