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In the titration of a solution of a weak acid `HX` with `NaOH`, the `pH` is `5.8` after `10mL` of `NaOH` solution has been added and `6.40` after `20mL` of the `NaOH` has been affed. What is the ionisation constant of the acid `HX`? |
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Answer» Let the mormality of `HX` and `NaOH = 1`. Let `x mEq` or mmol of `HX` is present initially. When `pH = 5.8`. `["Salt"] = 10 xx 1 = 10 mmol` `["Acid"] = (x - 10) mmol` When `pH = 6.4` `["Salt"] = 20 xx 1 = 20 mmol` `["Acid"] = (x - 20) mmol` `5.8 = pK_(a) +log (10//x - 10) ..(i)` `6.4 = pK_(a) +log (20//x - 20)...(ii)` SOlving equaitons (i) and (ii), calculate the value of `x` and `pK_(a)`. `:. pK_(a) = 6.1` |
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