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Titration of strong base + weak acid

• Thread starter Thread starter UKWildcat
• Start date Start date Nov 14, 2009
• Tags Tags Acid Base Titration Weak

Click For Summary In a titration of 50.00 mL of 0.450 M benzoic acid with 0.2000 M Ca(OH)2, the equivalence point is reached at 56.25 mL of base. After adding 60.00 mL of Ca(OH)2, the discussion focuses on calculating the pH, noting that there is no unneutralized benzoic acid left. Participants emphasize the importance of determining the moles of excess hydroxide to find the pOH and subsequently the pH. One contributor arrives at a pH of 12.13, indicating 0.0015 moles of Ca(OH)2 remain after the equivalence point. The conversation also touches on the solubility limits of calcium hydroxide and benzoic acid, suggesting practical considerations in the calculations. UKWildcat Messages 4 Reaction score 0

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. Physics news on Phys.org

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symbolipoint Homework Helper Education Advisor Gold Member Messages 7,616 Reaction score 2,056 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? UKWildcat Messages 4 Reaction score 0

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? symbolipoint Homework Helper Education Advisor Gold Member Messages 7,616 Reaction score 2,056 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? Borek Mentor Messages 29,155 Reaction score 4,605 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. -- UKWildcat Messages 4 Reaction score 0 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. symbolipoint Homework Helper Education Advisor Gold Member Messages 7,616 Reaction score 2,056 Yes.

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