# PH and buffers

• jsmith613
In summary, the conversation discusses the process of creating a buffer solution with a 1:2 ratio of acid to base using a 0.1M acid and 0.25M salt. The solution requires 333.3 cm3 of acid and 666.6 cm3 of base, with no additional water needed. However, after considering the initial molar ratio of 1:2.5, the correct ratio is determined to be 1:0.8, requiring 555.6 cm3 of acid and 444.4 cm3 of base. The correctness of this solution is not confirmed, as the conversation is not focused on checking homework but providing guidance.

## 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?

## 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?

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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?