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calculate the ph 0f final solution of following 20 ml of 0.1M ch3cooh,Ka 1.8*10^-5
the a Ph becomes: CH3COOH (aq) + H2O (l) CH3COO-(aq)+ H3O+(aq) Ions 0.1M -G +G +G 0.1 M -G +G +G dissociation constant : Ka= 1.8*10^-5 [CH3COOH][H3O+][CH3COO]=1.8x10^5[CH3COOH]Replacing with G We have [G][G]/0.1-G =1.8x10^-5Solving using the quadratic formula : G = 3.399 x 10^-5 pH = -log(G) = 4.4686
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