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Derive the Henderson-Hasselbalch equation. |
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Answer» Solution :Acetic ACID and sodium ACETATE mixture act as acidic buffer solution. And its pH is approximate 4.75 The relation between pH, `K_a` , HA and `A^-` is given as follows : The equilibrium of ionization of WEAK acid HA in water. `HA+H_2O hArr H_3O^(+) + A^(-)` `K_a=([H_3O^+][A^-])/([HA])` `therefore [H_3O^+]=K_a ([HA])/([A^-])` taking log both the side log `[H_3O^+]=log K_a+ "log" "[HA]"/([A^-])` taking minus sign both the side, `-log [H_3O^+]=-log K_a-log ([HA])/([A^-])` `therefore pH=pK_a + "log" ([A^-])/([HA])`....(Eq.-i) `pH=pK_a + "log" "[Conjugate base]"/"[Acid [HA]]"`....(Eq.-ii) This equation -(ii) is called Hendrson-Hasselbalch equation In this equation of value : (i)`([A^-])/([HA])` is the ratio of concentration of conjugate base concentration of weak acid (NEGATIVE ion). (ii)[HA] is weak acid . So, it undergo very LESS ionization so the concentration of [HA] is negligible compare to initial concentration . (iii) The around of conjugate base is due to ionization of salt . So the concentration of conjugate base is also differ from of initial concentration . So equation can be , pH=`pK_a+"log" "[Salt]"/"[Acid]"`.....(Eq.-iii) |
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