Calculate the pH of 100mL HOAc + 100mL Ba(OH)2

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    Acids Bases Ph
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Discussion Overview

The discussion revolves around calculating the pH of a solution resulting from mixing 100 mL of 0.20 M acetic acid (HOAc) with 100 mL of 0.10 M barium hydroxide (Ba(OH)2). Participants explore the implications of the chemical properties of the substances involved and the appropriate methods for calculating pH in this context.

Discussion Character

  • Homework-related
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant calculates the pH based on the moles of acetic acid and barium hydroxide, suggesting a pH of 12.7.
  • Another participant challenges the initial calculation, stating that Ba(OH)2 is dibasic rather than diprotic and emphasizes the importance of using equivalents instead of moles.
  • A further response questions the clarity of the initial calculation, suggesting that the acetic acid is actually in excess and that the resulting pH should be less than 7.
  • Another participant reiterates the dibasic nature of Ba(OH)2 and implies that language barriers may contribute to misunderstandings in the discussion.
  • There is a suggestion to consider the conjugate base of acetic acid and its significant Kb in further calculations to determine the pH accurately.

Areas of Agreement / Disagreement

Participants express disagreement regarding the characterization of Ba(OH)2 and the implications for pH calculation. There is no consensus on the correct approach to determining the pH, and multiple competing views remain regarding the interpretation of the chemical interactions involved.

Contextual Notes

Participants highlight the need for clarity in calculations and the importance of using the correct chemical terminology. There are unresolved aspects regarding the assumptions made in the calculations, particularly concerning the definitions of diprotic versus dibasic and the use of equivalents.

Joules23
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Homework Statement


find the pH of:
100mL x .20M HOAc + 100mL x.10M Ba(OH)2



The Attempt at a Solution



100x.2 = 20mmol HOAc
100x.1 = 10mmol Ba(OH)2
20-10=10
10/200=.05M
pOH = -log(.05) = 1.3
pH=14-1.3=12.7

Is this correct?
 
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No. Ba(OH)2 is 'diprotic'.
 
Ba(OH)2 is not diprotic. Dibasic.

Think in terms of equivalents instead of moles (or mmol) with the barium hydroxide...

BTW, the way you did the work, it wasn't clear (to you) whether the "10" left over in the reaction was barium or acetic acid. Using the logic you provided, the acetic acid was actually present in excess and the pH should have been less than 7.

Using equivalents in place of moles should "completely neutralize" any misunderstanding you may have.
 
Joules23 said:

Homework Statement


find the pH of:
100mL x .20M HOAc + 100mL x.10M Ba(OH)2



The Attempt at a Solution



100x.2 = 20mmol HOAc
100x.1 = 10mmol Ba(OH)2
20-10=10
10/200=.05M
pOH = -log(.05) = 1.3
pH=14-1.3=12.7

Is this correct?

hint, hint, the situation is going to be "neutralized" as chemisttree has said

However, the conjugate of the acetic acid is a base with a significant Kb, you're going to need to use the Kb equation to find the [OH-], then the pH through further calculations.
 
chemisttree said:
Ba(OH)2 is not diprotic. Dibasic.

That's what happens when English is not your first language...



 

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