Calculating Buffer Volumes for pH and Concentration Ratios

  • Thread starter Thread starter jsmith613
  • Start date Start date
  • Tags Tags
    Buffers Ph
Click For Summary
SUMMARY

The discussion focuses on calculating the required volumes of a buffer solution made from a 0.1M acid and a 0.25M salt to achieve a 1 dm³ solution with a specific acid-to-base ratio of 1:2. The correct volumes determined are 555.6 cm³ of acid and 444.4 cm³ of base, which maintains the initial molar ratio of 1:2.5. The conversation emphasizes that no additional water is needed for dilution, as mixing the correct volumes achieves the desired concentration and pH balance.

PREREQUISITES
  • Understanding of buffer solutions and their components
  • Knowledge of pH calculations and the Henderson-Hasselbalch equation
  • Familiarity with molarity and volume calculations
  • Basic algebra for solving equations with two variables
NEXT STEPS
  • Study the Henderson-Hasselbalch equation in detail
  • Learn about buffer capacity and its implications in solution preparation
  • Explore titration techniques for acid-base reactions
  • Investigate the effects of dilution on pH and buffer effectiveness
USEFUL FOR

Chemistry students, laboratory technicians, and anyone involved in preparing buffer solutions or studying acid-base equilibria.

jsmith613
Messages
609
Reaction score
0

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:
Physics news on Phys.org
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?
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
6K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 7 ·
Replies
7
Views
4K
Replies
5
Views
3K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 23 ·
Replies
23
Views
3K