Titration of strong base + weak acid

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Discussion Overview

The discussion revolves around a titration problem involving the reaction of a strong base, calcium hydroxide (Ca(OH)2), with a weak acid, benzoic acid (C6H5COOH). Participants explore the calculation of pH after adding a specific volume of the base, particularly focusing on the conditions after surpassing the equivalence point.

Discussion Character

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

Main Points Raised

  • One participant calculates the equivalence point at 56.25 mL of Ca(OH)2 and expresses confusion about the remaining moles of acid and base after this point.
  • Another participant suggests focusing on the moles of excess Ca(OH)2 added beyond the equivalence point to determine pOH.
  • A participant questions the use of an ICF chart, proposing to use fundamental principles instead to find the concentration of excess hydroxide.
  • Concerns are raised about the solubility of Ca(OH)2 and benzoic acid, with one participant stating that the concentration of Ca(OH)2 may not be feasible due to its solubility limits.
  • One participant reports calculating a pH of 12.13 and mentions the remaining moles of Ca(OH)2, but expresses uncertainty about the correctness of their answer.
  • A later reply simply affirms the previous participant's calculation without further elaboration.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct approach to the problem, with differing opinions on the use of ICF charts versus fundamental principles, and varying views on the solubility of the reactants. The discussion remains unresolved regarding the final pH calculation and the interpretation of remaining moles of reactants.

Contextual Notes

Participants express uncertainty about the assumptions related to the solubility of Ca(OH)2 and benzoic acid, as well as the application of the ICF chart in this context. There are unresolved mathematical steps in determining the remaining moles of reactants after the equivalence point.

UKWildcat
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Homework Statement


Consider a titration in which 50.00 mL of .450 M benzoic acid, C6H5COOH, is reacted with 0.2000 M Ca(OH)2. What will the pH be when 60.00 mL of Ca(OH)2 has been added?

Homework Equations


2C6H5COOH + Ca(OH)2 -> 2 H2O + Ca(C6H5COO)2

The Attempt at a Solution


I have determined that the equivalence point is at 56.25 mL, so this is past that. The thing that is throwing me off is the mol ratio of acid:base. I try doing an ICF chart and keep getting that there is still acid left and no base. I thought that after the equiv. point, there is no acid left and you determine the pH based on the remaining base.

Can anyone help me fill in the ICF chart so that I can find the pH?
I need moles of acid and base to do this.
 
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Your first calculation is correct, that equivalence point needs 56.25 ml. of 0.2000 Ca(OH)2.

Now think in steps: How many moles of extra Ca(OH)2 was added beyond equivalence?
How many moles of hydroxide is this? What is the concentration of excess hydroxide and therefore what is the pOH?
 
How many moles of extra Ca(OH)2 was added beyond equivalence?

This is the part I am stuck on. When I plug into my ICF chart I get that there are .0225 moles of C6H5COOH.
Then I calculate the moles of Ca(OH)2 using .060L X 0.200mol/L and get .012 moles of Ca(OH)2.
Do I divide that by 2 because it is a 2:1 reaction?
Even if I don't, when I go to subtract the smaller amount of moles, the Ca(OH)2 comes out as the limiting reagent, which means it is all used up and I know this is not right.
What am I missing?
 
Forget your ICF chart. Use fundamental principles. You know you passed the equivalence point for neutralization so you have no unneutralized C6H5COOH. How many moles beyond the equivalence point have you gone, and what is the concentration of the hydroxide (in moles per liter), and from this, what is pOH?
 
Think in terms of limiting reagents only. And compare with this pH of acid/base mixture calculation example.

In a way question is idiotic - there is no such thing as 0.2M calcium hydroxide, its solubility is about 10 times smaller if memory serves me well. No idea about benzoic acid solubilityk, but with bulky phenyl it won't be easily soluble as well.

--
 
I am not sure if I came up with the right answer, but I got a pH of 12.13. I calculated that there were .0015 mols of Ca(OH)2 remaining.
 
Yes.
 

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