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0.01M acetic acid 5% ionization occurs. So find its dissociation immediately. |
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Answer» We have given, Concentration of acetic acid (c) = 0.01 M Percent ionization = 5% CH3COOH(l) + H2O(l) ⇌ CH3COO⊖(aq) + H3O⊕(aq) We know that, Ka = α2c and percentionization = α x 100 where, ka = dissociation constant α = degree of dissociation ∴ degree of dissociation α = 5/100 ∴ ka = (5/100)2 x 0.01 ka = 2.5 x 10-5 Hence, dissociation constant ka = 2.5 x 10-5. |
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