Calculating Buffer Volumes for pH and Concentration Ratios

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Discussion Overview

The discussion revolves around calculating the volumes of acid and base required to create a buffer solution with a specific pH and concentration ratio. The context includes a homework-like scenario where participants explore the implications of the given concentrations and ratios.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant proposes that to achieve a 1:2 ratio of acid to base, 333.3 cm3 of acid and 666.6 cm3 of base are needed, but adjusts the volumes based on initial concentrations.
  • Another participant questions the need for additional water for dilution, suggesting that correct volumes of the solutions alone can yield the desired total volume.
  • Further contributions suggest that the initial molar ratio of acid to base is 1:2.5, leading to a recalculated ratio of 1:0.8 based on the pH equation.
  • One participant asserts that the solution is not homework but rather a revision question, seeking confirmation of their calculations.

Areas of Agreement / Disagreement

Participants express differing views on whether additional water is necessary and how to calculate the correct volumes based on the given concentrations and desired ratios. No consensus is reached regarding the correct volumes or the necessity of water.

Contextual Notes

There are unresolved assumptions regarding the interpretation of the problem, particularly concerning the inclusion of water in the final volume and the implications of the initial concentrations on the final acid-base ratio.

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Homework Statement


An buffer is made from a 0.1M acid and a 0.25 M salt.
According to the equation
pH = pKa - log[acid]/[base]
the ratio acid:base is 1:2

What volumes are required to make a 1 dm3 solution?

Homework Equations


The Attempt at a Solution


According to the 1:2 ratio we would need
333.3 cm3 of acid and 666.6 cm3 of base
However as the initial concentration of acid to base is 0.1:0.25 we instead say
volume acid = 333.3cm3
volume base = 666.6 cm3/2.5 = 266.6 cm3
volume water = 400 cm3

(all of these volumes are rounded to the nearest 0.1 cm3)

is this correct?
 
Last edited:
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Question doesn't say anything about using additional water for dilution, so I don't think that's the correct answer.
 
yes but as I need to make up a litre the rest of the solution must be water
 
No, you don't have to add water. If you mix correct volumes of both solutions you will have a liter of solution with a correct ratio of acid and conjugate base. Try to write two equations describing both ratio and volume - that will give you two equations in two unknowns, easy to solve.
 
666.67 cm3 acid
and
333.333 cm3[/base]
 
Have you checked if the ratio is correct if you use these volumes?
 
ratio acid:base assuming 1M:1M initial ratio gives 1:2
BUT initial molar ratio is 1:2.5
therefore new acid:base ratio from pH = pKa - log[acid]/[base] is 1:(2/2.5) = 1:0.8
...oh
acid: 555.6 cm3
base: 444.4 cm3

right?
 
Sorry, we are not checking homework, we are helping solve it.
 
This is not homework
this is a revision question I can't find an answer to
am i right?
 

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