Solving Acids & Bases Questions - pH, Ionisation Constant

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SUMMARY

The discussion focuses on calculating the pH and hydroxide ion concentration of a 1.0M solution of propanoic acid, which has an ionization constant (Ka) of 1.26x10^-5 mol dm^-3. Participants clarify that the ionization constant is indeed a specific type of equilibrium constant, and they discuss the implications of this for calculating pH and hydroxide ion concentration using the formula pH + pOH = 14. Additionally, the color change of picric acid, with a Ka of 5.0x10^-3 mol dm^-3, in a solution of pH 5 is analyzed, concluding that it will appear yellow due to the acidic conditions.

PREREQUISITES
  • Understanding of acid-base chemistry and ionization constants
  • Familiarity with the pH scale and its calculations
  • Knowledge of equilibrium constants and their significance in chemical reactions
  • Basic concepts of acid-base indicators and their color changes
NEXT STEPS
  • Learn how to calculate pH from ionization constants using the formula K = α²/(1-α)
  • Explore the relationship between pH, pOH, and hydroxide ion concentration in aqueous solutions
  • Investigate the properties and applications of acid-base indicators like picric acid
  • Study the differences between various types of equilibrium constants, including Ka and Kb
USEFUL FOR

Chemistry students, educators, and professionals involved in acid-base chemistry, particularly those focusing on pH calculations and the behavior of acid-base indicators.

Jack16
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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?
 
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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, pH+pOH=14 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...
 
I don't think the ionization constant is the same as the equilibrium constant.

C2H5COOH <---> C2H5COO(-) + H(+)
At equil : (1-\alpha)~moles~C_2H_5COOH~&lt;---&gt;~\alpha~moles~~C_2H_5COOH^-~+~\alpha~moles~H^+

\alpha =~ ionization~(dissociation)~~constant
K = \frac{\alpha ^2}{1-\alpha}
 
Yes, Gokul, you are right. The \alpha 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 mol.dm^{-3}, I thought \frac{\alpha^{2}}{(1-\alpha)} has the same dimension, and just said it.
 
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.
 
The "ionization constant", "dissociation constant", Ka of such acids are simply special equilibrium constants...yes they are equilibrium constants.
 
I can do the first two but what about the 3rd?
Jack16 said:
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?
I can't undertand this part?
 
Gokul43201 said:
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.

I thought units for K_p and K_c (equilibrium constants) change for each reaction
Eg:units for 2NO_2&lt;--&gt;N_2O_4 is different from that for
C_2H_5COOH&lt;---&gt;C_2H_5COOH+H^+
 
Last edited:
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:

Pic-H \rightarrow Pic^-~+~H^+

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.
 
  • #10
The Last Question...

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?
 
  • #11
Jack, you've seen our various discussions and concerns on this matter. What do you think ?
 
  • #12
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.
 
  • #13
Thanks but Sorry I was asking about red and yellow part
 
  • #14
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 :smile:)
 

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