How do you calculate moles and neutralization in a buffer solution?

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Discussion Overview

The discussion revolves around calculating moles and neutralization in a buffer solution, specifically involving the components sodium acetate trihydrate (NaC2H3O2*3H2O) and acetic acid (HC2H3O2). Participants address a homework problem that includes determining moles of acetic acid in a buffer, the amount of sodium hydroxide (NaOH) needed for complete neutralization, and the volume of hydrochloric acid (HCl) required to neutralize the buffer system.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant calculates the concentration of HC2H3O2 and finds that there are approximately 9.12 x 10^-3 moles in 19 mL of the buffer.
  • Another participant suggests that the moles of HCl needed for complete neutralization should equal the moles of NaC2H3O2*3H2O present in the solution.
  • A calculation is presented estimating that 12 mL of 2.0 M HCl would be required to neutralize the buffer, based on the moles of sodium acetate calculated.
  • One participant expresses uncertainty about the correctness of their answers for parts a and b of the problem.
  • A later reply indicates that while the numbers have not been verified, the logic behind the calculations appears sound.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correctness of the calculations for parts a and b, and there is uncertainty regarding the approach to part c. Multiple viewpoints and calculations are presented without resolution.

Contextual Notes

Participants have not confirmed the molar mass of NaC2H3O2*3H2O, which is necessary for accurate calculations. There is also a lack of clarity on the stoichiometry involved in the neutralization reactions.

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Homework Statement


this is a part of a buffer solutions problem

3.3g of solid NaC2H3O2*3H2O added to 46mL of deionized water and 4.0 mL of 6.0M HC2H3O2.

a. How many moles of HC2H3O2 are contained in a 19 mL of the above buffer?
b. How many moles of NaOH would need to be added to the 19 mL of buffer to destroy the system? (complete neutralization)
c. How many mL of 2.0 M HCL would be needed to destroy 19 mL of the buffer system?

Homework Equations





The Attempt at a Solution


a. The concentration of HC2H3O2 will be 4x6/50 = 0.48M
moles of HC2H3O2 are contained in a 19ml will be 0.48M x 0.019mL = 9.12 x 10^-3 mols
b. same amount of moles as the acid?

c. clueless.


thanks for your time
 
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Think what would be the reaction in c. Question is very similar to b, it is just a matter of realizing what is going on and following the stoichiometry.
 
so to completely destroy the buffer system, the amount of HCl should be equal to the number of moles of NaC2H3O2*3H2O?
3.3/(MM of NaC2H3O2*3H2O) = 0.024 moles?
the volume will then be
0.024 mole / 2.0 M = 0.012L = 12mL

is this correct and also were the answer for a and b correct as well?

thanks for your time.
 
I have not checked the numbers, but the logic looks OK.
 

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