Henderson hassel


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basic knowledge..also go thro buffer nd ph solutin

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Henderson hassel

  1. 1. HENDERSON HASSELBALCHEQUATION Prepared by :-  Rabita Maharjan  Rajina Shakya
  2. 2. The Henderson–Hasselbalch EquationDescribes the derivation of pH as a measure ofacidity 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 noryield 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 anequation, in 1908, describing the use of carbonic acid as abuffer 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 acidsand bases, an acid (HA) is capable of donating aproton (H+) and a base (B) is capable of accepting aproton.-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 bedescribed by an equilibrium expression: HA + H20 H3O+ + A-Consider the case of acetic acid(CH3COOH) and acetate anion (CH3COO-): CH3COOH + H2O CH3COO- + H3O+
  8. 8. Acetate is the conjugate base of acetic acid.Acetic acid and acetate are a conjugate acid/basepair. We can describe this relationship with anequilibrium constant: Ka = [H3O+][A-] [HA] Taking the negative log of both sides of theequation gives -logKa = -log [H3O+][A-] [HA] or, -logKa = -log [H3O+] + (-log [A-] )
  9. 9. By definition,pKa = -logKa and pH = -log[H3O+], so pka=pH – log [A-] [HA]This equation can then be rearranged togive the Henderson-Hasselbalch equation:pH = pKa + log [A-] = pKa + log [conjugatebase] [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 dissociationconstant of carbonic acid. It is equal to 6.1.[HCO3-] is the concentration of bicarbonate in theblood[H2CO3] is the concentration of carbonic acid in theblood
  11. 11. Limitation :-The most significant is the assumption that the concentration of the acid and its conjugate baseat 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 aswell.-These approximations will fail when dealing with: relatively strong acids or bases dilute or very concentrated solutions (less than 1mM or greater than1M),