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| Salt of Weak Acid and a Weak Base |
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| In this case both the cation and anion undergo hydrolysis to the same or different extents. The resulting solution may be neutral, acidic or |
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| basic depending upon the relative strengths of acids and bases. The hydrolysis may be written as: |
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| Some common examples are CH3COONH4, (NH4)2CO3, AlPO4, etc. |
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| The aqueous solution of a salt of strong acid and weak base is acidic. For the general reaction: |
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| or |
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| This involves anionic and cationic hydrolysis. |
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| Hydrolysis constant |
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| The hydrolysis constant may be written as: |
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 (vi) |
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| for weak acid, |
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| For weak base BOH, |
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| Also, Kw = [H+] [OH-] |
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| Multiplying equation (vi) by equation [H+] [OH-], we get |
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| Degree of hydrolysis |
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| If the original concentration of the salt in the solution is 'c' mol/litre and 'h' is the degree of hydrolysis at that concentration, then |
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| Kh = h2 |
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or
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| Here the degree of hydrolysis is independent of the concentration of the solution. The weaker the acid and the base, the greater is the degree of hydrolysis of the salt. |
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| pH of the hydrolysed salt solution |
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| Now, pH = -log [H+] |
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| Knowing the molar concentration 'c' of the solution, Ka, Kb and Kw, the pH of the solution can be calculated. |
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| It is clear from the above equation that pH of the solution will depend upon the pK value of the acid and the base. |
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 If pKa < pKb, pH of the solution will be less than 1/2 pKW and consequently the solution will be acidic. |
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 If pKa > pKb, then pH of the solution will be more than 1/2 pKW and hence the solution will be alkaline. |
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 If pKa = pKa, pH of the solution will be equal to ½ pKw and hence the solution will be neutral. |
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