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Explain about the ionisation of weak acid and how K_a is derived? |
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Answer» Solution :(i) WEAK ACID are partially dissociated in water and there is an equilibrium between the undissociated acid and its dissociated IONS. (ii) Consider the following of weak monobasic acid HA in water `HA+H_2O leftrightarrow H_3O^(+)+A^-`………(i) (iii) Applying law of chemical equilibrium , the equilibrium constant `K_c` is given by the expression `K_c=([H_3O^+][A^-])/([HA][H_2O])`........(2) (iv) In dilute solutions, water is presentin large excess,hence its concentration may be taken as constant say K. Further `H_3O^+` indicates hydrogen ions,for SIMPLICITY it may be replaced by `H^+`. So the EQUATION (2) becomes `K_c=([H^+][A^-])/([HA])`........4 The constant `K_a` is called dissociation constant of weak acid. |
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