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Define Henderson- Hasselbalch equation |
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Answer» Solution :(i) the concentration of hydronium ION in acidic buffer solution DEPENDS on the ratio of concentration of the weak acid to the concentration of its conjugate base present in the solution i.e., `[H_3O^+]=K_a.([acid]_(aq))/([Base]_(aq))`..........(1) (ii) The weak acid is dissociated only to a small extent. moreover due to common ion effect, the dissociation is further suppressed and HENCE the equilibrium concentration of the acid is nearly equal to the initial concentration of the unionised acid. Similarly the concentration of the conjugate base is nearly equal to the initial concentration of the added salt. `[H_3O^+]=K_a.([acid]_(aq))/([Base]_(aq))`..........(2) (iii) [Acid] and [salt] REPRESENT the initial concentration of the acid and salt respectively used to prepare the buffer solution. (iv) Taking logarithm on both sides of the equation `log[H_3O^+]=log K_a-log([acid])/(["salt"])` ...........(3) (v) reverse the sign on both sides `-log[H_3O^+] and pK_a=-log K_a`...........(4) we know that `ph=-log[H_3O^+] and pK_a=-log k_a` `therefore pH=pK_a-log ([acid])/(["salt"])`..........(5) `pH=pK_a-log ([acid])/(["salt"])`......(6) Similarly for basic buffer, `pOH=pK_a+log([acid])/([base])`......(7) Equation (6) & (7) are called Henderson- Hasselbalch equation. |
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