Estimating pH Without [H+][A-]: Is There Another Way?

  • Biology
  • Thread starter TytoAlba95
  • Start date
  • Tags
    Ph
In summary, the pH of a concentrated solution of weak acid is approximately -2log3 + 3log10 (rounded off 9.1 into 9).
  • #1
TytoAlba95
132
19
Homework Statement
The acidic nature of orange juice is mainly due to citric acid. What would be the pH of 0.1M citric acid, if Ka is 8.4 x 10^-4?
Relevant Equations
pH= pKa + log ([H+][A-]/[HA])
How can we estimate the pH when the [H+][A-] is not provided. Is there some other formula?
 
Physics news on Phys.org
  • #2
Are you using all the relevant information you have, even if you are not retold it when the question is asked?
Electroneutrality?
It also helps if you know that a concentrated solution of weak acid will be little dissociated which allows an easy approximation though if you have any qualms about that you can still calculate it more accurately as a check.
 
Last edited:
  • #3
The equation you wrote is wrong (check the part under log) and - assuming you meant Henderson-Hasselbalch equation - not applicable to this question.

This is just about a solution of a weak acid, not about a buffer solution.
 
  • Like
Likes epenguin
  • #4
Borek said:
The equation you wrote is wrong (check the part under log) and - assuming you meant Henderson-Hasselbalch equation - not applicable to this question.

This is just about a solution of a weak acid, not about a buffer solution.

Thank you for the pointer.
pH= pKa + log ([H+][A-]/[HA]) -> pH= pKa + log ([A-]/[HA]) .

New attempt:

Ka= x2 / C-x => 10-6 = x2 / 0.01-x
=> x2 + 10-6 - 10-8 = 0
I'm stuck at this point. Please help.
 
  • #5
Ka is not 10-6, C is not 0.01. Not only you started with wrong values, you also made several simple math mistakes while trying to solve.

At least it looks like you are trying to start from a correct equation this time. I wonder if you know what x is and where does this equation come from.
 
  • #6
Yes the first equation of your new attempt looks as if it could be right if you used the proper brackets. And then the right numerical values as Borek says.
Alternatively or additionally even your corrected equation before that (which is now the Henderson-Hasselbalch equation) can be used together with the information I mentioned previously.
You seem to have used one of my suggestions, not the other.
 
  • #7
Start from first principles.
What is the equilibrium expression for

HA <——> H+ + A-
 
Last edited:
  • #8
epenguin said:
You seem to have used one of my suggestions, not the other.

It also helps if you know that a concentrated solution of weak acid will be little dissociated which allows an easy approximation though if you have any qualms about that you can still calculate it more accurately as a check.

You meant, excluding the subtraction of x from C in the denominator, right?

Ka is not 10-6, C is not 0.01. Not only you started with wrong values, you also made several simple math mistakes while trying to solve.

At least it looks like you are trying to start from a correct equation this time. I wonder if you know what x is and where does this equation come from.

Oh! I'm sorry. I hurriedly solved the sum, which is why it was full of mistakes.
Ka= x2 / C
=> 8.4 x 10-4 = x2 / 0.01
=> 8.4 x 10-6 = x2
=> 2.8 x 10 -3= x
I think this time I nailed it!
 
  • #9
Yes that was what I meant.
But you are not finished.
Everyone makes a % of mistakes in arithmetic etc. or can mix one figure with another, I am never that confident I haven't, :angel: , but your mistake rate is at alarm level! :oldsurprised: You have now made the citric acid concentration 0.01 M, wasn't it 0.1 M?
And you haven't answered the question which was what is the pH?! So you give your evil examiners every excuse to fail you!

If you complete the answer I have a couple of other things to tell you.
 
Last edited:
  • #10
SanjuktaGhosh said:
Ka= x2 / C-x

SanjuktaGhosh said:
Ka= x2 / C

Either - either. These are two different equations. Each has its applicability, but you don't use them at random, you need to check which one can be used.
 
  • #11
SanjuktaGhosh said:
Ka= x2 / C
=> 8.4 x 10-4 = x2 / 0.01
=> 8.4 x 10-6 = x2
=> 2.8 x 10 -3= x
I think this time I nailed it!

I'm sorry, for making such silly mistakes again. Doing maths on computer is really not my thing.

Ka= x2 / C
=> 8.4 x 10-4 = x2 / 0.1
=> 8.4 x 10-5 = x2
=> 9.1 x 10 -3= x
Therefor pH= -log [H] = -log(9.1 x 10 -3)
= -2log3 + 3log10 (rounded off 9.1 into 9)
= -2x 0.477 + 3
= 2.046

@Borek
These are two different equations. Each has its applicability, but you don't use them at random, you need to check which one can be used.
Yup. By now I know where to use them. Thanks for the pointer.

@epenguin
If you complete the answer I have a couple of other things to tell you.

I'm done. Please share.
 
Last edited:
  • #12
SanjuktaGhosh said:
= 2.046

@epenguinI'm done. Please share.

It was just to say that from the HH equation in #4, electroneutrality condition [A-] = [H+], definition of pH, and approximation (C - [H+]) ≈ C you can get the equation, often quoted

pH ≈ (pK - log C)/2

This is completely equivalent to what you did. From that I get pH = 2.033 but the small difference is likely due to your rounding.

It's also convenient for quick mental calculation, useful in laboratory thinking. The pH of a 1M solution of weak acid equals half the pKa; then for every 10 fold dilution it increases by 0.5 (for as long as the approximation you made holds good, which it doesn't very by the time the concentration gets to the order of magnitude of Ka).

The other thing is that citric acid has three acid dissociations from its three carboxylic groups, in your solution only one of these is significantly dissociated, so you don't have to worry here about the others whose pK's are about 4.76 and 6.4
 
Last edited:
  • Like
Likes TytoAlba95
  • #13
Untitled-1.png


(pKa2 = 4.761, pKa3 = 6.396, by default I am using a bit different value for Ka1, but modified it here to the value from the initial post)
 
  • Like
Likes TytoAlba95

1. How can pH be estimated without using [H+][A-]?

There are several methods for estimating pH without using [H+][A-]. One common method is using indicators, which are substances that change color in response to changes in pH. Another method is using a pH meter, which measures the electrical potential difference between a reference electrode and a sensing electrode. Additionally, pH can be estimated using titration, where a known amount of acid or base is added to a solution until it reaches a neutral pH.

2. Why is estimating pH without [H+][A-] important?

Estimating pH without using [H+][A-] is important because it allows for faster and more convenient measurements of pH. Additionally, it can be useful in situations where [H+][A-] is not readily available, such as in natural water sources. It also provides an alternative method for estimating pH, which can be helpful in cases where the traditional methods may not be accurate or feasible.

3. Are there any limitations to estimating pH without [H+][A-]?

Yes, there are some limitations to estimating pH without using [H+][A-]. One limitation is that the accuracy of the measurement may vary depending on the method used. Additionally, some methods may only provide an estimation of pH within a certain range, and may not be able to accurately measure very high or low pH levels. It is important to consider these limitations when choosing a method for estimating pH without [H+][A-].

4. What are some common applications of estimating pH without [H+][A-]?

There are many common applications of estimating pH without using [H+][A-]. Some examples include testing the pH of household products, such as cleaning solutions or cosmetics, monitoring the pH of water sources for environmental purposes, and measuring the pH of food and beverages in the food industry. It is also commonly used in laboratory experiments and research studies.

5. Can estimating pH without [H+][A-] be used in both acidic and basic solutions?

Yes, estimating pH without using [H+][A-] can be used in both acidic and basic solutions. The methods for estimating pH, such as using indicators or a pH meter, are applicable to a wide range of pH levels. However, it is important to choose the appropriate method and indicators for the specific pH range being measured in order to obtain accurate results.

Similar threads

Chemistry How much NaOH I have to add to increase pH?
    • Biology and Chemistry Homework Help
Replies
3
Views
3K
Calculating pH with and without ionic strength
    • Biology and Chemistry Homework Help
Replies
1
Views
1K
Determining if the pH changes when a chemical is added
    • Biology and Chemistry Homework Help
Replies
2
Views
2K
The relationship between pH and pOH at 70 ◦C/ 343.15 K
    • Biology and Chemistry Homework Help
Replies
5
Views
2K
Calculating the pH of a Buffer Solution
    • Biology and Chemistry Homework Help
Replies
7
Views
3K
How a pH Probe Works: A Comprehensive Guide
    • Biology and Chemistry Homework Help
Replies
2
Views
2K
Henderson Hasselbalch Equation - Request for Clarification
    • Chemistry
Replies
8
Views
858
Determining pH Fluctuations at Maximum Dosage of 50mg/L of 93% Sulphuric Acid
    • Biology and Chemistry Homework Help
Replies
3
Views
2K
Concentration of both buffers is the same, only the pH is different
    • Biology and Chemistry Homework Help
Replies
4
Views
2K
Calculating pKa from pH: Understanding Titration
    • Biology and Chemistry Homework Help
Replies
3
Views
2K

Hot Threads

Recent Insights

Back
Top