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Calculate the pH of 0.08 M solution of hypochlorous acid, HOCl. The ionization constant of the acid is `2.5xx10^(-5)`. Determine the percent dissociation of HOCl. |
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Answer» Correct Answer - `2.85, 1.76%` `{:(,HOCl,+,H_(2)O,hArr,H_(3)O^(+),+,ClO^(-)),("Initial conc.",0.08,,,,,,),("At. Eqm.",0.08-x,,,,x,,x" (x=Amount of HOCl dissociated)"),(,~= 0.08,,,,,,):}` `K_(a)=(x^(2))/(0.08) or x=sqrt((2.5xx10^(-5))(0.08))=1.41xx10^(-3)` `pH = - log [H_(3)O^(+)]=-log (1.41xx10^(-3))=2.85` Alternatively, directly `pH = (1)/(2) [pK_(a) - log C]` % dissociation = `("Amount dissociated")/("Amount taken")xx100=(1.41xx10^(-3))/(0.08) xx100=1.76%` |
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