Solving System of Reactions: .450 M HCN & .230 KCN

[pooker]

Homework Statement

Our teacher gave us a bunch of problems, and this is one of the ones I am having problems with.

What are the values of a syste that contains .450 M HCN and .230 KCN

Homework Equations

Ka = 4.9 x 10^-10

The Attempt at a Solution

Well I tried a few different ones.

HCN + KCN --- H + CN + K + CN

I tried solving the answer that way since it was the first ionization step, but got a wrong answer.

Then I tried the second ionization step caused by hydrolosis

CN + CN + H2O -- HCN + HCN + OH-




I still got the wrong answer . I do not need help solving the equation I just need help figuring out how to break it down. I am just really confused by this one for some reason :(

 

[epenguin]

Firstly Ka = 4.9 x 10^-10 M I think. Not just pedantry but will remind you what K_a really is, and a bit what to expect.

Your equations are no use, I don't know what they mean.

I suggest you write out the equation of electroneutrality for your problem

After you mix the HCN and KCN, which latter is all dissociated, K⁺ and CN^- what is going to happen? Well some HCN will dissociate and give you some extra CN^-. So you might worry how am I going to calculate that. A bit difficult. The good news is you don't need to calculate that! Because the amount of that happening is very small compared with the other charged species you already have and have mixed, K⁺ and CN^-. You can simplify the electroneutrality equation . I hope you can see, in other words, the ratio of CN^- to HCN will remain practically the same as of the amounts of those two you mixed in! Now you know the equilibrium equation (what K_a is) - use it.

Not all pH calculations are so simple!

I guess the twist that often is stopping people is that you want to calculate [H⁺] or [OH^-] and at some point in the calculation you have to say that these things you eventually want to calculate are negligible! Well they are when they come in as a sum with much larger things, but not when they are a factor.

Below is where I tried a more complete guide.

I got pH 9.01, but my arithmetic is not good certain times of day. modesthttps://www.physicsforums.com/showpost.php?p=1754973&postcount=3
https://www.physicsforums.com/showpost.php?p=2029975&postcount=2

 

[Borek]

My bet is that your teacher expects the answer that can be calculated by direct application of the Henderson-Hasselbalch equation. This will be different from the result given by epenguin (which is much better, although still wrong ).

 

[epenguin]

My bet is that your teacher expects the answer that can be calculated by direct application of the Henderson-Hasselbalch equation. This will be different from the result given by epenguin (which is much better, although still wrong ).

Isn't it the same thing? I guess I prefer to think from fundamentals than remember an equation - my reasoning is about what the terms in the HH equation are, and why it gives the right (enough) answer.

edit. I think though it will be quite helpful in getting confident grasp, to also spend the time to calculate say [H⁺] of .450 M HCN (weak acid) and .230 M KCN (salt of weak acid and strong base).