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

moleculues that donate protons H+ HA ↔ H+ + A- (dissociation of an acid)

Buffer solution Flashcards by Jessica Hartmann

Defintion of buffer solution

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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Important buffer systems of the human body

carbonic acid / sodium bicarbonate

phosphate buffer

carbonic acid / Potassium bicarbonate

kaliumhydrogenphosphat / dikaliumhydrogenphosphat

acetic acid / sodium acetate

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Henderson - Hasselsbach equation

equation to calculate the ph of the buffer

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

= pKa + log [salt / acid]

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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).

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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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)

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Physiological pH of the blood

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

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

carbonic acid / sodium bicarbonate system (H₂CO₃/NaHCO₃)

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Acidosis

blood ph < 7.35

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Metabolic acidosis

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

↓ bicarbonate [HCO₃^-]

causes:

↑ of acid production (HCO_{3^-}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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Acid - Base imbalance

↓ 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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Respiratoray acidosis

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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Metabolic alkalosis

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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Respiratory alcalosis

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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Acid

moleculues that donate protons H⁺

^{HA ↔ H+ + A- (dissociation of an acid)}

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Bases

molecules that can accept protons H+

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

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ph defintion

power of hydrogen

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

pH scale

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

Strong bases or strong acids..

completely dissociated in a solution

Weak acids or bases

partially dissociated in a solution

Calculation of strong acid

pH = -lgC = pC

C = molar concentration of the acid (HA)

Calculation of strong base

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

C = molar concentration of the base (B)

Calculation of weak acid

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

C = molar concentration of the acid (HA)

Ka = dissociation constant of the acid

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

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

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

acetic acid = weak acid; pKa = 4,73 ## Footnote **pH (O,1N)** = 1/2 (-lgC -lgK_a) = 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

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.

**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**

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

sulfuric acid = strong acid **1) Transform normal to molar concentration:** C_M = C_N /n = 0,02N / 2 = 0,01M pH (strong acid) = -lgC = -log(0,01) = -log(10^-2) = -(-2) = 2

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

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

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.

1. Transformig percentage into molar concentrarion: CM = (C% x 10)/MW = (6,2 x 10) /62 = 1 M carbonic acid = weak acid 2. pH (weak acid) = 1/2 (-log(1) -log(3,47x10^-7)) = 1/2 (0 -0,54 +7) = **3,23**

Machanism that reduce the concentration of hydrogen ions

Buffer systems lungs - CO2 elimination kidneys - bicarbonate absorbtion

The way to calculate the pH

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?

Henderson-Hasselbalch equation

**Extracellular buffers**

**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

**Intracellular buffers**

**Potassium dihydrogen phosphate** (weak acid)/**Dipotassium hydrogen phosphate** (KH2PO4/ K2HPO4) **Carbonic acid** / **potassium bicarbonate** (H2CO3 / KHCO3) **K+Protein** / **H+Protein** **KHb/ HHb**

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

C_acidx V_acid= V_basex C_base =\> C_acid= V_e x C_NaOH / V_e x 1/100

pKa

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.

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

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

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