Henderson Hasselbalch Equation

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Henderson Hasselbalch Equation

  1. 1. HENDERSON HASSELBALCHEQUATION Prepared by :- Rajina Shakya Bachelor In Pharamcy 4th year, 7th semester
  2. 2. The Henderson–Hasselbalch Equation  EquationDescribes the derivation of pH as a measure of acidity in biological and chemical systems. The equation is also useful for estimating the pH of a buffer solution.it is widely used to calculate the isoelectric point of proteins( point at which protein neither accept nor yield proton) .
  3. 3. The Henderson hasselbalch equation for acid is :- pH = pKa + log [ Aˉ ] [HA] Here, pKa= -log(Ka)   where Ka  is the acid dissociation constant, that is pKa= -log [H3O+][A-] [HA] for the non-specific Brønsted acid-base reaction:  HA + H20 A- + H3O+ ( Acid ) ( Conjugate base )
  4. 4. The Henderson Hasselbalch Equation for base is : pOH = pKb + log [ BH+ ] [B] where BH+ denotes the conjugate acid of thecorresponding base B. B + H2O+ BH + OH- (Base ) (Conjugate acid)
  5. 5. History- Lawrence Joseph Henderson wrote an equation, in 1908, describing the use of carbonic acid as a buffer solution. - Karl Albert Hasselbalch later re-expressed that formula in logarithmic terms, resulting in the Henderson–Hasselbalch equation.- Hasselbalch was using the formula to study metabolic acidosis.
  6. 6. Henderson-Hasselbalch EquationDerivation:-According to the Brønsted-Lowry theory of acids and bases, an acid (HA) is capable of donating a proton (H+) and a base (B) is capable of accepting a proton.-After the acid (HA) has lost its proton, it is said to exist as the conjugate base (A-).-Similarly, a protonated base is said to exist as the conjugate acid (BH+).
  7. 7. The dissociation of an acid can be described byan equilibrium expression: HA + H20 H3O+ + A-Consider the case of acetic acid (CH 3COOH)and acetate anion (CH3COO-): CH3COOH + H2O CH3COO- + H3O+
  8. 8. Acetate is the conjugate base of acetic acid. Acetic acidand acetate are a conjugate acid/base pair. We candescribe this relationship with an equilibrium constant: Ka = [H3O+][A-] [HA]Taking the negative log of both sides of the equationgives -logKa = -log [H3O+][A-] [HA] or, -logKa = -log [H3O+] + (-log [A-] ) [HA]
  9. 9. By definition,pKa = -logKa and pH = -log[H3O+], so pka=pH – log [A-] [HA] This equation can then be rearranged to givethe Henderson-Hasselbalch equation:pH = pKa + log [A-] = pKa + log [conjugate base] [HA] [acid]
  10. 10. Estimating blood pHA modified version of the Henderson–Hasselbalchequation can be used to relate the pH of blood toconstituents of the bicarbonate buffering system. pH = pKaH2CO3 + log [HCO3-] [H2CO3], where:-pKa H2CO3 is the acid dissociation constant of carbonicacid. It is equal to 6.1.[HCO3-] is the concentration of bicarbonate in theblood[H2CO3] is the concentration of carbonic acid in the blood
  11. 11. Limitation :-The most significant is the assumption that the concentration of the acid and its conjugate base at equilibrium will remain the same.-This neglects the dissociation of the acid and the hydrolysis of the base.-The dissociation of water itself is neglected as well.-These approximations will fail when dealing with: relatively strong acids or bases dilute or very concentrated solutions (less than 1mM or greater than1M),

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