At equilibrium, it was found that [CH3COOH] = 0.8537 M, [CH3COO-] = 0.0040 M, and [H3O+] = 0.0040 M. Calculate the acidity constant of CH3COOH and the pH value of the solution. - Please show how you solve the problem.
You're asking some very basic questions here and previously. What do you not understand about this problem?
I don't understand it. To you is simple and basic but to me is very hard subject to understand and solve.
Thank you anyways for your help.
I didn't say I wouldn't help. I asked what you don't understand. If I know what your problem is I can help. If I only provide an answer I've not helped you. I know chemistry can be difficult for some and I have trouble with parts of it too. We all do. With your help I can help. But I don't want to work problems without knowing what is wrong. Getting defensive won't get the trouble solved. So I invited you to pour out your heart.
I'm not trying to be defensive, even though that's how I came across. I simply don't understand nothing of it. I don't know how or what to answer the question. I look at it and don't know how to solve, in other words it's like another language that I have or don't know how to approach it.
I too am trying to understand.
So far I have figured this but do not know if correct.
Acidity constant=Ka=[A^-][H3O]/[HA]
A^-= [CH3OO^-]=0.0040M
[H3O]=0.0040
[HA]=0.8537
So plug them in and you get 1.8*10^-5
im lost at the pH value help with that please.
So you're totally confused. There should be some solace in that you're not along AND there is hope. I might suggest that you get a tutor if you are just starting out in chemistry. It's important to get the basics. Here is how you do the problem along with some of the rationale that goes with it.
Acetic acid is a weak acid. Weak acids in chemistry, by definition, are those that do not ionize 100% (strong acids ionize 100%; weak acids less than 100%). We have developed a system for determining just how weak a weak acid is. That's called the dissociation constant. For acids it is Ka; for bases it is Kb. For acetic acid, it ionizes slightly (only about 1% or so) like this.
CH3COOH + H2O ==> CH3COO^- + H3O^+.
By definition, Ka = product of ions/unionized.
Ka = (H3O^+)(CH3COO^-)/(CH3COOH)
The problem give you
(H3O^+) = 0.004 M
(CH3COO^-) = 0.004 M
(CH3COOH) = 0.8537 M
We simply substitute those numbers into the Ka expression like this.
Ka = (0.004)(0.004)/(0.8537). Plug those numbers into your calculator and solve for Ka. I obtained 1.9E-5
Your text probably has a list in th appendix that will list the Ka values for various weak acids and weak bases.
Follow up questions are welcome but be sure and explain what you don't understand.
I think the answer is 1.874E-5 which rounds to 1.9E-5 to two significant figures. The pH = -log(H^+) = -log(1.9E-5) = -(-4.72)= 4.72
DrBob22 please help with the pH.
I am not sure if I would use pKa=-logKa=-log(1.9*^-5)=4.74 or do I use
[H3O]= 10^-pH
=.0040=10*^-ph
never mind u have already responded thanks I was right the first time.
OOPS. I gave you the pKa value which is not what the problem wanted.
pH = -log(H^+) = -log(0.004) = -(-2.3979) = 2.3979 which I would round to 2.40 to two s.f. (The 2 doesn't count in significant figures since the 2 is the characteristic.)