How do you calculate the pH of 35.0-mL of 0.15M acetic acid?

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Discussion Overview

The discussion revolves around calculating the pH of a 35.0-mL sample of 0.150 M acetic acid during a titration with 0.150 M NaOH. Participants explore various methods for determining pH at different volumes of base added, addressing both theoretical and practical aspects of the calculations involved.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents a method involving the use of the acid dissociation constant (Ka) and a quadratic equation to find the concentration of hydrogen ions, leading to a calculated pH of 2.069, which differs from the expected value of 2.78.
  • Another participant challenges the initial approach, suggesting that equilibrium calculations should maintain concentration units rather than converting to moles, and questions the validity of the method used.
  • A later reply suggests that the original poster may have made an arithmetic mistake in their calculations and proposes using an approximation method to simplify the calculations, noting that the concentration of hydrogen ions is much smaller than the total acid concentration.
  • Further discussion highlights the importance of using appropriate approximations in equilibrium problems and critiques the reliance on quadratic equations when simpler methods may suffice.
  • Another participant agrees with the need for corrections in the calculations and emphasizes the importance of maintaining proper units throughout the process.
  • One participant reports success using the "x-is-small" approximation, leading to a pH calculation that aligns with the expected value.

Areas of Agreement / Disagreement

Participants express differing opinions on the initial approach to the problem, with some supporting the use of moles and others advocating for concentration-based calculations. The discussion remains unresolved regarding the best method to use, as multiple perspectives on the calculations and their validity are presented.

Contextual Notes

Limitations include potential arithmetic errors, the need for proper unit consistency, and the appropriateness of using approximations in equilibrium calculations. The discussion reflects various assumptions about the applicability of different mathematical approaches to the problem.

Eclair_de_XII
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Homework Statement


"A 35.0-mL sample of 0.150 M acetic acid (CH3COOH) os titrated with 0.150 M NaOH solution. Calculate the pH after the following volumes of base have been added: (a) 0 mL, (b) 17.5 mL, (c) 34.5 mL, (d) 35.0 mL, (f) 50.0 mL."

Homework Equations


Ka = [H+][A-]/[HA]
Ka = 1.8 ⋅ 10-5
x = ½(-b ± √(b2 - 4ac)
pH = -log[H+]
pH = 2.78 (from back of book)

The Attempt at a Solution


0.15 mol/L ⋅ 1 L/1000 mL ⋅ 35.0 mL = 5.25 ⋅ 10-3 mol HAc
1.8 ⋅ 10-5 = (x)(x)/(5.25 ⋅ 10-3 - x)
x2 + 1.8 ⋅ 10-5x - 9.45 ⋅ 10-8 = 0
x = ½(-1.8 ⋅ 10-5 ± √3.24 ⋅ 10-10 - 37.8 ⋅ 10-8)
x = 2.99 ⋅ 10-4 mol H+

[H+] = (2.99 ⋅ 10-4 mol H+)/(0.0350 L) = 8.54 ⋅ 10-3 M
-log(8.54 ⋅ 10-3) = 2.069 ≠ 2.78

This is rather an unorthodox, and thus, unreliable method of finding the initial pH of the acetic acid solution. But I couldn't find any other way to do it... Could someone tell me if I'm doing this problem right?
 
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You can't do equilibrium calculations using Ka (which is a concentration constant) and ignoring the concentrations. Why do you convert to number of moles, instead of using concentrations throughout the calculations?
 
Eclair_de_XII said:

Homework Statement


"A 35.0-mL sample of 0.150 M acetic acid (CH3COOH) os titrated with 0.150 M NaOH solution. Calculate the pH after the following volumes of base have been added: (a) 0 mL, (b) 17.5 mL, (c) 34.5 mL, (d) 35.0 mL, (f) 50.0 mL."

Homework Equations


Ka = [H+][A-]/[HA]
Ka = 1.8 ⋅ 10-5
x = ½(-b ± √(b2 - 4ac)
pH = -log[H+]
pH = 2.78 (from back of book)

The Attempt at a Solution


0.15 mol/L ⋅ 1 L/1000 mL ⋅ 35.0 mL = 5.25 ⋅ 10-3 mol HAc
1.8 ⋅ 10-5 = (x)(x)/(5.25 ⋅ 10-3 - x)
x2 + 1.8 ⋅ 10-5x - 9.45 ⋅ 10-8 = 0
x = ½(-1.8 ⋅ 10-5 ± √3.24 ⋅ 10-10 - 37.8 ⋅ 10-8)
x = 2.99 ⋅ 10-4 mol H+

[H+] = (2.99 ⋅ 10-4 mol H+)/(0.0350 L) = 8.54 ⋅ 10-3 M
-log(8.54 ⋅ 10-3) = 2.069 ≠ 2.78

This is rather an unorthodox, and thus, unreliable method of finding the initial pH of the acetic acid solution. But I couldn't find any other way to do it... Could someone tell me if I'm doing this problem right?

I don't think the approach is unorthodox except as below. I think there is only an arithmetical/algebraic mistake. Your square root is written without right bracketing and you have got wrong sign in the -4ac term. Probably you will get the right answer if you correct that.

Where it is not wrong but unorthodox is there is a bit of unnecessary calculation.

Orthodox would be to say that [H+] is very much smaller than total acid, and therefore you can approximate the denominator of your equilibrium equation by total acid concentration [HAc] = 0.15. Then your calculation reduces to taking a square root rather than solving a quadratic. Do it both ways and you will see they are close. You can if you wish, with benefit, look into the algebra and see why.

The point is this: all this family of problems (the most frequent single one on this part of this forum) are solved using a handful of principles: mass conservation, charge conservation, equilibrium laws, I think that's all... plus use of approximations. Don't think of it as Important Laws and then, oh yes, also get rough answer quicker if approximate. No, the appropriate approximations, understanding them, and how to use them and why and when and how good and sufficient they may be depending on situations, is an essential part of what you have to think about, grasp and learn!

In most cases if you have a quadratic or higher equation you have missed a possibility of approximation. (Of course if the physical nature of the problem requires it, e.g. some equilibria with two protonation etc. polynomials may be of the essence.)
 
epenguin said:
Probably you will get the right answer if you correct that.

I think he will not - units are wrong. Using number of moles instead of concentrations works when the volumes cancel out, like \frac {C_1}{C_2} = \frac {n_1}{n_2}, but here we have \frac {C_1^2}{C_2} = \frac {n_1^2}{V\times n_2} \neq \frac {n_1^2}{n_2}.

I admit I have not checked numbers, but the logic looks faulty.
 
Okay, I'll try using concentration instead of moles.
 
So I thought the answer was more complicated than it seemed. I just used the initial concentration, and used the "x-is-small" approximation.

1.8 ⋅ 10-5 = (x)(x)/(0.15 - x)
1.8 ⋅ 10-5 = (x)(x)/(0.15)
2.7 ⋅ 10-6 = x2
x = 0.00164
-log[H+] = 2.78

Now to figure out the rest of the problem...
 

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