# Preparing 0.3M Phosphate Buffer at pH 7.9: c), d), e), f)

• MusicMonkey
M NaH2PO4 and 125 ml of 0.3M Na2HPO4 , so i can say that c) is the correct answer.In summary, the correct amounts of NaH2PO4 and Na2HPO4 needed to prepare 150 ml of a 0.3 M phosphate buffer at pH 7.9 are 25 ml and 125 ml respectively. This is determined by using the Henderson-Hasselbach equation and considering the molecular weights of the compounds.
MusicMonkey
According to theory, which amounts of the following acid and its conjugate base would you need to prepare 150 ml of a 0.3 M phosphate buffer at pH 7.9? Use the pKa table below to help you in your answer.

H3PO4 <–> H2PO4- + H+ pKa = 2.3

H2PO4- <–> HPO4-2 + H+ pKa = 7.2

HPO4-2 <–> PO4-3 + H+ pKa = 12.1

Molecular weights: PO4-3 = 95, Na = 23, H = 1

a) 7.5 ml of 0.3M NaH2PO4 and 37.5 ml of 0.3M Na2HPO4 plus water

b) 37.5 ml of 0.3M NaH2PO4 and 7.5 ml of 0.3M Na2HPO4 plus water

answer:--> c) 25 ml of 0.3M NaH2PO4 and 125 ml of 0.3M Na2HPO4

d) 125 ml of 0.3M NaH2PO4 and 25 ml of 0.3M Na2HPO4

e) 0.05 ml of 0.3M NaH2PO4 and 0.25 ml of 0.3M Na2HPO4 plus water

f) 0.25 ml of 0.3M NaH2PO4 and 0.05 ml of 0.3M Na2HPO4 plus water

Can you should show some of your reasoning people might tell you what is wrong and point you toward the right direction

Well, I am not sure how to get c as the answer because that is the answer that was given. This question appeared on a test.
I thought the answer should be b because after using the equation pH=pKa+log(A/HA) I got the results of 37.5 and 7.5, but then in theory water is not added to the solution to create a buffer unless I believe a titration is made. I may be mixing this information up with something else. In theory no titration is made and the Henderson-Hasselbach equation is used. That is why I do not understand how the values for C are correct. Thank you.

Hellp
I don't understand why c) would be the correct answer.
can you give me the lows please

MusicMonkey said:
According to theory, which amounts of the following acid and its conjugate base would you need to prepare 150 ml of a 0.3 M phosphate buffer at pH 7.9? Use the pKa table below to help you in your answer.

H3PO4 <–> H2PO4- + H+ pKa = 2.3

H2PO4- <–> HPO4-2 + H+ pKa = 7.2

HPO4-2 <–> PO4-3 + H+ pKa = 12.1

Molecular weights: PO4-3 = 95, Na = 23, H = 1

a) 7.5 ml of 0.3M NaH2PO4 and 37.5 ml of 0.3M Na2HPO4 plus water

b) 37.5 ml of 0.3M NaH2PO4 and 7.5 ml of 0.3M Na2HPO4 plus water

answer:--> c) 25 ml of 0.3M NaH2PO4 and 125 ml of 0.3M Na2HPO4

d) 125 ml of 0.3M NaH2PO4 and 25 ml of 0.3M Na2HPO4

e) 0.05 ml of 0.3M NaH2PO4 and 0.25 ml of 0.3M Na2HPO4 plus water

f) 0.25 ml of 0.3M NaH2PO4 and 0.05 ml of 0.3M Na2HPO4 plus water

This question looks hard, but in fact is pretty obvious.

First of all - out of 6 answers listed, 4 are wrong, because they don't give 0.3M buffer solution. That leaves only two answers to select from. pKa2 is 7.2 - so to have buffer with pH lower than that you need a solution which contains more acid than base. That leaves only one answer.

Sure, that requires assumption that one of the answers given is correct

Buffer Maker - the ultimate buffer calculator

V=150 ml= 0.15 L
C=0.3 M
pH= 7.9
pKa= 7.2
MW Na2HPO4=(23 x 2) + 1+ 95 = 142
MW NaH2PO4= 23 + (1 x 2) + 95= 120
……..
H2PO4- <–> HPO4-2 + H+
pH = pKa +log conjugated base/ acid
7.9=7.2 + log Na2HPO4/ NaH2PO4
0.7= log Na2HPO4/ NaH2PO4
1) Na2HPO4/ NaH2PO4 = 5.011
2) Na2HPO4 + NaH2PO4= 0.3 M
From 1 & 2 : Na2HPO4= 0.245 NaH2PO4= 0.049

G (wight) [gr]= C(concentration) [M] x mw (molecular weight) X v (volume) [L]
Na2HPO4 G= 0.245 x 0.15 x 142= 5.218 gr in 0.15 L water
NaH2PO4 G= 0.049 x 0.15 x 120= 0.882 gr in 0.15 L water
Or
V= g/C x MW
Na2HPO4 V= 5.218/ 0.3 x 142 = 0.122 L = 122 ml of 0.3M Na2HPO4
NaH2PO4 V = 0.882 / 0.3 x 120= 0.0245 = 24.5 ml of 0.3M NaH2PO4

These answers are close enought to C)

## What is the purpose of preparing a 0.3M Phosphate Buffer at pH 7.9?

The purpose of preparing a 0.3M Phosphate Buffer at pH 7.9 is to create a solution with a specific pH level that can be used in various scientific experiments and procedures. A phosphate buffer is a type of solution that helps maintain a constant pH level even when small amounts of acid or base are added.

## What materials are needed for preparing a 0.3M Phosphate Buffer at pH 7.9?

The materials needed for preparing a 0.3M Phosphate Buffer at pH 7.9 include phosphoric acid, sodium phosphate, and water. You will also need a graduated cylinder or volumetric flask for accurate measurements, and a pH meter to ensure the desired pH level is achieved.

## Can any type of phosphate be used for preparing a 0.3M Phosphate Buffer at pH 7.9?

No, only specific types of phosphate can be used for preparing a 0.3M Phosphate Buffer at pH 7.9. The most commonly used types are potassium dihydrogen phosphate (KH2PO4) and disodium hydrogen phosphate (Na2HPO4). These chemicals are selected because they can easily be converted to their conjugate acid or base form, allowing for the maintenance of a stable pH level.

## What is the process for preparing a 0.3M Phosphate Buffer at pH 7.9?

The process for preparing a 0.3M Phosphate Buffer at pH 7.9 involves calculating the amount of phosphoric acid and sodium phosphate needed based on the desired volume and concentration. The chemicals are then added to a volumetric flask or beaker of water, mixed well, and the pH level is measured and adjusted if necessary.

## Why is it important to prepare a 0.3M Phosphate Buffer at pH 7.9 accurately?

It is important to prepare a 0.3M Phosphate Buffer at pH 7.9 accurately because the pH level of a solution can greatly affect the outcome of a scientific experiment or procedure. If the pH level is not accurate, it can result in inaccurate or unreliable data. Additionally, preparing the buffer accurately ensures that the solution will maintain its pH level over time, allowing for consistent and reproducible results.

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