Class 12: Biological Applications of Buffers Flashcards
To predict the ratio of [conjugate base]/[acid] will be present in a solution buffered to a particular pH.
- Based on the Henderson-Hasselbalch equation: pH = pKa + log([base]/[acid])
- Rearrange to: [base]/[acid] = 10^(pH - pKa)
- So the ratio depends on:
- The desired buffer pH
- The pKa of the weak acid
- Higher pH > pKa means [base] > [acid]
- Lower pH < pKa means [acid] > [base]
- At pH = pKa, [base] = [acid]
- The further pH deviates from pKa, the larger the ratio difference
- Good buffer range is typically pKa ± 1 pH unit
- Choose a weak acid with pKa close to desired pH
To apply the concept in #1 to biological molecules, particularly the amino acid side chains.
- Amino acids have ionizable side chains that can act as weak acids/bases
- Side chains contain groups like: -COOH, -NH2, -SH, imidazole, phenol
- Can identify conjugate acid-base pairs:
- COOH/COO-
- NH3+/NH2
- SH/S-
- Protonated/deprotonated imidazole or phenol groups
- The pKa values determine the predominant form at a given pH
- At pH = pKa, equal amounts of acid and conjugate base
- Amino acids are zwitterions with both positive and negative charges
- The isoelectric point (pI) is when charges cancel out
To define what a “salt bridge” is in a biological system.
- A salt bridge is an ionic bond formed between two charged/ionized groups
- It occurs between a positively charged group (e.g. -NH3+) and a negatively charged group (e.g. -COO-)
- These oppositely charged groups are often found on amino acid side chains
- Salt bridges help stabilize the 3D structure of proteins and nucleic acids
- They form between charged residues that are relatively close in space
- The strength depends on the distance between groups and surrounding environment
- Salt bridges are one type of non-covalent interaction along with hydrogen bonds, hydrophobic effects
- They play a role in protein folding, binding interactions, and structural stability
To use pH and LeChatelier’s Principle as predictive tools to explain shifts in equilibria.
- Le Chatelier’s Principle states that equilibria shift to counteract applied stresses
- Adding H3O+ (lowering pH) is a stress on equilibria involving H+
- Pushes reactions to the left (reactant side)
- Favors reactants that don’t consume H+
- Adding OH- (increasing pH) is a stress that removes H+
- Pushes reactions to the right (product side)
- Favors products that don’t produce H+
- So pH changes cause equilibrium shifts by altering [H+]
- Equilibria also shift to counteract concentration changes
- Adding a reactant shifts toward products
- Adding a product shifts back toward reactants
- Temperature changes also cause equilibrium shifts (exothermic vs endothermic)
To predict the ratio of [conjugate base]/[acid] will be present in a solution buffered to a particular pH.
Henderson-Hasselback equation:
pH=pka + log (prob deprotonated/prob protonated)
A buffer is a strong acid reacted with a weak base, or a strong base reacted with a weak acid and there must be both weak conj acid and weak conj base left
Or just weak acid and weak conj base
USE MOLE TABLE
To apply the concept in #1 to biological molecules, particularly the amino acid side chains.
When there is a high or low pH, it affects whether or not the acids or bases get protonated or deprotonated and can change AA properties
Destroy salt bridges, mess up shape, and prevent binding to receptors
Can harm affinity with binding (EX: hemoglobin)
To define what a “salt bridge” is in a biological system.
Forms between deprotonated aspartic acid and protonated histidine in hemoglobin and acts to change affinity of the active site
To use pH and LeChatelier’s Principle as predictive tools to explain shifts in equilibria.
Increase in products: backwards
Decrease in reactants: backwards
Increase in reactants: forwards
Increase in products: forwards
pH determines the amount of H+ concentration that is available to react with the relevant acids and bases
Acetic acid has Ka of 1.8x10^-5
. What would the ratio of acetate to acetic acid be in these
different solutions buffered to the pH indicated?
The key points are:
- The pH of a solution is related to the ratio of [CH3COO-]/[CH3COOH] by the equation: pH = pKa + log([CH3COO-]/[CH3COOH])
- The pKa of acetic acid is -log(1.8 x 10^-5) = 4.74
- At pH = pKa (4.74), the ratio of [CH3COO-]/[CH3COOH] is 1 (equal concentrations).
- At pH < pKa, there is more acetic acid (CH3COOH) than acetate ion (CH3COO-).
- At pH > pKa, there is more acetate ion (CH3COO-) than acetic acid (CH3COOH).
The answers provided are:
a) At pH 2, the ratio [CH3COO-]/[CH3COOH] is 0.0018, meaning there is about 550 times more acetic acid than acetate ion.
b) At pH 4.74 (pKa), the ratio is 1, meaning equal concentrations of acetic acid and acetate ion.
c) At pH 7.00, the ratio is 182, meaning there is 182 times more acetate ion than acetic acid.
d) At pH 11.00, the ratio is 1 x 10^6 (1 million), meaning there is 1 million times more acetate ion than acetic acid.
Methyl amine has a Kb of 4.4 X 10-4
. What would the ratio of methyl amine to methyl
ammonium ion be in these different solutions buffered to the pH indicated?
(i) 2.00 (ii) 4.74 (iii) 7.00 (iv) 11.00
The key points are:
- Methylamine is a weak base, and its conjugate acid is the methylammonium ion.
- The Ka (acid dissociation constant) of the conjugate acid is calculated using the relationship: Ka = Kw/Kb, where Kw is the ion product constant of water (1.0 x 10^-14 at 25°C).
- The pKa of the conjugate acid is calculated as -log(Ka) = 10.64.
- The pH of the solution is related to the ratio of [CH3NH2]/[CH3NH3+] by the equation: pH = pKa + log([CH3NH2]/[CH3NH3+])
- At pH = pKa (10.64), the ratio of [CH3NH2]/[CH3NH3+] is 1 (equal concentrations).
- At pH < pKa, there is more methylammonium ion (CH3NH3+) than methylamine (CH3NH2).
- At pH > pKa, there is more methylamine (CH3NH2) than methylammonium ion (CH3NH3+).
The answers provided are:
e) At pH 2, the ratio [CH3NH2]/[CH3NH3+] is 2.3 x 10^-9, meaning there is a very low concentration of methylamine compared to methylammonium ion.
f) At pH 4.74, the ratio is 1.3 x 10^-6, meaning there is a low concentration of methylamine compared to methylammonium ion.
g) At pH 7.00, the ratio is 2.3 x 10^-4, meaning there is a slightly higher concentration of methylamine than methylammonium ion.
h) At pH 11.00, the ratio is 3.3, meaning there is about 3 times more methylamine than methylammonium ion.
Amines are very often used as bases. Based on your answers in part 2, is methyl amine
likely to act as a base at a pH of 2.00? Why or why not?
Based on the calculated ratio of [CH3NH2]/[CH3NH3+] at pH 2, which is 2.3 x 10^-9, methylamine is highly unlikely to act as an effective base at this pH.
At pH 2, the vast majority of methylamine molecules exist in the protonated form (CH3NH3+) rather than the free base form (CH3NH2). The protonated form cannot accept additional protons, which is the defining characteristic of a base.
For an amine to effectively act as a base, it needs to have a significant concentration of the free base form present in solution. However, at pH 2, which is well below the pKa of 10.64 for the methylammonium ion, the equilibrium heavily favors the protonated cationic form.
The structure of aspartic acid has a side chain (highlighted in a box), with a carboxylic
acid. Based on what you learned in Part I, would you expect the carboxylic acid side
chain to be mostly protonated or deprotonated at a normal physiological pH of 7.4?
Explain.
pH = 5 + log ([CH3COO-]/[CH3COOH])
7.4 = 5 + log ([CH3COO-]/[CH3COOH])
([CH3COO-]/[CH3COOH]) = 251
This ratio indicates that at 7.4, most of the aspartic acid will be in its deprotonated
(carboxylate form)
Here are some notes on how to approach questions like this:
- Identify the acid/base species involved and its pKa value.
- For carboxylic acids like in aspartic acid, a typical pKa is around 4-5.
- Use the Henderson-Hasselbalch equation to relate the pH to the ratio of conjugate acid/base:
pH = pKa + log([base]/[acid]) - Rearrange the equation to solve for the ratio [base]/[acid]:
[base]/[acid] = 10^(pH - pKa) - Substitute the given pH (7.4 for physiological conditions) and the pKa of the carboxylic acid group.
- Assume a pKa of 5 for the carboxylic acid side chain.
- [base]/[acid] = 10^(7.4 - 5) = 10^2.4 = 251
- Interpret the ratio:
- A ratio > 1 indicates more deprotonated (base) form than protonated (acid) form.
- The ratio of 251 means there are 251 times more deprotonated carboxylate (-COO-) than protonated carboxylic acid (-COOH) at pH 7.4.
- Conclude:
- At physiological pH 7.4, which is well above the pKa of ~5, the carboxylic acid side chain of aspartic acid will be mostly deprotonated and exist as the carboxylate (-COO-) form.
In summary, the key steps are:
1) Identify the pKa of the acid/base group.
2) Use the Henderson-Hasselbalch equation to calculate the ratio of base/acid forms at the given pH.
3) Interpret the ratio to determine which form (protonated or deprotonated) is favored.
- Most Basic Nitrogen:
- This nitrogen is part of the imidazole ring and is sp2 hybridized.
- It has a lone pair of electrons in an sp2 hybridized orbital, which is more accessible for protonation.
- The lone pair is localized on this nitrogen, making it more basic.
- Least Basic Nitrogen:
- This nitrogen is also part of the imidazole ring and is sp2 hybridized.
- It has a lone pair of electrons in a p orbital, which is less accessible for protonation due to its delocalized nature.
- The lone pair is delocalized through resonance structures, making it less available for protonation and, therefore, less basic.
In summary:
- The most basic nitrogen is the one with the localized lone pair in an sp2 hybridized orbital.
- The least basic nitrogen is the one with a delocalized lone pair in a p orbital due to resonance.
The higher basicity of the first nitrogen is reflected in the pKa value of 6.00 for the protonated histidine, which corresponds to the deprotonation of this more basic nitrogen.
The buffer system in the blood is regulated by both the pH and CO2 according to the equilibria shown.
(i) CO2 (aq) + H2O (l) ⇆ H2CO3 (aq)
K = 1.7x10-3
(ii) H2CO3 (aq) + H2O (l) ⇆ HCO3-(aq) + H3O+ (aq)
Ka = 4.3x10-7
- ## What is the healthy [HCO3]/[H2CO3] ratio at a pH of 7.4? Is this ratio consistent with
what you know about effective buffers and buffer capacity?
7.4 = -log (4.3x10-7) + log ([HCO3-]/[H2CO3])
([HCO3-]/[H2CO3]) = 10.8 - After seeing their teaching evaluations, a faculty member has a panic attack and starts
to hyperventilate. This is a medical condition caused by the rapid expulsion of CO2 from
the lungs.
a) Why does hyperventilation cause an increase in blood pH? Use the equilibria provided
to explain your reasoning.
If the [CO2] in the blood drops because the faculty member is breathing it out rapidly, the
equilibrium (i) will shift to the left and H2CO3 will decrease. This will cause equilibrium (ii)
to shift to the left also. This will decrease the concentration of H3O+ and the pH will go up.
b) What would the ratio that you calculated in #7 become if the blood pH rose to 7.7?
7.7 = 6.37 + log ([HCO3-]/[H2CO3])
([HCO3-]/[H2CO3]) = 22
c) A student EMT arrives to the office of the panicking professor by providing them with a
paper bag and tells them to breathe into it. Why would this be a remedy for hyperventilation?
As the professor breathes out his or her own CO2, they will be able to breathe it back in
because it would be trapped in the bag. This will shift the equilibria back to the right and
restore the ideal blood pH.
b) If you wish to study ovalbumin, then you are going to need a solution that is buffered to the pH you determined above. On the shelf you find a 0.75 M solution of aqueous NH4Cl and some solid NaOH. What mass of NaOH do you need to add to 500.0 mL of the ammonium chloride solution to make a buffer with the appropriate pH? You may assume that the volume change of the solution is negligible when you add the solid. Show your work clearly and include an appropriate chemical equation.
The buffer has to include some mixture of NH4+ and NH3 that will lead to a pH of 9. The
NH3 will have to be formed from the reaction below. (Remember that Cl- is a spectator.)
Also, the OH- added will have to be the limiting reagent.
NH4+ + OH- ->. NH3 + H2O
You’re correct, my previous solution was incorrect. Let me rework this from the beginning:
Given:
- 0.75 M solution of aqueous NH4Cl
- Volume = 500.0 mL
- Need to prepare a buffer with pH = 7.0 by adding solid NaOH
- Weak base: NH3
- Conjugate acid: NH4+
Step 3: Calculate initial amounts:
Initial moles of NH4+ = 0.75 M x 0.5 L = 0.375 mol
Step 4: At the desired pH of 9.0, we want [NH3] = [NH4+]
Let x = moles of OH- added
= moles of NH3 formed
Then,
[NH4+] = 0.375 - x
[NH3] = x
Step 5: Use Henderson-Hasselbalch:
pH = pKa + log([base]/[acid])
9.0 = 9.25 + log(x/(0.375 - x))
Step 6: Solve for x:
x/(0.375 - x) = 10^(-2.25) = 0.005618
x = 0.005618(0.375) / (1 - 0.005618)
x = 0.135 mol
Step 7: Amount of NaOH needed = x = 0.135 mol
Mass of NaOH = 0.135 mol x 40.00 g/mol = 5.40 g
Therefore, 5.40 g of solid NaOH needs to be added to 500.0 mL of 0.75 M NH4Cl to make a buffer with pH = 9.0.
