Acids, Bases, & Buffers Flashcards
- Describe the law of mass action.
AKA Le Chatlier’s principle. For reversible reactions, forward and reverse rates are dependent on concentrations of reactants and products, respectively. If you dump in lots of product, the reaction goes in the reverse direction to establish balance.
Keq = Kf/Kr = [C][D]/[A][B]
its the rubber duck in the tub example.
- Define pH and pKa.
-Strength of an acid or base is measure of its propensity to either donate or accept protons, measured as it’s pKa = -logKa.
pH = -log[H+] = a measure of the acidity. Physiological pH cannot be changed all that much or it can be life-threatening. Venous blood slightly more acidic because it’s deoxygenated and more associated with carbonic acid.
- In neutral solutions, [H+][OH-] = 1e-14, so pH = 7.
- Strong acids and bases are completely dissociated in H2O whereas weak acids/bases are reversibly protonated and deprotonated.
Ka = [H+][A-]/[HA] and pKa = -log ([H+][A-]/[HA])–> lower pKa = strong acid.
-High pKa = stronger base.
pH strongly affects enzyme activity –> can affect amino acids (charge) and drugs.
- Write the Henderson-Hasselbalch (H-H) equation for any given weak acid or base.
pH = pKa + log ([A-]/[HA])
- When acid or base is 50% deprotonated, 50% protonated, then [A-] = [HA], so log([A-]/[HA]) = 0, so pH = pKa.
- If < 50% dissociated, pH < pKa. Means that less half of the chains will be deprotonated.
- If > 50% dissociated, pH > pKa.
-A buffer is a mixture of weak acid HA and conjugate base A-.
HA –> A- + H+.
- Maximum buffering capacity is when pH = pKa. pKa is a speed-bump in the titration curve; buffering region resists changes in the pH and maximum buffering capacity at pH = pKa (and range = plus or minus 1 unit around that value).
- Effective buffering occurs in range from [A-]/[HA] = 0.1 to 10.
- Define the H-H equation for the bicarbonate buffer system in extracellular fluid and use it to evaluate clinical lab data.
Bicarbonate buffer system = extremely important regulator extracellular pH.
-Aqueous phase = blood in capillaries; gas phase = lung air space.
pH = 6.1 + log([HCO3- mM]/0.03 PCO2 mmHg)
Normal HCO3- = 24 mM
Normal pCO2 = 40 mmHg.
pH = 6.1 + log (24/(.03 * 40))
pH= 7.4
If [HCO3-] drops below 24 mM –> pH decreases = metabolic acidosis.
-If it rises above –> pH increases = metabolic alkalosis.
- If pCO2 increases above 40 mmHg, blood pH decreases, = respiratory acidosis.
- If pCO2 decreases below 40 mmHg, increased pH = respiratory alkalosis.
- Define normal blood pH, [HCO3-] and pCO2.
- Normal arterial blood pH = 7.34-7.44
- Normal venous blood pH = 7.28-7.42 (due to higher [CO2])
- Normal [HCO3-] in blood = 24 mM
- Normal PCO2 in blood = 40 mmHg.
- Describe how weak acids and bases work to buffer pH and define the pH range of maximal buffering capacity.
If you have a mix of weak bases and acids, the weak acids neutralize the bases without changing the pH much (resistant to change), and vice versa. Optimal buffering capacity determined by pKa. There’s a pH at which buffer system can neutralize added acid/base with little change in overall pH (helps avoid big jumps in physiological pH) = optimal buffering pH is plus or minus 1 pH unit from the pKa of the acid/base.
- Use the H-H equation to solve problems of how pH changes in defined buffers, i.e., determine how many equivalents of acid or base are needed to titrate the ionizable groups(s) of a weak acid or base from a starting pH to a final pH, given its concentration and pKa.
SEE PRACTICE PROBLEMS.
Hints from Practice problems:
1)
To deprotonate the side chain on all molecules, add an equal molar amount of NaOH.
For example, when starting with 10 mmoles of acid HA (all molecules protonated) you will add
exactly 10 mmoles of strong base (NaOH) to deprotonate all of the molecules of to A- (base).
Hints from Practice problems:
2)
Remember if [A-]/[HA] = 2, then there are 2 molecules A- for every 1 molecule HA
Therefore 2/3 or 66.6% of the total (HA + A- ) is deprotonated
Hints from Practice problems:
3)How many moles of acid H2CO3 and base HCO3
- are present in 20 mMoles at pH 2.8
(pKa = 3.8)?
Without any calculation, since the pH is lower than the pKa, less than 1/2 of the side chains will
be deprotonated.
(remember the other card in this deck)
What is the ratio of [A-]/[HA] or [HCO3-]/[H2CO3]?
pH = pKa + log [HCO3-]/[H2CO3] ;
2.8 = 3.8 + log [HCO3-]/[H2CO3];
- 1.0 = log [HCO3-]/[H2CO3];
or 10^—1 = [HCO3-]/[H2CO3] =0.1 i.e. there is 1 deprotonated HCO3 - for every 10 protonated H2CO3 Finally then, 1/11 (9%) of molecules are deprotonated and 10/11 (91%) are protonated. Remember H2CO3 + HCO3 - = 20 mMoles so moles H2CO3 = 20 mmoles X .91 = 18.2 moles HCO3 - = 20 mmoles X .09 = 1.8
(In words)
So when doing a problem like this, set up the H-H equation.
Plug in the variables you know.
Look at whats left, and see what number needs to be plugged in to make both sides of the equation equal.
Recognize that this number has to be composed of a log ratio. In this case we knew that -1.0 = log[base]/[acid]. We must undo the log; which is a base 10 system. So take the number, in this case -1; and put 10 to the power of it –>10^-1. plug number into calculator; it will give you what the number actually means in this case (1/10)
so ratio of acid to base is 1:10; which means 1/11 total are deprotonated (.09). .91 or rather 10/11 are protonated. multiply these decimals by 20mmoles to get the actual amout of acid and base in solution.
Hints from Practice problems:
4)
If given a choice of pKa’s for a compound with multiple ionizable groups like an amino acid.
(e.g. 1.8, 6.0, 9.2) Use the pKa within the range of the indicated pH
i.e. if titrating from pH 5.0
⇒ 7.0 only the pKa of 6.0 would be relevant your calculations.
What is the normal molar ratio of bicarbonate to dissolved CO2 in blood?
20
bc 24/1.2 = 20
A short period of rapid ventilation (panting) temporarily would do what to the pH of
blood?
raise it.
b/c you are getting rid of co2, thus getting less acidic, which increases pH
A brief period of breathing into a paper bag would do what to the pH of blood?
lower it.
you are breathing in what you exhale, so more co2
makes blood more acidic; decreases pH
What is the pI for a drug with the following 2 ionizable groups:
imidazole pKa=5.9; sulfonic acid pKa=2.1
pI is the average of the two pkas so in this case 4
