steps of how to prepare the solution What compounds make this solution a buffer?

Question

steps of how to prepare the solution What compounds make this solution a buffer? Draw their structures. 10pts 6. Buffer solutions are very important in research laboratories. i. Discuss the importance of buffer solutions in biochemistry experiments ii. What compounds would you use to prepare buffer solutions and what property of these compounds would determine the pH range of their buffers? Give two examples of each of these compounds. Write Henderson-Hasselbach equation. Write the equation that shows the use of the property you mentioned in ii above to determine the range of pH you would use the compounds for buffers. How would you prepare IL of IX Tris buffered saline (TBS) that is 20mM Tris and 0.15M NaCl at pH 7.4? Include calculations to iv. v. e the amount of salt needed, pH adjustment and the detailed steps involved to prepare the buffer (FW NaCl- 58.44g/mol; Tris- 121.14g/mol) 20pts

Explanation / Answer

i)

Buffer is a system which avoids drastic changes of H+ concentration, that is, the pH won't change drastically as we add H+ and OH- to the acid/bases.

We need to keep pH since plenty of metabolic reactions, such as enzymes, cells, etc... will depend direclty on H+

ii)

we must use a conjugate acid + weak base or vice versa, weak acid and its conjguate base.

It is important, so:

A buffer is any type of substance that will resist pH change when H+ or OH- is added.

This is typically achieved with equilibrium equations. Both type of buffer will resist both type of additions.

When a weak acid and its conjugate base are added, they will form a buffer

The equations:

The Weak acid equilibrium:

HA(aq) <-> H+(aq) + A-(aq)

Weak acid = HA(aq)

Conjugate base = A-(aq)

Neutralization of H+ ions:

A-(aq) + H+(aq) <-> HA(aq); in this case, HA is formed, H+ is neutralized as well as A-, the conjugate

Neutralization of OH- ions:

HA(aq) + OH-(aq) <-> H2O(l) + A-(aq) ; in this case; A- is formed, OH- is neutralized as well as HA.

iii)

The equatino cna be derived as:

Recall that, in equilibrium

Ka = [H+][A-]/[HA]

Multiply both sides by [HA]

[HA]*Ka = [H+][A-]

take the log(X)

log([HA]*Ka) = log([H+][A-])

log can be separated

log([HA]) + log(Ka) = log([H+]) + log([A-])

note that if we use "pKx" we can get:

pKa = -log(Ka) and pH = -log([H+])

substitute

log([HA]) + log(Ka) = log([H+]) + log([A-])

log([HA]) + -pKa = -pH + log([A-])

manipulate:

pH = pKa + log([A-]) - log([HA])

join logs:

pH = pKa + log([A-]/[HA])

which is Henderson hasselbach equations.

iv)

the equation that allows us to choose the acid/base pairs is:

pH = pKa + log(A-/HA)

since we must choose to aim a A-/HA ratio = 1

so

pH = pKa

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