Acids And Bases...

Jack16

Hi People ,
I had a problem solving this question,could you please help me:D
The question:
a)The ionisation constant for propanoic acid is 1.26x10^-5 mol dm^-3.Calculate pH of a 1.0M solution
b)What is the hydroxide ion concentration of this solution?
c)Pocroc acid has an ionisation constant of 5.0x10^-3 mol dm^-3. It can act as an acid base indicator,the anions being red,and the free acid yellow.What colour will it show in a solution of pH 5?

chem_tr

Hello,
Ionization constant is the same as equilibrium constant, I think. Write the equilibrium constant for propanoic acid's ionization with the help of its reaction, and find the unknown values. Omit some, and find hydrogen concentration, and then, pH value.

Hydroxide ion concentration is linked to hydrogen ion concentration with the equation, [tex]pH+pOH=14[/tex] for normal aqueous systems. Do something with this equation to learn the desired value.

Picric acid (2,4,6-trinitrophenol) can show the acidity, it's correct. Just find its ionization constants with this simplification, pic-H is picric acid, and pic^- is picrate ion. You know the hydrogen concentration (pH). Just do it...

Gokul43201

I don't think the ionization constant is the same as the equilibrium constant.

C2H5COOH <---> C2H5COO(-) + H(+)
At equil : [tex](1-\alpha)~moles~C_2H_5COOH~<--->~\alpha~moles~~C_2H_5COOH^-~+~\alpha~moles~H^+[/tex]

[tex] \alpha =~ ionization~(dissociation)~~constant[/tex]
[tex] K = \frac{\alpha ^2}{1-\alpha}[/tex]

chem_tr

Yes, Gokul, you are right. The [tex]\alpha[/tex] shows how much of it is ionized. I'd better freshen my knowledge. So ionization constant IS the same as hydrogen ion concentration, right? I made the error since it's expressed as [tex]mol.dm^{-3}[/tex], I thought [tex]\frac{\alpha^{2}}{(1-\alpha)}[/tex] has the same dimension, and just said it.

Gokul43201

Yes, I too was concerned about the units. I'm sure that the ionization constant should be unitless.

So, either that is the equilibrium constant (and NOT the ionization constant) or there should be no units.

Can't have both.

GCT

The "ionization constant", "dissociation constant", Ka of such acids are simply special equilibrium constants...yes they are equilibrium constants.

poolwin2001

I can do the first two but what about the 3rd?
I cant undertand this part?

poolwin2001

I thought units for [tex]K_p[/tex] and [tex]K_c[/tex] (equilibrium constants) change for each reaction
Eg:units for [tex]2NO_2<-->N_2O_4[/tex] is different from that for
[tex]C_2H_5COOH<--->C_2H_5COOH+H^+[/tex]

chem_tr

Dear poolwin2002,

You are right about changeable-unit-equilibrium constants. About picric acid's ionization, it is very simple, just look at the following reaction:

[tex]Pic-H \rightarrow Pic^-~+~H^+[/tex]

Here, we know the pH, that is, the hydrogen ion concentration, from the added acid. Then we may use the ionization constant's formula to obtain the initial hydrogen concentration from ionization of picric acid. But there may be a problem, that the initial picric acid concentration is not given. If the problem was correct, then the acidity is fully supplied by picric acid, and the initial concentration can easily be found. But then the problem is invalid, since the pH is given to be 5, and the solution is acidic. I'm confused indeed. Please, jack16, check the last question and post a message.

Jack16

c)Picric acid has an ionisation constant of 5.0x10^-3 moldm^-3.It can act as an acid base indicator,the anions being red,and the free acid yellow.What colour will it show in a solution of pH5?

Gokul43201

Jack, you've seen our various discussions and concerns on this matter. What do you think ?

chem_tr

Hello, since GeneralChemTutor explained that they are different expressions about equilibrium constant, you can calculate the ionization constant as hydrogen ion concentration. Picric acid is dissolved in water at pH=5, add some hydrogen coming from pH=5.

poolwin2001

Thanks but Sorry I was asking about red and yellow part

chem_tr

Hello,

Since the medium is acidic, I presume that yellow color will be formed. Just compare the two acidities; the first, I mean, coming from the ionization is even enough to produce yellow color. Then you put more enough to bring the acidity of pure water to pH 5. It's your task to find the final pH and the correct color.

Regards,
chem_tr (my 100th message )