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Biochemistry > Buffer solution > Flashcards

Buffer solution Flashcards

1
Q

Defintion of buffer solution

A

in compostion, buffer systems are mixtures containing:

  • weak acid and its salt (with a strong base) eg. acetic acid + sodium acetate
  • weak base and its salt (with a strong acid) eg. ethyl amine + ethyl ammonium chloride

Buffer solutions are cabable to resist changes after addition of a limited amount of acids or bases

in case of weak acid and its salt

  • the salt is completly dissociated: NaA -> Na+ + A-
  • the acid is slightly dissociated: HA + H2O -> H3O+ + A-
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2
Q

Important buffer systems of the human body

A

carbonic acid / sodium bicarbonate

phosphate buffer

carbonic acid / Potassium bicarbonate

kaliumhydrogenphosphat / dikaliumhydrogenphosphat

acetic acid / sodium acetate

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3
Q

Henderson - Hasselsbach equation

A

equation to calculate the ph of the buffer

pH = pKa + lg [proton acceptor]/[proton donor]

= pKa + log [salt / acid]

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4
Q

Calculate the pH of a mixture of 0.25 M acetic acid and 0.1 M sodium acetate, knowing that the pKa of acetic acid is 4.7 (log 4 = 0.6).

A

pH = pKa + log (salt/acid)

= 4,7 + log (0,1 / 0,25)

= 4,7 + log (0,4)

= 4,7 + log (4 x 10-1)

= 4,7 + 0,6 + (-1)

4,3

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5
Q

Calculate:

a) The initial pH of the CH3-COOH 0,5N, knowing that pKa=4,7;
b) pH after 1⁄2 of acid was neutralized with NaOH (log5 = 0,7)

A
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6
Q

Physiological pH of the blood

A

7,4 (7,35 - 7,45)

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7
Q

Which is the most important buffer system of the blood that opposes (widersetzen) pH variations ?

A

carbonic acid / sodium bicarbonate system (H2CO3/NaHCO3)

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8
Q

Acidosis

A

blood ph < 7.35

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9
Q

Metabolic acidosis

A

represents the pathological situation when the blood pH <7.35 due to

↓ bicarbonate [HCO3-]

causes:

↑ of acid production (HCO3<span>- </span>is consumed) in case of:

– tissue hypoxia → ↑ lactic ac.

– diabetic ketoacidosis, alcoholic ketosis

↑ exogenous intake of acids (HCO3- is consumed)

↓ of acids renal excretion
Digestive or renal HCO3-losses:

–digestive fistula

–profuse diarrhea

–renal tubular acidosis

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10
Q

Acid - Base imbalance

A

↓ blood pH under 7,35-7,45 = acidosis

can be produced by

↓ [HCO3-] metabolic acidosis

↑ [CO2] respiratory acidosis

↑ blood pH above 7,35-7,45= alkalosis
can be produced by

↑ [HCO3-] metabolic alkalosis

↓ [CO2] respiratory alkalosis

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11
Q

Respiratoray acidosis

A

Represents the pathological conditions when the blood ph is < 7,35 due to ↑ [CO2] (hypoventilation)

causes:

Lesions of respiratory center

disrupted ventilation

obstructive/ restrictive respiratory dieases

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12
Q

Metabolic alkalosis

A

represents the pathological situation when the blood pH > 7,45 due to ↑ [HCO3-]

caused by:

  • Excessive bicarbonate administration (Verwaltung)

=> wrong parenteral treatment with bicarbonatesor reduced glomerular filtration

  • Severe K+depletion (Erschöpfung)

=> K+ is eliminated through kidneyby a compensatory mechanism with H+

  • Cl-losses

=> pyloric stenosis, incoercible vomiting, nasogastric aspiration → increased bicarbonate generation

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13
Q

Respiratory alcalosis

A

represents the pathological situation when the blood ph > 7.45 due to ↓ CO2

causes: hyperventilation (loss of CO2)

  • mechanical ventilation
  • altitude (in Höhe)
  • fever
  • hyteria crisis, panic attacks
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14
Q
A
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15
Q

Acid

A

moleculues that donate protons H+

<span><strong>HA ↔ H+ + A- (dissociation of an acid)</strong></span>

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16
Q

Bases

A

molecules that can accept protons H+

XOH ↔ X+ + OH-(dissociation of a base)

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17
Q

ph defintion

A

power of hydrogen

indicates the acitvity of a solution and represents the negative logarithm of the molar concentration of hidrogen ions (H+)

18
Q

pH scale

A

The pH scale extends from 0 to 14:

At pH=7 the solution is neutral: H+ = OH-

At pH<7 the solution is acidic :H+ > OH-

At pH>7 the solution is basic (alkaline): H+ < OH

H+ = hydrogen ion

OH- = hydroxid ions

19
Q

Strong bases or strong acids..

A

completely dissociated in a solution

20
Q

Weak acids or bases

A

partially dissociated in a solution

21
Q

Calculation of strong acid

A

pH = -lgC = pC

C = molar concentration of the acid (HA)

22
Q

Calculation of strong base

A

pOH = -lgC = pC; pH = 14 -pOH

C = molar concentration of the base (B)

23
Q

Calculation of weak acid

A

pOH = 1/2(-lgC -lgKa) = 1/2(pC + pKa)

C = molar concentration of the acid (HA)

Ka = dissociation constant of the acid

24
Q

Calculate the pH of the hydrochloric acidsolutions (HCl) with a concentration of 0.1N and 1N.

A

HCI = strong acid; pKa = -7

ph (strong acid) = -logC

pH (0,1N)= -log10(0,1) = -log10(10-1) = -(-1) = 1

pH (1) = -log10(1) = 0

25
Q

Calculate the pH of acetic acid (CH3COOH) solutions with a concentration of 0.1N and 1N, knowing that pKa for CH3COOH is 4.73

A

acetic acid = weak acid; pKa = 4,73

pH (O,1N) = 1/2 (-lgC -lgKa) = 1/2 (0,1) + 4,73) = 1/2 (-log (10-1) + 4,73 = 1/2 (-(-1) + 4,73) = +1 + 4,73 = 1/2 x 5,73 = 2,865

pH (1N) = 1/2 (-lg (1) + 4,73 = 1/2 (0 + 4,73) = 2,365

26
Q

Calculate the pH of a β-hydroxybutyric acid solution [CH3-CH(OH)-CH2-COOH] with a concentration of 0.1M, knowing that Ka = 4.07 x 10-5 and log 4.07 = 0.61.

A

β-hydroxybutyric acid = weak acid

pH (weak acid) = 1/2 (-logC -logKa) = 1/2 (-log (0,1) -log(4,07 x 10-5)

= 1/2 (-log(10-1) -log(4,07) - log(10-5)

= 1/2 (1 -0,61 + 5) = 1/2 (5,39 = 2,65

27
Q
  1. Calculate the pH of a sulfuric acid solution (H2SO4) with a concentration of 0.02N.
A

sulfuric acid = strong acid

1) Transform normal to molar concentration: CM = CN /n = 0,02N / 2 = 0,01M

pH (strong acid) = -lgC = -log(0,01) = -log(10-2) = -(-2) = 2

28
Q

Calculate the pH of a sodium hydroxidesolution (NaOH) with a concentration of 0.001M.

A

sodium hydroxide (NaOH) = strong base

pOH (strong base) = -logC; pH = 14 - POH

pOH = -log(0.001) = -log(10-3) = -(-3) = 3; pH = 14 - 3 = 11

29
Q

Calculate the pH of a carbonic acid solution (H2CO3) with a percentage concentration of 6.2%, knowing that:

Molecular weight (MW) H2CO3= 62; Ka (H2CO3)= 3.47 x 10-7log 3.47 = 0.54.

A
  1. Transformig percentage into molar concentrarion:

CM = (C% x 10)/MW = (6,2 x 10) /62 = 1 M

carbonic acid = weak acid

  1. pH (weak acid) = 1/2 (-log(1) -log(3,47x10-7)) = 1/2 (0 -0,54 +7) = 3,23
30
Q

Machanism that reduce the concentration of hydrogen ions

A

Buffer systems

lungs - CO2 elimination

kidneys - bicarbonate absorbtion

31
Q

The way to calculate the pH

A
  1. Which equation is needed?
  2. Do I have a buffer solution or only a acid or base?
  3. Is a strong/ weak base or a strong acid/weak acid given?
32
Q

Henderson-Hasselbalch equation

A
33
Q

Extracellular buffers

A

carbonic acid (weak acid) / sodium bicarbonate (H2CO3 / NaHCO3)

acetic acid (weak acid) / sodium acetate (CH3COOH / CH3COONa)

Sodium dihydrogen phosphate (weak acid)/Disodium hydrogen phosphate

(NaH2PO4 / Na2HPO4)

Na+Protein / H+ Protein

34
Q

Intracellular buffers

A

Potassium dihydrogen phosphate (weak acid)/Dipotassium hydrogen phosphate (KH2PO4/ K2HPO4)

Carbonic acid / potassium bicarbonate (H2CO3 / KHCO3)

K+Protein / H+Protein

KHb/ HHb

35
Q
A
36
Q

Knowing the concentration of the acetic acid we can: Establish the concentration of the acetic acid

A

Cacid x Vacid = Vbase x Cbase

=> Cacid = Ve x CNaOH / Ve x 1/100

37
Q
A
38
Q

pKa

A

represents the pH that corresponds to half equivalence volume, at which only half ot the acetic acids has been converted into sodium acetate. So at the half-equivalence volume, at which only half of the acetic acid has been converted into sodium acetate and the pH will be equal to the pKa of the acid.

39
Q

Ve

A

equivalence volume

= represents the volume added base (NaOH) necessary to convert all the acetic acid into sodium acetate.

40
Q

Calculate the concentration of CH3COOH knowing that for titration of 100ml acid we are used 1N NaOH solution.

A

CM = 1N/1 = 1M

MW = 60 g/mol

C% = Cm x MW /10 = (1 M x 60g/mo)/ 10 = 6 g

Biochemistry (15 decks)

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